Congruence & Similarity — Revision Notes

The question	Where it is answered	The tool
Where is the point equidistant from two others, or from two lines?	7.1 Bisectors	the perpendicular bisector and the angle bisector
A drawing says 1 : 25 000. How far is that really?	7.1 Scale drawings	multiply by the scale — after making both units the same
Are these two triangles exactly the same? ✕ Not in G2	7.2 Congruence tests	SSS, SAS, AAS or RHS — and never SSA
Are these two triangles the same shape? ✕ Not in G2	7.3 Similarity tests	AA, SSS or SAS Similarity Test
One length is missing, and the triangles are similar.	7.3 Finding a length	one equation of matching ratios
The lengths are in the ratio 2 : 3. What about the areas? ✕ Not in G2	7.4 Area ratio	`k`² — here 4 : 9
… and the volumes, or the masses? ✕ Not in G2	7.5 Volume ratio	`k`³ — here 8 : 27
I am given an area or a volume ratio and want a length. ✕ Not in G2	7.5 Working backwards	square root for an area, cube root for a volume

The perpendicular bisector of a line segment

Cut a line segment in half, at right angles, and you get a line with a surprising property: every point on it is the same distance from the two ends. That is the whole reason the perpendicular bisector is on the syllabus — it is the answer to "where are all the points equidistant from A and B?"

The perpendicular bisector of a line segment AB is the line that cuts AB in half and is perpendicular to it. If XY is the perpendicular bisector of AB, then

XY ⊥ AB;
XY passes through the midpoint M of AB, so AM = MB;
any point P on XY is equidistant from the two endpoints, i.e. PA = PB;
△PAM ≡ △PBM.

Fig. 7.1

The construction, and the property in one picture. Open the compasses to more than half of AB and draw an arc from A and an arc from B, above and below; the arcs cross at X and Y, and the line XY is the perpendicular bisector. Every point P on it satisfies PA = PB, because △PAM and △PBM are congruent.

Open the compasses to more than half of AB, or the two arcs do not cross and there is nothing to join. Leave the arcs on the page: they are the marked evidence of the construction.

Walkthrough 1 — constructing the perpendicular bisector Basic

The line segment AB, of length 7.4 cm, is drawn below. Using a pair of compasses and a straight edge only, construct the perpendicular bisector of AB. Mark the point where it crosses AB as M, and state the length of AM.

Fig. 7.2

Step the walkthrough and the construction is drawn one compass act at a time; nothing is rubbed out afterwards. All four arcs are struck with the same radius, 5 cm, so X and Y are each 5 cm from A and 5 cm from B; the line ruled through them is the perpendicular bisector, cutting AB at right angles at its midpoint M.

Open the compasses to 5 cm — more than half of 7.4 cm — and, with the point on A, draw an arc above AB and an arc below it.

Half of AB is 3.7 cm, so the radius has to beat 3.7 cm or the arcs struck from the two ends do not meet. 5 cm also brings them across each other at close to a right angle, which is easy to read.
Without changing the radius, put the point on B and draw two more arcs, cutting the first two at X and at Y.

Keeping the radius fixed is what makes X exactly as far from A as from B, and the same for Y. Re-setting the compasses between the two centres loses that, and the line then misses the midpoint.
Rule the line through X and Y. It crosses AB at M.

Both X and Y are equidistant from A and B, so both lie on the perpendicular bisector — and two points fix a line. Lay the straight edge along the two crossings themselves.
AM = MB = 3.7 cm, and ∠AMX = 90°.

That is what the construction has produced, and it is the check worth making: half of 7.4 cm is 3.7 cm, and a set square laid along AB sits flush against XY.
XY is the perpendicular bisector of AB, and AM = 3.7 cm.

The question wanted the construction and a length, so give both. Leave the arcs on the page: they are the evidence that the line was constructed rather than measured, and the method mark sits on them.

Walkthrough 2 — using the perpendicular-bisector property Basic

The point P lies on the perpendicular bisector of the line segment AB. Given that PA = 5x − 3 and PB = 2x + 9, find the value of x and the length of PA.

PA = PB (P lies on the perpendicular bisector of AB)

This is the property doing all the work. Write the reason in brackets after PA = PB; the reason is what the mark is for.
5x − 3 = 2x + 9

Substitute the two given expressions into that equation. Nothing here is about geometry any more — it is a linear equation in one variable.
3x = 12

Take 2x from both sides and add 3 to both sides. Collecting the x terms on the side where the coefficient is bigger keeps the number in front of x positive.
x = 4

Divide both sides by 3. The question asked for x first, so this is already half the answer — but only half.
PA = 5(4) − 3 = 17

Substitute back into the expression the question gave for PA. Check it against the other one: PB = 2(4) + 9 = 17 as well, which is exactly what "equidistant" promised, and is a free verification.

More detail

Why the property holds: M is the midpoint, so AM = MB; PM is shared; and ∠PMA = ∠PMB = 90°. That is two sides and the angle between them, so △PAM ≡ △PBM and the matching sides PA and PB are equal. It works for every P on XY, which is why the line is the answer to "where are the points equidistant from A and B?" On the radius: any opening greater than half of AB works, and 5 cm for a 7.4 cm segment is a comfortable choice because it brings the arcs across each other at close to a right angle — a nearly square crossing is one a pencil point can be put on exactly. A radius that is re-set between the two centres gives a point that is no longer equally far from both ends, and the ruled line then misses the midpoint. Real papers print the instruction not to rub out the construction arcs, and the reason is that the arcs are the only record of how the line was found. If the finished drawing measures wrong — AM is not half of AB, or the set square does not sit flush — an earlier compass act is the place to look.

The angle bisector

The perpendicular bisector answers "equidistant from two points". The angle bisector answers the other question: "equidistant from two lines".

The angle bisector of ∠BAC is the line through A that cuts the angle into two equal halves. If AX is the angle bisector of ∠BAC, then

∠BAX = ∠CAX;
any point P on the angle bisector AX is equidistant from the two sides BA and CA.

Fig. 7.3

An arc centred on A cuts the two arms at D and E; equal arcs from D and from E cross at F; the ray AF is the bisector. The distance from a point to a line always means the perpendicular distance — that is why the two dashed lines from P meet the arms at right angles.

"Distance from a line" means the perpendicular distance. "P is 4 cm from AB" gives the length of the perpendicular from P to AB. Draw that perpendicular in and mark the right angle.

Walkthrough 3 — constructing the bisector of an angle Basic

The diagram below shows ∠PQR = 74°. Using a pair of compasses and a straight edge only, construct the bisector of ∠PQR, and state the size of each of the two angles it makes.

Fig. 7.4

Step the walkthrough and the construction is drawn one compass act at a time; every arc stays on the finished diagram. The first arc gives QS = QT = 4.5 cm and the second pair gives SU = TU = 3.5 cm, so the ray from Q through U splits the 74° angle into two equal halves of 37°.

With the point on Q and any convenient radius — 4.5 cm here — draw an arc cutting QP at S and QR at T.

This radius really is free; it only has to be short enough that the arc cuts both arms. What it buys is QS = QT, and that equality is what will make the two halves of the angle come out the same size.
Open the compasses to 3.5 cm — more than half of ST — and, with the point on S, draw an arc inside the angle.

ST measures about 5.4 cm here, so half of it is under 2.8 cm. A radius bigger than that meets the matching arc struck from T — the "more than half" rule again, now applied to ST.
Keeping the same radius, put the point on T and draw an arc cutting the last one at U.

Same radius again, so SU = TU = 3.5 cm. U is now exactly as far from the mark on one arm as from the mark on the other, which is what puts it on the bisector rather than beside it.
Rule the line from Q through U. Then ∠PQU = ∠RQU = 37°.

QS = QT, SU = TU, and QU is shared, so the two halves of the picture are mirror images of each other and the two angles at Q must be equal. 74° ÷ 2 = 37° is then the protractor check.
QU produced is the bisector of ∠PQR, and each half is 37°.

Both halves of the question are answered: the line, and its size. Leave all three arcs on the paper — they are what shows the bisector was constructed rather than measured off with a protractor.

More detail

"P is 4 cm from AB" never means the length of some slanting line from P to the arm: the distance from a point to a line is the perpendicular distance, which is why the two dashed lines in the figure meet the arms at right angles. Once that perpendicular is drawn in and its right angle marked, the rest of the question is usually ordinary geometry. The first radius in the construction is genuinely free — it only has to be short enough for the arc to cut both arms. What it buys is QS = QT. The second radius has to beat half of ST for the same reason the perpendicular bisector needs more than half of AB: below that the two arcs never cross. With QS = QT, SU = TU and QU shared, the two halves of the picture are mirror images (SSS), so the two angles at Q are equal.

Which bisector does the question want?

Nearly every exam use of these two constructions is one of the four rows below. Read the question for the words in the left-hand column.

The question says…	Construct	Because
"equidistant from `A` and `B`", "the same distance from the two towns"	the perpendicular bisector of `AB`	every point on it satisfies `PA` = `PB`
"equidistant from the two roads", "the same distance from `AB` and `AC`"	the angle bisector of the angle between them	every point on it is the same perpendicular distance from both arms
"equidistant from three towns `A`, `B` and `C`"	two perpendicular bisectors — of `AB` and of `BC` — and take where they cross	the first gives `PA` = `PB`, the second gives `PB` = `PC`; at the crossing all three are equal
"nearer to `A` than to `B`"	the perpendicular bisector of `AB`, then shade `A`'s side of it	the bisector is the boundary; either side of it one distance beats the other

Past-paper questions

G3 N2017 · Paper 1 · Q10 3 marks

The diagram represents a plot of land, ABCD, which is to be used for a park.

Construct the perpendicular bisector of BC. [1]
Construct the bisector of angle ADC. [1]
A café is to be built in the park, nearer to C than to B and nearer to AD than to CD. Shade the region where the café is to be built. [1]

Worked solution

(a) Set the compasses to more than half of BC. With the point at B draw an arc on each side of BC, and without changing the setting repeat from C. Join the two crossing points: that line is the perpendicular bisector of BC, and every point on it satisfies PB = PC. (b) With the point at D draw one arc cutting DA and DC. From those two cuts draw two arcs of equal radius that cross inside the angle, and join D to the crossing: that line is the bisector of angle ADC, and every point on it is the same perpendicular distance from DA as from DC. (c) "Nearer to C than to B" is everything on C's side of the line from (a); "nearer to AD than to CD" is everything on A's side of the line from (b). The café goes in the overlap — the triangle bounded by the perpendicular bisector on the left, the angle bisector above and AD below, with D as its third corner (shaded green above). Why this works. Three marks: two constructions and one region. The arcs are the marked evidence of each construction, so they stay on the page. For (c), "nearer to C than to B" is about two points — perpendicular bisector — while "nearer to AD than to CD" is about two lines, and the shading is the overlap of the two regions.

G3 N2019 · Paper 1 · Q12 3 marks

The diagram shows a quadrilateral ABCD.

On the diagram,

construct the bisector of angle ADC, [1]
construct the perpendicular bisector of AB, [1]
shade the region inside ABCD that is closer to DC than to DA and closer to A than to B. [1]

Worked solution

(a) With the compass point at D, draw an arc cutting DA and DC; from those two cuts draw two equal arcs that cross, and join D to the crossing. That is the bisector of angle ADC (pink above), and every point on it is the same perpendicular distance from DA as from DC. (b) With the compasses set to more than half of AB, draw arcs from A and from B above and below AB and join the two crossings. That is the perpendicular bisector of AB (blue above), and every point on it satisfies PA = PB. (c) "Closer to DC than to DA" is the DC side of the bisector in (a) — above it. "Closer to A than to B" is the A side of the bisector in (b) — to its left. The region asked for is the overlap: the triangle with corners at D, where the vertical bisector cuts DC, and where the two constructed lines cross. Why this works. The same three marks as N2017/P1/Q10 with the two constructions swapped, so what is tested is reading points (perpendicular bisector) apart from lines (angle bisector). Mark which side of each bisector you want before shading anything. The region wanted is inside ABCD, so the shading stops at DC.

Enlargement and reduction ⓘ Recap

A scale drawing is nothing more than an enlargement (or a reduction) of the real thing, so it is worth being clear about what "scale factor" does before the units get involved.

To enlarge a figure by a scale factor k, multiply every length by k. Every angle stays exactly the same. If k > 1 the figure gets bigger (an enlargement); if 0 < k < 1 it gets smaller (a reduction). The new figure and the old one are similar.

Angles do not scale. Doubling every length of a triangle leaves all three angles untouched, which is why a photocopy at 200% still looks like the same triangle, and why two equal angles are enough to prove similarity.

Scale drawings, and the 1 : `n` form

A map, a floor plan and an architect's elevation are all the same object: a drawing in which one length on the paper stands for a fixed multiple of itself in real life. The scale can be written two ways, and questions use both.

A scale may be given

in words — "1 cm represents 20 m"; or
as a ratio 1 : n — "1 : 2000", meaning 1 unit on the drawing stands for n of the same units in reality.

To turn one into the other, put both sides into the same unit first: 1 cm to 20 m is 1 cm to 2000 cm, i.e. 1 : 2000.

Fig. 7.5

A triangular plot drawn at 1 cm to 20 m, i.e. 1 : 2000. Every length on the drawing is the real length divided by 2000, and every angle is the real angle — which is why a scale drawing can be measured to answer a question the numbers do not give you directly. (Shown here at page size, not at true 6 cm.)

A ratio 1 : n has no units. Both sides are in the same unit, so convert before reading n. Keep three conversions ready: 1 m = 100 cm, 1 km = 1000 m = 100 000 cm.

Walkthrough 4 — both directions on a 1 : n map Basic

A map is drawn to a scale of 1 : 25 000. (i) Two towns are 14.4 cm apart on the map. Find the actual distance between them, in km. (ii) A straight canal is 9 km long. Find its length on the map, in cm.

1 cm on the map represents 25 000 cm in reality.

Say the ratio out loud in one unit before doing anything else. "1 : 25 000" is unitless, so it is true of centimetres, metres or miles — choosing centimetres here matches the units the map lengths come in.
Actual distance = 14.4 × 25 000 = 360 000 cm

Going from the map to reality means the length grows, so multiply. The answer is still in centimetres at this point — the conversion is a separate step, and merging the two is where the zeros go missing.
= 360 000 ÷ 100 000 = 3.6 km

100 000 cm make 1 km (100 cm in a metre, 1000 m in a kilometre). The question named the unit it wanted, so the answer is not finished until it is in that unit.
9 km = 9 × 100 000 = 900 000 cm

Part (ii) runs the other way, so convert the real length into the map's unit first. Dividing kilometres by 25 000 would give an answer in kilometres, which no map length ever is.
Map length = 900 000 ÷ 25 000 = 36 cm

Going from reality to the map means the length shrinks, so divide by the scale. Sanity-check the direction every time: a map distance must come out smaller than the real one.

Walkthrough 5 — finding the scale of a drawing, then using it Intermediate

On a scale drawing of a coastline, the towns A and B are 7.5 cm apart, and the actual distance AB is 45 km. (i) Find the scale of the drawing in the form 1 : n. (ii) A lighthouse L is 5.2 cm from A on the drawing. Find the actual distance AL, in km.

7.5 cm represents 45 km = 45 × 100 000 = 4 500 000 cm

A ratio needs one unit on both sides, and the drawing is measured in centimetres, so the kilometres have to become centimetres before the ratio is written at all.
Scale = 7.5 : 4 500 000

Write it as a plain ratio first, in the order drawing : real. That order is what keeps every line after it the right way up.
= 1 : 600 000 (dividing both sides by 7.5)

The 1 : n form always comes from dividing both parts by the left-hand one. 4 500 000 ÷ 7.5 = 600 000.
AL = 5.2 × 600 000 = 3 120 000 cm

Now the scale is known, part (ii) is Walkthrough 4 again: drawing to reality, so multiply. Keep the working in centimetres until the last line.
= 3 120 000 ÷ 100 000 = 31.2 km

Convert once, at the end. A quick check: 5.2 cm is a bit under 7.5 cm, and 31.2 km is a bit under 45 km — the two lengths are in the same proportion, as they must be.

More detail

Almost everything that goes wrong in a scale question is a factor of 100 or 1000 rather than a wrong method, which is why the conversion deserves its own line of working. "1 : 25 000" is unitless, so it is equally true of centimetres, metres or miles; choosing the unit the drawing is measured in keeps the arithmetic small. The direction is worth checking on every line. Drawing to reality makes the length bigger, so it multiplies; reality to drawing makes it smaller, so it divides. A map distance that comes out larger than the real one has been multiplied the wrong way, and the fastest fix is to reread the ratio in the order drawing : real. A scale written n : 1 rather than 1 : n says the drawing is the bigger of the two — a microscope drawing rather than a map. The arithmetic is the same; only the direction swaps.

Past-paper questions

G3 N2023 · Paper 2 · Q1(c) 3 marks

The diameter of a red blood cell is

7.8 \times 10^{−6}

m.
In a scale drawing, the diameter of a red blood cell is 3.9 cm.

Find the scale used for the drawing.
Give your answer in the form n : 1. [2]
Drawn to the same scale as the red blood cell, the diameter of a blood platelet is 8 mm.
Find the actual diameter of the blood platelet.
Give your answer in millimetres in standard form. [1]

Worked solution

(a) Put both lengths in the same unit first. In centimetres, $7.8 \times 10^{−6}$ m $= 7.8 \times 10^{−6} \times 100 = 7.8 \times 10^{−4}$ cm.

drawing : actual = 3.9 : $7.8 \times 10^{−4}$

Divide both parts by $7.8 \times 10^{−4}$ : $\frac{3.9}{7.8 \times 10^{−4}} = 0.5 \times 10^{4} = 5000$ , so the scale is 5000 : 1.

(b) The drawing is 5000 times the real thing, so divide by 5000:

actual diameter $= \frac{8}{5000} = 0.0016$ mm = $1.6 \times 10^{−3}$ mm

Why this works. A scale question is one ratio written twice, and the marks sit on the conversion: the cell is given in metres and the drawing in centimetres. A microscope drawing is n : 1 because the drawing is the bigger one, so 5000 enlarges here. The check on (b) is that a platelet comes out smaller than the 8 mm on the paper, so 5000 divides.

Area on a plan ⓘ Recap

If every length is multiplied by 2000, the area is multiplied not by 2000 but by 2000². A plan is an enlargement, and an enlargement scales area by the square of the scale factor.

For a scale of 1 : n,

a length on the drawing × n = the real length;
an area on the drawing × n² = the real area.

Remember the unit conversions square too: 1 m² = 100² = 10 000 cm² and 1 km² = 100 000² = 10 000 000 000 cm².

Walkthrough 6 — the real area from a plan Intermediate

A rectangular garden is drawn on a plan of scale 1 : 400. On the plan the garden measures 6 cm by 4.5 cm. Find the actual area of the garden, in m².

Actual length = 6 × 400 = 2400 cm = 24 m

The safest route is always to scale the two lengths first and only then multiply. Convert to metres straight away, because the answer is wanted in m².
Actual width = 4.5 × 400 = 1800 cm = 18 m

Both dimensions are scaled by the same factor — that is what "similar" means. A plan that stretched one direction more than the other would not be a scale drawing at all.
Area = 24 × 18 = 432 m²

Area of a rectangle = length × width, using the real lengths. Units: metres times metres gives m², which is what was asked for.
Check, the fast way. Area on the plan = 6 × 4.5 = 27 cm²

The alternative route is to scale the area itself. Work out the plan's area first — this is a genuine area, but of the drawing, not of the garden.
27 × 400² = 27 × 160 000 = 4 320 000 cm² = 4 320 000 ÷ 10 000 = 432 m² ✓

Areas scale by n², not n: multiplying by 400 once gives 10 800 cm², which is 400 times too small. The two routes agree, and that agreement is the check.

More detail

The last line carries a second squared conversion, and it is the one most often halved: 1 m² is 10 000 cm², not 100, because both the length and the width are converted. The same doubling-up gives 1 km² = 10 000 000 000 cm².

This is the same fact that runs the whole of 7.4, met a year early on a map: an area ratio is the square of a length ratio, whatever the two figures are.

Check yourself

The point P lies on the perpendicular bisector of MN, and PM = 3y + 2, PN = 5y − 8. Find y and PM.
Answer
PM = PN (P is on the perpendicular bisector of MN), so 3y + 2 = 5y − 8, giving 2y = 10 and y = 5. Then PM = 3(5) + 2 = 17 (and PN = 5(5) − 8 = 17, as a check).
AX is the angle bisector of ∠BAC, and ∠BAC = 78°. The point Q lies on AX and its distance from AB is 4.6 cm. Write down ∠BAX and the distance from Q to AC.
Answer
∠BAX = 78° ÷ 2 = 39°. Every point on the angle bisector is equidistant from the two arms, so the distance from Q to AC is also 4.6 cm — and both of those distances are measured perpendicular to the arm.
A map has a scale of 1 : 50 000. Two towns are 8.6 cm apart on the map. Find the actual distance between them, in km.
Answer
8.6 × 50 000 = 430 000 cm, and 430 000 ÷ 100 000 = 4.3 km.
On a plan of scale 1 : 250, a room measures 3.2 cm by 2.4 cm. Find the actual area of the room, in m².
Answer
3.2 × 250 = 800 cm = 8 m and 2.4 × 250 = 600 cm = 6 m, so the area is 8 × 6 = 48 m². (Or: 3.2 × 2.4 = 7.68 cm² on the plan, × 250² = 480 000 cm², ÷ 10 000 = 48 m².)
A scale drawing uses 1 cm to represent 8 m. Write the scale in the form 1 : n, and find the length on the drawing of a wall that is 68 m long.
Answer
8 m = 800 cm, so the scale is 1 : 800. The wall is 68 m = 6800 cm, so on the drawing it is 6800 ÷ 800 = 8.5 cm.
A pole is to be put up so that it is the same distance from each of three towns A, B and C. Explain, in one sentence, which constructions you would draw.
Answer
Construct the perpendicular bisector of AB (every point on it satisfies PA = PB) and the perpendicular bisector of BC (every point on it satisfies PB = PC); the pole goes where the two bisectors cross, because there PA = PB = PC.

Past-paper questions

G3 N2016 · Paper 1 · Q20 4 marks

A map of Western Europe has a scale of 1 : 2 500 000.

The length of the river Rhine on the map is 49.3 cm.
Calculate the actual length, in kilometres, of the river Rhine. [2]
The area of Switzerland is 41 285 km².
Calculate the area, in square centimetres, of Switzerland on the map. [2]

Worked solution

(a) 1 : 2 500 000 means 1 cm on the map is 2 500 000 cm on the ground.

actual length = 49.3 × 2 500 000 = 123 250 000 cm

1 km = 100 000 cm, so 123 250 000 ÷ 100 000 = 1232.5 km

(b) First put the scale in the units the question uses: 1 cm : 2 500 000 cm = 1 cm : 25 km.

Square both sides: 1 cm² : 25² km² = 1 cm² : 625 km².

area on the map $= \frac{41 285}{625}$ = 66.056 cm² (exactly; 66.1 cm² to 3 s.f.)

Why this works. The two parts are the same scale used in opposite directions. Map to ground multiplies, and the conversion comes once at the end, because 2 500 000 counts centimetres. For the area, square the converted scale first: 1 cm : 25 km becomes 1 cm² : 625 km², so 625 is what divides 41 285.

G3 N2020 · Paper 1 · Q17 4 marks

A map has a scale of 1 : 2 000 000.

The distance between Singapore and Phuket is 950 km.
Calculate the distance, in centimetres, between Singapore and Phuket on the map. [2]
The area of Malaysia is 330 803 km².
Calculate the area, in square centimetres, of Malaysia on the map. [2]

Worked solution

(a) 1 cm : 2 000 000 cm = 1 cm : 20 000 m = 1 cm : 20 km.

This time the journey is ground to map, so divide:

distance on the map $= \frac{950}{20}$ = 47.5 cm

(b) Squaring the scale: 1 cm² : 20² km² = 1 cm² : 400 km².

area on the map $= \frac{330 803}{400}$ = 827.0075 cm² (exactly; 827 cm² to 3 s.f.)

Why this works. Each part turns on one decision, multiply or divide, and asking which of the two numbers has to be bigger settles it: the ground is bigger than the map, so going to the map divides. Part (b) is the same square as N2016/P1/Q20(b) — convert 1 : 2 000 000 into 1 cm : 20 km first, then square the 20. Nothing here needs rounding.

G3 N2024 · Paper 1 · Q6 4 marks

A map of Singapore has a scale of 1 : 200 000.

The scale can be written in the form 1 cm : n km.
Find the value of n. [1]
The distance on the map from Changi Airport to Bukit Panjang is 18.9 cm.
Calculate the actual distance, in kilometres, between these two places. [1]
The area of Singapore is 728.6 km².
Calculate the area, in square centimetres, of Singapore on the map. [2]

Worked solution

(a) 1 cm : 200 000 cm = 1 cm : 2000 m = 1 cm : 2 km, so n = 2.

(b) Map to ground, so multiply: 18.9 × 2 = 37.8 km.

(c) Square the scale: 1 cm² : 2² km² = 1 cm² : 4 km².

area on the map $= \frac{728.6}{4}$ = 182.15 cm² (exactly)

Why this works. Part (a) carries the rest: once the scale reads 1 cm : 2 km, (b) is a multiplication and (c) is that same 2 squared. Getting there is two conversions — 200 000 cm is 2000 m, and 2000 m is 2 km — or one division by 100 000. In (c) the number that divides is 4, and every answer is exact.

What congruent means, and why the order of the letters matters

Two figures are congruent if they have exactly the same shape and exactly the same size. One can be picked up, turned over and put down on top of the other so that it fits perfectly. We write

△ A B C \equiv △ X Z Y

and the order of the letters is part of the statement: it says A ↔ X, B ↔ Z, C ↔ Y.

Fig. 7.6

The same triangle twice, reflected and turned. Matching sides carry the same number of ticks and matching angles the same number of arcs, so the correspondence can be read straight off the marks: A ↔ X, B ↔ Z, C ↔ Y, and therefore △ABC ≡ △XZY — not △XYZ.

The vertex order is part of the statement. From △ABC ≡ △XZY:
AB = XZ, BC = ZY, CA = YX, ∠A = ∠X, ∠B = ∠Z, ∠C = ∠Y.

Walkthrough 7 — reading a congruence statement Basic

Given that △DEF ≡ △PQR, DE = 7.4 cm and EF = 5.9 cm, and ∠D = 63° and ∠E = 48°, write down PQ, QR, ∠Q and ∠R.

D ↔ P, E ↔ Q, F ↔ R

Unpack the statement before touching a number. The congruence sign is written between two ordered lists of letters, and this pairing is the whole content of it.
PQ = DE = 7.4 cm

The side PQ is made of the 1st and 2nd letters of △PQR, so it matches the side made of the 1st and 2nd letters of △DEF, which is DE.
QR = EF = 5.9 cm

Same rule, 2nd and 3rd letters. Notice you never needed the diagram — and never needed to know which side is longest.
∠Q = ∠E = 48°

Q is the 2nd letter of △PQR and E is the 2nd letter of △DEF, so those are the matching angles.
∠F = 180° − 63° − 48° = 69°

∠R matches ∠F, but ∠F was not given, so find it first from the angle sum of △DEF. Angles in a triangle add to 180°.
∠R = ∠F = 69°

State every part the question asked for, each with its unit or degree sign. Four things were asked for; four things are written down.

More detail

The six facts are read positionally: the first letter of one triangle pairs with the first letter of the other, the second with the second, the third with the third. So a congruence statement handed to you in a question can be unpacked without looking at the diagram at all — and when you write the statement yourself, the order is the first thing the marker reads.

A tip the textbook offers for getting the order right from a picture: match the largest angle of one triangle to the largest angle of the other (the angle opposite the longest side), then the second largest, then the smallest. Tick marks and arc marks on a diagram do the same job, which is why the figure above can be read straight off the marks.

The four congruence tests ✕ Not in G2

Two triangles have six measurements each — three sides and three angles — and not all six are needed to pin a triangle down. Each of the four tests below says: "these three facts are already enough".

Test	What you must show	In symbols, for △`ABC` and △`XYZ`
SSS Congruence Test	3 corresponding sides are equal	`AB` = `XY`, `BC` = `YZ`, `AC` = `XZ`
SAS Congruence Test	2 corresponding sides and the included angle are equal	`AB` = `XY`, `BC` = `YZ`, ∠`ABC` = ∠`XYZ`
AAS Congruence Test	1 corresponding side and 2 corresponding angles are equal	`BC` = `YZ`, ∠`ABC` = ∠`XYZ`, ∠`ACB` = ∠`XZY`
RHS Congruence Test	in two right-angled triangles, the hypotenuse and one other side are equal	∠`ABC` = ∠`XYZ` = 90°, `AC` = `XZ`, `BC` = `YZ`

Fig. 7.7

The four congruence tests. In each pair the second triangle has been reflected and turned, because congruent triangles almost never arrive in the same orientation — matching marks, not matching positions, are what you read.

How to write a congruence proof, in four lines.
1. Three lines of facts, each with a reason in brackets — "(given)", "(common side)", "(vertically opposite ∠s)", "(alt. ∠s, AB // CD)", "(opp. sides of a parallelogram)".
2. A conclusion line: "∴ △… ≡ △…", with the vertices in corresponding order.
3. The name of the test in brackets after it.
4. Only then, whatever the question actually wanted.

The A in SAS is the included angle — the one between the two sides you quoted. An angle anywhere else does not pin the triangle down, which is the SSA warning below.

There is no AAA congruence test. Three equal angles fix the shape but say nothing about the size — that is similarity, not congruence.

Walkthrough 8 — proving a congruence with SSS Basic ✕ Not in G2

In △ABC, AB = 8 cm, BC = 11 cm and CA = 6 cm. In △XYZ, XY = 11 cm, YZ = 6 cm and ZX = 8 cm. Prove that the two triangles are congruent, and write the congruence statement with the vertices in the correct order.

AB = ZX = 8 cm

Pair the sides by length, in any order, and write each pair as an equation: those three equations are the proof. Only one side in each triangle is 8 cm, so this pairing is forced.
BC = XY = 11 cm

The 11 cm sides. Notice that B appears in both AB and BC, and the letter appearing in both ZX and XY is X — so B must pair with X. The correspondence is being built as you go.
CA = YZ = 6 cm

The 6 cm sides, and the last pairing. A is in AB and CA; the letter in both ZX and YZ is Z, so A ↔ Z, and what is left gives C ↔ Y.
∴ △ABC ≡ △ZXY (SSS Congruence Test)

Write the vertices in the order the pairing produced: A ↔ Z, B ↔ X, C ↔ Y. The alphabetical-looking "△ABC ≡ △XYZ" would be wrong here, and the test name in brackets is what finishes the proof.

Fig. 7.8

Two configurations that keep coming back. Left: in an isosceles triangle, the line from the apex to the midpoint of the base splits it into two congruent triangles (SSS). Right: a diagonal splits a parallelogram into two congruent triangles — the diagonal is a common side, which is the fact students most often forget to write down.

Walkthrough 9 — a congruence proof that earns a right angle Intermediate ✕ Not in G2

In △ABC, AB = AC, and D is the midpoint of BC. (i) Prove that △ABD ≡ △ACD. (ii) Hence prove that AD is perpendicular to BC.

AB = AC (given)

First of the three facts. Even a fact copied straight out of the question needs its reason written — "(given)" is a perfectly good one, and its absence is what turns a proof into a list.
BD = CD (D is the midpoint of BC)

Second fact. "Midpoint" is exactly the statement that the two halves are equal; say so, rather than assuming the marker will supply it.
AD = AD (common side)

Third fact, and the one that gets left out. AD belongs to both triangles, so it is equal to itself — a free side, and without it there are only two facts and no test.
∴ △ABD ≡ △ACD (SSS Congruence Test)

Three pairs of equal sides, so SSS. The order is set by the three facts: A ↔ A, B ↔ C, D ↔ D. Part (i) is finished.
∠ADB = ∠ADC (corresponding ∠s of congruent triangles)

Part (ii) starts by cashing in the congruence: once two triangles are congruent, every pair of corresponding angles is equal. Pick the pair you need — here the two angles at D.
∠ADB + ∠ADC = 180° (adj. ∠s on a straight line, BDC)

B, D and C lie on a straight line, so the two angles at D are adjacent angles on a straight line and add to 180°. This is the second fact the conclusion needs.
2∠ADB = 180°, so ∠ADB = 90° ∴ AD ⊥ BC

Equal and adding to 180° forces each to be 90°. Finish with the sentence the question asked for — "∴ AD is perpendicular to BC" — because a "prove that" question is only answered when the statement it named has been written down.

Walkthrough 10 — RHS, and the length that follows Intermediate ✕ Not in G2

In △PQR, ∠PQR = 90°, PR = 17 cm and QR = 8 cm. In △XYZ, ∠XYZ = 90°, XZ = 17 cm and YZ = 8 cm. (i) Prove that the two triangles are congruent. (ii) Find XY.

∠PQR = ∠XYZ = 90° (given)

The R of RHS. Both triangles must be right-angled before the test may be used at all, so this line comes first.
PR = XZ = 17 cm (hypotenuse, given)

The hypotenuse is the side opposite the right angle: with the right angle at Q, that is PR, the side made of the other two letters. Naming it as the hypotenuse in the reason is what makes the test RHS.
QR = YZ = 8 cm (given)

The S — one other pair of equal sides. Note that QR and YZ are in the same position relative to the right angle in each triangle.
∴ △PQR ≡ △XYZ (RHS Congruence Test)

Right angle, hypotenuse, side — RHS. The correspondence is P ↔ X, Q ↔ Y, R ↔ Z, which the three lines above already set out.
PQ² = 17² − 8² = 289 − 64 = 225, so PQ = 15 cm

Pythagoras' Theorem in the triangle you have three facts about. The hypotenuse is the largest number, so it is the one being squared and subtracted from.
XY = PQ = 15 cm (corresponding sides of congruent triangles)

This is the point of part (i): the congruence transfers the length. X ↔ P and Y ↔ Q, so XY matches PQ. Quote the reason — without it, the 15 looks as though it came from nowhere.

More detail

AAS and ASA are the same test. Once two angles of one triangle match two angles of the other, the third pair matches automatically (the angle sum of a triangle is 180°), so it makes no difference whether the equal side lies between the two angles or not. think! Mathematics prints the test as AAS; either name is accepted.

The four diagrams above show the second triangle reflected and turned on purpose: congruent triangles almost never arrive in the same orientation, so it is the matching marks — ticks for equal sides, arcs for equal angles — that are read, not matching positions on the page.

Past-paper questions

G3 N2016 · Paper 2 · Q3(a) 2 marks

The diagram shows part of a regular n-sided polygon, ABCDE….

Explain why triangles ABC and CDE are congruent. [2]

Worked solution

AB = CD (all sides of a regular polygon are equal)

BC = DE (all sides of a regular polygon are equal)

∠ABC = ∠CDE (all interior angles of a regular polygon are equal)

In each triangle the equal angle is the one between the two equal sides, so △ABC ≡ △CDE (SAS Congruence Test).

Why this works. Two marks: one for the three matching facts, one for naming a test those facts justify. Both triangles use two sides and the angle squeezed between them, which is SAS; SSS is not available, because nothing so far gives AC = CE. Write the vertices in matching order, A ↔ C, B ↔ D, C ↔ E.

G3 N2018 · Paper 2 · Q7(a) 3 marks

AB is a diameter of the large circle, centre O.
CD is a diameter of the small circle, centre O.
AC and BD are tangents to the small circle.
Show that triangle OAC is congruent to triangle OBD.
Give a reason for each statement you make. [3]

Worked solution

∠OCA = ∠ODB = 90° (a tangent is perpendicular to the radius at the point of contact)

∠AOC = ∠BOD (vertically opposite angles: AOB and COD are both straight lines through O)

OC = OD (radii of the small circle)

In each triangle OC, OD is the side joining the two equal angles, so △OAC ≡ △OBD (AAS Congruence Test). (shown)

The same three marks can be earned with RHS instead: the right angles at C and D, the hypotenuses OA = OB (radii of the large circle) and the sides OC = OD (radii of the small circle).

Why this works. "Give a reason for each statement" puts the marks in the brackets: three statements, three reasons, then the test. "Diameter" makes AOB and COD straight lines, which turns the angles at O into vertically opposite ones; "tangent" plants the right angles at the points of contact. AC = BD follows from the congruence, so it is not one of the reasons.

G3 N2019 · Paper 1 · Q16 4 marks

TA and TB are tangents to a circle, centre O.
Show that triangle OAT and triangle OBT are congruent.
Give a reason for each statement you make. [3]
Point P lies on the circumference of the circle so that triangle APT is congruent to triangle BPT.
On the diagram, mark the two possible positions for the point P. [1]

Worked solution

(a) ∠OAT = ∠OBT = 90° (a tangent is perpendicular to the radius at the point of contact)

OA = OB (radii of the circle)

OT is common — and it is the hypotenuse of both right-angled triangles.

Right angle, hypotenuse and one side, so △OAT ≡ △OBT (RHS Congruence Test).

(b) PT is a side of both △APT and △BPT, and TA = TB already (from part (a)), so the triangles are congruent exactly when PA = PB — that is, when P lies on the perpendicular bisector of AB. Both T (because TA = TB) and O (because OA = OB) lie on that bisector, so the bisector is the line TO itself.

The two positions of P are the two points where the line TO, produced, cuts the circle — the near end of that diameter and the far end, marked below.

Why this works. Part (a) is the tangent-pair congruence: two right angles and a shared hypotenuse name RHS, and TA = TB is what (a) proves rather than a reason inside it. Part (b) asks what is missing: PT is shared, so the last condition is PA = PB, the perpendicular bisector. It cuts the circle twice, which is why two positions are wanted.

G3 N2022 · Paper 1 · Q15 4 marks

In the diagram, AB and DE are equal in length and parallel.
ACE and BCD are straight lines.
Show that BD bisects AE.
Give a reason for each statement you make. [4]

Worked solution

Work in the two triangles △ABC and △EDC.

∠ABC = ∠EDC (alternate angles, AB // ED, with BCD the transversal)

∠BCA = ∠DCE (vertically opposite angles)

AB = ED (given)

An angle, an angle and a matching side, so △ABC ≡ △EDC (AAS Congruence Test).

Hence AC = EC (corresponding sides of congruent triangles), so C is the midpoint of AE. C lies on BD, so BD bisects AE. (shown)

Why this works. Four marks, and the last is for using the congruence: AC = EC, so BD cuts AE in half. AB // ED with ACE and BCD as transversals gives alternate angles, and the two angles at C are vertically opposite. Letter order matters: △ABC ≡ △EDC pairs A with E and B with D.

G3 N2024 · Paper 2 · Q8(a)(i) 3 marks

A, B, C and D are points on the circle, centre O.
M is the midpoint of AB and N is the midpoint of CD.
AB = CD.

Show that triangle BMO is congruent to triangle CNO.
Give a reason for each statement you make. [3]

Worked solution

∠OMB = ∠ONC = 90° (the line from the centre to the midpoint of a chord is perpendicular to that chord)

OB = OC (radii of the circle) — and each is the hypotenuse of its right-angled triangle

BM = ½AB = ½CD = CN (M and N are the midpoints of equal chords)

Right angle, hypotenuse and one side, so △BMO ≡ △CNO (RHS Congruence Test). (shown)

Why this works. The circle is scenery; the marks are for three labelled statements. The third is where the given AB = CD is spent: halve both chords and the halves are the sides you need, which is shorter than quoting "equal chords are equidistant from the centre". The paper's order △BMO ≡ △CNO pairs B with C, so BM matches CN.

SSA is not a congruence test ✕ Not in G2

The textbook gives this its own section. Two sides and an angle that is not between them does not fix a triangle — there are usually two different triangles with those measurements.

Fig. 7.9

∠B = 60°, AB = 8 cm and AC = 7 cm — three facts, but the angle is not between the two sides. Swinging a 7 cm arc from A meets the other arm twice, at C₁ and C₂. So BC is 3 cm or 5 cm: two triangles, both fitting the data, and not congruent to each other.

The one exception is the right angle. Two sides and a non-included angle do pin the triangle down when that angle is the right angle — and that case has its own name, the RHS Congruence Test. Outside it, "SSA" is not a reason.

Walkthrough 11 — spotting the SSA trap Advanced ✕ Not in G2

In △ABC, ∠ABC = 52°, AB = 9 cm and AC = 8 cm. In △PQR, ∠PQR = 52°, PQ = 9 cm and PR = 8 cm. A student writes "△ABC ≡ △PQR (SAS)". (i) Explain what is wrong with the student's answer. (ii) State one extra piece of information that would make the two triangles congruent, and name the test it would use.

In △ABC, the angle 52° is at B, so it lies between AB and BC.

Locate the given angle before judging the test. An angle "at B" is the angle between the two sides that contain the letter B — here AB and BC.
The two given sides are AB and AC. The angle between those is ∠BAC, which is not what was given.

This is the test. For SAS the angle must sit between the two sides you quoted. AB and AC meet at A, so SAS would need ∠A; the question gave ∠B.
So the data is side, side, non-included angle — SSA, not SAS, and SSA is not a congruence test.

Name it precisely. The answer wanted is "SSA, which is not a congruence test", rather than "the angle is in the wrong place".
Two different triangles can have ∠B = 52°, AB = 9 cm and AC = 8 cm, so the two triangles need not be congruent.

Back the claim with the reason SSA fails: an arc of 8 cm swung from A can cut the arm from B in two places. One sentence is enough; a sketch like the one above is worth adding if there is room.
Any one of: BC = QR → SSS; ∠BAC = ∠QPR → SAS; ∠ACB = ∠PRQ → AAS.

Each extra fact repairs the proof a different way: a third pair of sides gives SSS; the included angle for AB and AC gives SAS; a second pair of angles turns AB into AAS. One of them, with its test named, is enough.

Check yourself

Given that △LMN ≡ △RST, LM = 9.3 cm, ∠M = 71° and ∠N = 44°, write down RS, ∠S and ∠L.
Answer
L ↔ R, M ↔ S, N ↔ T. So RS = LM = 9.3 cm and ∠S = ∠M = 71°. For the last one, ∠L = 180° − 71° − 44° = 65°.
Name the congruence test proved by each set of facts, or write "none". (a) three pairs of equal sides; (b) two pairs of equal sides and a pair of equal angles that are not between them; (c) a right angle, equal hypotenuses and one other pair of equal sides; (d) three pairs of equal angles. ✕ Not in G2
Answer
(a) SSS.
(b) None — that is SSA, which is not a congruence test.
(c) RHS.
(d) None — equal angles fix the shape but not the size, so the triangles are similar, not necessarily congruent.
ABCD is a parallelogram and the diagonal AC is drawn. Prove that △ABC ≡ △CDA. ✕ Not in G2
Answer
AB = CD (opp. sides of a parallelogram)
BC = DA (opp. sides of a parallelogram)
AC = CA (common side)
∴ △ABC ≡ △CDA (SSS Congruence Test). Watch the order: A ↔ C, B ↔ D, C ↔ A.
ABCD is a kite with AB = AD and CB = CD. Prove that △ABC ≡ △ADC, and hence that AC bisects ∠BAD. ✕ Not in G2
Answer
AB = AD (given)
CB = CD (given)
AC = AC (common side)
∴ △ABC ≡ △ADC (SSS Congruence Test).
Hence ∠BAC = ∠DAC (corresponding ∠s of congruent triangles), so AC bisects ∠BAD — which is why the angle-bisector construction works: the compass arcs build exactly this kite.
In △ABC, ∠BAC = 90°, AB = 20 cm and BC = 29 cm. In △PQR, ∠QPR = 90°, PQ = 20 cm and QR = 29 cm. Name the test that proves the triangles congruent, and find AC. ✕ Not in G2
Answer
The right angles are at A and P, so the hypotenuses are BC and QR: equal at 29 cm. With AB = PQ = 20 cm as well, the test is RHS. Then AC² = 29² − 20² = 841 − 400 = 441, so AC = 21 cm.

Similar figures, and the scale factor `k`

Two figures are similar if they have exactly the same shape, though not necessarily the same size. For two triangles this means both of the following:

all their corresponding angles are equal, and
all their corresponding sides are proportional — every ratio of matching sides is the same number.

That common number is the scale factor k. Congruence is the special case k = 1, so congruence is a special case of similarity.

Fig. 7.10

△ABC and △PQR at a genuine scale factor of 1.5: every length in the second is 1.5 times its match in the first (6.6 = 1.5 × 4.4, 9 = 1.5 × 6, 7.8 = 1.5 × 5.2), while the three angles are untouched. Matching angles carry matching arcs, so the correspondence reads off the picture: △ABC is similar to △PQR.

The corresponding order runs the ratios too: "△ABC is similar to △PQR" fixes A ↔ P, B ↔ Q, C ↔ R, and therefore

\frac{A B}{P Q} = \frac{B C}{Q R} = \frac{A C}{P R}

Every fraction has the same triangle on top. Mixing them — AB/PQ on one side and QR/BC on the other — turns a scale factor into its own reciprocal.

Walkthrough 12 — scale factor, then the missing sides Basic

△ABC is similar to △PQR, with AB = 6 cm, BC = 9 cm and CA = 7.5 cm and PQ = 9 cm. Find the scale factor and the lengths of QR and RP.

A ↔ P, B ↔ Q, C ↔ R, so AB matches PQ, BC matches QR and CA matches RP.

Unpack the statement first, exactly as with a congruence. Which side matches which is fixed by the letter order, not by which looks longest in the diagram.
k = $\frac{P Q}{A B}$ = $\frac{9}{6}$ = 1.5

The only matching pair whose lengths are both known is AB and PQ, so that pair supplies k. Putting △PQR on top makes k multiply from △ABC to △PQR, the direction the rest of the question needs.
QR = k × BC = 1.5 × 9 = 13.5 cm

Multiply the matching side of the smaller triangle by k. A quick sanity check every time: k > 1, so every answer must come out bigger than the side it came from.
RP = k × CA = 1.5 × 7.5 = 11.25 cm

Same factor, every time — that is what "similar" means. One k does the whole figure; there is never a second scale factor for a second side.
k = 1.5, QR = 13.5 cm, RP = 11.25 cm

Collect the three answers the question asked for, with units. Notice that 11.25 is exact, so it stays as 11.25: nothing here asked for rounding.

More detail

The textbook writes similarity out in words — "△ABC is similar to △XYZ" — rather than with a symbol, and that is what these notes do too. Congruence keeps its sign, ≡.

Turning every fraction the other way up is allowed, as long as it is done to all of them at once: PQ/AB = QR/BC = PR/AC says exactly the same thing. What changes the meaning is mixing the two directions, because a scale factor and its reciprocal are different numbers.

The three similarity tests ✕ Not in G2

The definition of similar triangles lists six conditions — three equal angles and three equal ratios — and none of the tests below needs all six. Each one is a short cut, and each one is the mirror of a congruence test.

Test	What you must show	Mirrors
AA Similarity Test	2 corresponding angles are equal	the AAS Congruence Test
SSS Similarity Test	the 3 ratios of corresponding sides are equal	the SSS Congruence Test
SAS Similarity Test	2 ratios of corresponding sides are equal and the angle between those sides is equal	the SAS Congruence Test

Fig. 7.11

The three similarity tests, with the second triangle drawn at a real scale factor of 1.5. Lengths on the SSS and SAS diagrams are in centimetres.

AA is the whole test. Once two angles of one triangle equal two angles of another, the third pair matches automatically (angle sum 180°), so a third angle adds nothing.

Walkthrough 13 — proving similarity with the SSS Similarity Test Basic ✕ Not in G2

In △ABC, AB = 4.5 cm, BC = 6 cm and AC = 7.5 cm. In △XYZ, XY = 12 cm, YZ = 7.2 cm and XZ = 9.6 cm. Show that the two triangles are similar, and write the similarity statement with the vertices in the correct order.

Sorted by size: △ABC has 4.5, 6, 7.5 and △XYZ has 7.2, 9.6, 12.

Pair the sides shortest with shortest, longest with longest. The letters cannot be trusted to be in matching order — only the lengths can — and this ordering is what makes the three ratios comparable.
$\frac{Y Z}{A B}$ = $\frac{7.2}{4.5}$ = 1.6

First ratio: the shortest side of each. Write the bigger triangle on top so the answers come out greater than 1 and are easier to compare by eye.
$\frac{X Z}{B C}$ = $\frac{9.6}{6}$ = 1.6 and $\frac{X Y}{A C}$ = $\frac{12}{7.5}$ = 1.6

The other two ratios. All three must be shown — two out of three is not the SSS Similarity Test, and the third one is where a deliberately-not-similar pair gives itself away.
The three ratios are equal, so the sides are proportional. Matching sides give A ↔ Y, B ↔ Z, C ↔ X.

Read the correspondence off the pairs: AB matched YZ, so {A,B} pairs with {Y,Z}; BC matched XZ, so B, the letter in both, pairs with Z, the letter in both. The rest follows.
∴ △ABC is similar to △YZX (SSS Similarity Test)

Name the test, and put the vertices in the order the ratios produced — "△ABC is similar to △XYZ" would be a different, false, statement. The scale factor is 1.6, from △ABC to △YZX.

Why this works

Why AA and not AAA. If two angles of one triangle equal two angles of another, the third pair is equal automatically — the angles of a triangle add to 180°. So checking a third angle is wasted work, and AA is the whole test. There is no "AAS Similarity Test" for the same reason: once two angles match, the sides look after themselves.

And why SSA is missing here too. The similarity tests are the congruence tests with "equal" replaced by "in the same ratio", so the one arrangement that was never a congruence test is not a similarity test either.

The two configurations the paper keeps using

In an exam, similar triangles almost never arrive as two separate drawings. They arrive overlapping, in one of these two arrangements — and the whole skill is seeing which is which and untangling the correspondence.

Fig. 7.12

Left, the nested pair: PQ is parallel to BC, so ∠APQ = ∠ABC (corresponding angles) and ∠A is common to both triangles — two equal angles, so △APQ is similar to △ABC. Right, the butterfly: AB is parallel to DC and the lines AC and BD cross at O, so ∠AOB = ∠COD (vertically opposite angles) and ∠BAO = ∠DCO (alternate angles) — again two equal angles.

In the nested pair, AP matches the whole AB. The similar triangles are △APQ and △ABC, so the piece PB is not a matching side; if the question gives AP and PB, add them first.

More detail

The nested pair is where the correspondence is easiest to get wrong, because the picture offers PB and the similarity wants AB. Writing AP/PB where AP/AB belongs gives a plausible-looking answer from a method that was otherwise right, so the safe habit is to add AP + PB and label the whole side before any ratio is written.

Finding an unknown length

Once two triangles are known to be similar, one equation does the rest: pick the ratio containing the length you want, set it equal to a ratio you know completely, and solve.

Walkthrough 14 — the nested pair Intermediate

In △ABC, P lies on AB and Q lies on AC, and PQ is parallel to BC. Given that AP = 4 cm, PB = 6 cm and PQ = 5 cm, find the length of BC.

∠APQ = ∠ABC (corr. ∠s, PQ // BC)
∠PAQ = ∠BAC (common angle)
∴ △APQ is similar to △ABC (AA Similarity Test)

Never start calculating until the similarity has been established and named. The two angles come free: one from the parallel lines, one because the two triangles share the vertex A. Writing the reason for each is where the first marks are.
AB = AP + PB = 4 + 6 = 10 cm

The side of the big triangle that matches AP is the whole AB, not the piece PB. Doing this addition as its own line stops the two from being confused later.
$\frac{P Q}{B C}$ = $\frac{A P}{A B}$

Choose the ratio that contains what you want (BC) and set it equal to a ratio you know completely. Both fractions have the small triangle on top, which is the rule that keeps the equation the right way up.
$\frac{5}{B C}$ = $\frac{4}{10}$ = $\frac{2}{5}$

Substitute, and simplify the known ratio at once — 4/10 = 2/5 is the scale factor from △ABC down to △APQ, and small numbers make the last step easier.
BC = 5 × $\frac{5}{2}$ = 12.5 cm

Cross-multiply, or just multiply PQ by the reciprocal of the scale factor. Check the direction: △ABC is the bigger triangle, so BC must come out bigger than PQ = 5 cm — and 12.5 is.

Walkthrough 15 — the butterfly Intermediate

In the butterfly configuration, AB is parallel to DC, and the straight lines AC and BD cross at O. Given that AB = 6 cm, DC = 15 cm and OB = 4 cm, find OD and the ratio AO : OC.

∠AOB = ∠COD (vert. opp. ∠s)
∠BAO = ∠DCO (alt. ∠s, AB // DC)
∴ △AOB is similar to △COD (AA Similarity Test)

Two angles again, from the two facts the figure hands you: the lines cross, and the ends are parallel. The conclusion's order A ↔ C, O ↔ O, B ↔ D is set by those two angle statements.
k = $\frac{C D}{A B}$ = $\frac{15}{6}$ = 2.5

AB and CD are the matching pair (first-and-third letters of each triangle), and both lengths are given, so this pair supplies the scale factor from △AOB to △COD.
OD matches OB, so OD = k × OB = 2.5 × 4 = 10 cm

O is the 2nd letter of both triangles and B, D are the 3rd letters, so OB and OD correspond. Multiplying by k moves from the small triangle to the big one.
AO : OC = AB : CD = 6 : 15 = 2 : 5

AO matches CO, so their ratio is the same as every other matching pair's — it is 1 : k. Give the ratio in its simplest form: divide both parts by 3. A ratio question wants 2 : 5, not 0.4.

Walkthrough 16 — when the ratio hides an equation Advanced

In the figure, PXQ and PYR are straight lines and XY is parallel to QR. Given that PX = 6 cm, XQ = 9 cm, YR = 12 cm and XY = 8 cm, find (i) PY and (ii) QR.

Fig. 7.13

The two unknowns, and the four lengths that are given.

∠PXY = ∠PQR (corr. ∠s, XY // QR)
∠XPY = ∠QPR (common angle)
∴ △PXY is similar to △PQR (AA Similarity Test)

Sketch it before anything else — the nested configuration again, with P as the shared vertex. Establish and name the similarity first, whatever the question looks like.
PQ = PX + XQ = 6 + 9 = 15 cm, so $\frac{P X}{P Q}$ = $\frac{6}{15}$ = $\frac{2}{5}$

Add the two pieces to get the whole side, then simplify the ratio. Every matching pair in this figure is now in the ratio 2 : 5, small to large.
$\frac{P Y}{P Y + 12}$ = $\frac{2}{5}$

Here is the twist: PR is not given. But PR = PY + YR = PY + 12, so the ratio PY/PR can be written entirely in terms of the one unknown — and an equation in one unknown can always be solved.
5PY = 2(PY + 12) = 2PY + 24

Cross-multiply and expand the bracket. From here it is an ordinary linear equation; the geometry is finished.
3PY = 24, so PY = 8 cm

Collect and divide. Check it against the ratio: PR would be 8 + 12 = 20, and 8 : 20 is 2 : 5 — so the answer is consistent.
QR = XY × $\frac{5}{2}$ = 8 × 2.5 = 20 cm

XY and QR are a matching pair (2nd-and-3rd letters of each), so the same 2 : 5 applies — and going from small to large means multiplying by 5/2. State both answers: PY = 8 cm and QR = 20 cm.

Check yourself

△DEF is similar to △XYZ, with DE = 8 cm, EF = 6 cm and XY = 20 cm. Find the scale factor and the length of YZ.
Answer
D ↔ X, E ↔ Y, F ↔ Z, so DE matches XY and k = 20 ÷ 8 = 2.5. Then YZ = 2.5 × 6 = 15 cm.
Are two triangles with sides 3 cm, 4 cm, 6 cm and 7.5 cm, 10 cm, 15 cm similar? Show your working. ✕ Not in G2
Answer
Pair them shortest to shortest: 7.5 ÷ 3 = 2.5, 10 ÷ 4 = 2.5, 15 ÷ 6 = 2.5. All three ratios are equal, so yes — similar, by the SSS Similarity Test, with scale factor 2.5.
Are two triangles with sides 5 cm, 7 cm, 9 cm and 10 cm, 14 cm, 16 cm similar? ✕ Not in G2
Answer
10 ÷ 5 = 2 and 14 ÷ 7 = 2, but 16 ÷ 9 = 1.77… The third ratio is different, so no — the triangles are not similar. Two matching ratios are never enough on their own; that is why the test needs all three.
In △ABC, X lies on AB and Y on AC, with XY parallel to BC. Given AX = 5 cm, XB = 7 cm and XY = 4.5 cm, find BC.
Answer
△AXY is similar to △ABC (AA), and AB = 5 + 7 = 12 cm, so XY/BC = AX/AB = 5/12. Then BC = 4.5 × 12 ÷ 5 = 10.8 cm. The trap is using 5/7 instead of 5/12.
A vertical stick 1.4 m tall casts a shadow 2.1 m long. At the same moment a flagpole casts a shadow 13.5 m long. Find the height of the flagpole.
Answer
The sun's rays are parallel, so the two stick-and-shadow triangles are similar (AA: both have a right angle and the same angle of elevation). Then height ÷ 1.4 = 13.5 ÷ 2.1, so the height is 1.4 × 13.5 ÷ 2.1 = 9 m.
In △PQR, S lies on PQ and T on PR, with PS = 4 cm, SQ = 6 cm, PT = 5.2 cm and TR = 7.8 cm. Show that ST is parallel to QR. ✕ Not in G2
Answer
PQ = 10 cm and PR = 13 cm, so PS/PQ = 4/10 = 2/5 and PT/PR = 5.2/13 = 2/5. With ∠P common, △PST is similar to △PQR (SAS Similarity Test). Hence ∠PST = ∠PQR, and since these are corresponding angles, ST // QR.

Past-paper questions

G3 N2021 · Paper 1 · Q14 4 marks

In the diagram, ABC and ADE are triangles such that angle ABC = angle ADE.

Show that the two triangles are similar. [2]
Given that AC = 12 cm, AB = 10 cm and AE = 9 cm, show that CD = 4.5 cm.
Show all your working. [2]

Worked solution

(a) ∠ABC = ∠ADE (given)

∠BAC = ∠DAE (the angle at A is common to both triangles)

Two pairs of equal angles, so △ABC is similar to △ADE (AA Similarity Test).

(b) The similarity pairs A ↔ A, B ↔ D and C ↔ E, so AB matches AD and AC matches AE:

$\frac{A D}{A B} = \frac{A E}{A C}$ ⇒ $\frac{A D}{10} = \frac{9}{12}$

$A D = 10 \times \frac{9}{12} = 7.5$ cm

CD = AC − AD = 12 − 7.5 = 4.5 cm (shown)

Why this works. In the second standard configuration the correspondence comes first: the equal angles at B and D match, A matches itself, and C matches E. That fixes the ratio as $\frac{A D}{A B} = \frac{A E}{A C}$ : the sides cross over, so pairing AD with AC is the wrong match. The question asks for CD, so the subtraction is its own mark.

Everything in this tab is examined in G3 Mathematics only — "ratio of areas of similar plane figures" is G3 outcome 2.9, and the G2 syllabus has no matching item. ✕ Not in G2

Areas scale by `k`²

Double every length of a shape and the area does not double: two copies fit along the base and two more stack on top, so the area is multiplied by four. Triple the lengths and nine copies fit.

If two figures are similar with scale factor k, so that every length of the second is k times the matching length of the first, then

\frac{area of the second}{area of the first} = k^{2}

Equivalently, writing A₁, A₂ for the two areas and l₁, l₂ for any pair of matching lengths,

\frac{A_{2}}{A_{1}} = {(\frac{l_{2}}{l_{1}})}^{2}

Fig. 7.14

The original triangle, then the same triangle with every length × 2 and × 3, each cut by lines parallel to its own sides. Exactly 2² = 4 copies fill the first enlargement and 3² = 9 fill the second — one copy is shaded in each, drawn at its true size.

"Any pair of matching lengths" really is any pair. Heights, bases, diagonals, radii, slant heights — and perimeters, because a perimeter is a sum of lengths and scales like one. So a ratio of perimeters hands you k directly.

Same height, different bases is not a k² question. Those two triangles are not similar: their areas are in the ratio of their bases, because only one factor of ½ × base × height changes.

Walkthrough 17 — from a length ratio to an area Basic

Two similar triangles have corresponding sides of 5 cm and 8 cm. The area of the smaller triangle is 45 cm². Find the area of the larger triangle.

k = $\frac{8}{5}$ (larger ÷ smaller)

Write the scale factor as a fraction and say which way round it goes. Going from the small triangle to the large one puts 8 on top, and every later step depends on that choice being written down.
$\frac{larger area}{smaller area} = k^{2} = {(\frac{8}{5})}^{2} = \frac{64}{25}$

Quote the rule and then square the fraction — both the 8 and the 5. Squaring the top alone gives an answer 5 times too big.
Larger area = 45 × $\frac{64}{25}$

Multiply the known area by the area factor. The direction check: k² > 1, so the answer must come out bigger than 45.
= 115.2 cm²

45 ÷ 25 = 1.8 and 1.8 × 64 = 115.2 — cancelling before multiplying keeps the numbers small. Units are cm²: an area answer carries the squared unit.

Walkthrough 18 — and back again: from an area ratio to a length Basic

Two similar figures have areas of 48 cm² and 108 cm². The height of the smaller figure is 6 cm. Find the height of the larger figure.

$k^{2} = \frac{108}{48} = \frac{9}{4}$

The ratio of the areas is k², not k. Simplify it at once: 108 and 48 share a factor of 12, and the 9/4 that appears is a standing invitation to take a square root.
k = $\sqrt{\frac{9}{4}}$ = $\frac{3}{2}$

This is the reverse direction, and it is where the marks are: to get from an area ratio back to a length ratio, take the square root. Root the top and the bottom separately — √9 = 3 and √4 = 2.
Larger height = 6 × $\frac{3}{2}$

A height is a length, so it scales by k — not by k². The 9/4 was the area ratio; the height takes its square root, 3/2.
= 9 cm

6 × 3 ÷ 2 = 9, in cm, because this is a length again. Sense check: the larger figure has just over twice the area, so its height should be about 1.5 times as big — and 9 is 1.5 × 6.

Walkthrough 19 — a perimeter is a length Intermediate

Two similar polygons have perimeters of 84 cm and 105 cm. The area of the smaller polygon is 128 cm². Find the area of the larger polygon.

A perimeter is a sum of lengths, so it scales by k: $k = \frac{105}{84} = \frac{5}{4}$

Say why a perimeter may be used here: it is a length, not an area, so this ratio is k itself. 105 and 84 share 21, giving the tidy 5/4 — always simplify before squaring.
$k^{2} = {(\frac{5}{4})}^{2} = \frac{25}{16}$

Square the length ratio to get the area ratio. The figures are polygons, not triangles, but the rule does not care: k² holds for any pair of similar plane figures.
Larger area = 128 × $\frac{25}{16}$

Multiply the smaller area by the area factor, in that direction, because 105 > 84 means the second polygon is the larger one.
= 8 × 25 = 200 cm²

Cancel first: 128 ÷ 16 = 8, then 8 × 25 = 200. Answer in cm².

Walkthrough 20 — the triangle-and-trapezium split Advanced

In △ABC, P lies on AB and Q lies on AC, with PQ parallel to BC and AP : PB = 2 : 3. Given that the area of △APQ is 24 cm², find (i) the area of △ABC and (ii) the area of the trapezium PBCQ.

△APQ is similar to △ABC (AA: ∠APQ = ∠ABC, corr. ∠s, PQ // BC; ∠A common)

Establish and name the similarity first, exactly as in 7.3. The k² rule applies only to similar figures, so this line is what licenses everything below.
AP : AB = 2 : (2 + 3) = 2 : 5, so k = $\frac{5}{2}$ from △APQ to △ABC.

The question gave AP : PB — a part-to-part ratio. The similarity needs AP : AB, which is part-to-whole. Convert by adding: 2 + 3 = 5. Missing this turns every later answer into nonsense.
Area of △ABC = 24 × ${(\frac{5}{2})}^{2}$ = 24 × $\frac{25}{4}$

Areas scale by k², and k > 1 here because △ABC is the bigger triangle — so the answer must exceed 24 cm².
= 6 × 25 = 150 cm²

24 ÷ 4 = 6, then × 25. Part (i) is done, and it is worth boxing, because part (ii) depends on it.
Area of PBCQ = 150 − 24 = 126 cm²

The trapezium is what is left when the small triangle is cut out of the big one, so subtract. A trapezium and a triangle are not the same shape, so no scale factor connects them.

Where this comes from

The rule is the same for every shape, however complicated, because an area is a length times a length: multiply both of those lengths by k and the product is multiplied by k². The figure shows it for a triangle, where k = 2 lets exactly four copies of the original fill the enlargement and k = 3 lets nine.

The look-alike to watch for is two triangles that share a height but sit on different bases — two triangles on the same line with a common apex, for instance. Their areas are in the ratio of their bases, a plain ratio and not a squared one, because only one factor of ½ × base × height is changing. Ask first whether the two figures are actually the same shape; if they are not, k² does not apply at all.

Check yourself

Two similar triangles have corresponding sides of 6 cm and 15 cm. The area of the smaller is 32 cm². Find the area of the larger.
Answer
k = 15/6 = 5/2, so k² = 25/4 and the area is 32 × 25/4 = 200 cm².
Two similar figures have areas 27 cm² and 75 cm². A side of the smaller is 4.5 cm. Find the matching side of the larger.
Answer
k² = 75/27 = 25/9, so k = 5/3 and the side is 4.5 × 5/3 = 7.5 cm. Take the square root to come back from an area ratio to a length ratio.
Two similar polygons have perimeters 72 cm and 96 cm. The area of the larger is 144 cm². Find the area of the smaller.
Answer
A perimeter is a length, so k = 96/72 = 4/3 from small to large. Going the other way, area of the smaller = 144 × (3/4)² = 144 × 9/16 = 81 cm².
Every length of a figure is multiplied by 4. What happens to its area, and to its perimeter?
Answer
The area is multiplied by 4² = 16; the perimeter, being a length, is multiplied by 4.
In △ABC, PQ is parallel to BC with AP : AB = 3 : 5. The area of △ABC is 100 cm². Find the area of the trapezium PBCQ.
Answer
Here the ratio is already part-to-whole, so k = 3/5 from △ABC down to △APQ. Area of △APQ = 100 × 9/25 = 36 cm², so the trapezium is 100 − 36 = 64 cm².

Past-paper questions

G3 N2016 · Paper 2 · Q6 8 marks

The diagram shows two concentric circles, centre O.
A, B and C are points on the larger circle and D is a point on the smaller circle.
ADB is a tangent to the smaller circle and angle BCO = 65°.

Show that triangles ABC and ADO are similar.
Give a reason for each statement you make. [2]
Find the ratio area of triangle BOC : area of triangle ADO. [2]
The radius of the smaller circle is 3 cm.
Find the shaded area. [4]

Worked solution

(a) ∠ADO = 90° (ADB is a tangent to the smaller circle and OD is a radius: a tangent is perpendicular to the radius at the point of contact)

∠ABC = 90° (AC passes through the centre O, so it is a diameter of the larger circle, and the angle in a semicircle is a right angle)

∠DAO = ∠BAC (the angle at A is common to both triangles)

Two pairs of equal angles, so △ABC is similar to △ADO (AA Similarity Test). (shown)

(b) OD is perpendicular to the chord AB and OA = OB (radii), so D is the midpoint of AB; and O is the midpoint of the diameter AC. Every length of △ADO is therefore $\frac{1}{2}$ of the matching length of △ABC, so

area of △ADO : area of △ABC = 1² : 2² = 1 : 4.

Take area of △ABC = 4 units², so area of △ADO = 1 unit². Since O is the midpoint of AC, BO is a median of △ABC and splits it into two equal halves, so area of △BOC = 2 units².

area of △BOC : area of △ADO = 2 : 1

(c) ∠ABC = 90° and ∠BCA = 65°, so

∠BAC = 180° − 90° − 65° = 25° (angle sum of a triangle)

In the right-angled △ADO, OD = 3 cm and ∠DAO = 25°:

$\sin 25 ° = \frac{3}{O A}$ ⇒ $O A = \frac{3}{\sin 25 °} = 7.0986$ cm (5 s.f.)

shaded area = area of the larger circle − area of the smaller circle

$= π \times {7.0986}^{2} - π \times 3^{2} = π (50.390 - 9) = 130.03 \dots$

shaded area = 130 cm² (3 s.f.)

Why this works. "Tangent" gives the right angle at D; "concentric" makes AC a diameter and B a right angle. In (b), square k; D and O are midpoints, so △BOC is half of △ABC. Part (c) is the accuracy mark on the ring: ${(3 \div \sin 25 °)}^{2}$ stays whole: OA = 7 gives 40 $π$ = 126, not 130.

Everything in this tab is examined in G3 Mathematics only — "ratio of volumes of similar solids" is G3 outcome 2.10, and the G2 syllabus has no matching item. ✕ Not in G2

Volumes scale by `k`³

The argument that gave k² in two dimensions runs once more in three. Double every edge of a cube and eight copies of the original fit inside it — two along, two across, two up.

Two solids are similar if one is an enlargement of the other — same shape, every length multiplied by the same scale factor k. Then

\frac{surface area of the second}{surface area of the first} = k^{2} and \frac{volume of the second}{volume of the first} = k^{3}

In the textbook's subscript notation, with l₁, l₂ any pair of matching lengths,

\frac{A_{2}}{A_{1}} = {(\frac{l_{2}}{l_{1}})}^{2} \frac{V_{2}}{V_{1}} = {(\frac{l_{2}}{l_{1}})}^{3}

Fig. 7.15

A cube, and the same cube with every edge doubled. The larger one is ruled into a 2 × 2 × 2 grid: 2³ = 8 copies of the original. The surface area, meanwhile, has gone up by only 2² = 4.

Fig. 7.16

Two similar cones, drawn at their true ratio. Every length is in the ratio 2 : 3, so the surface areas are in the ratio 2² : 3² = 4 : 9 and the volumes in the ratio 2³ : 3³ = 8 : 27. Notice how much bigger 27 : 8 feels than the modest 3 : 2 the eye sees.

Mass, capacity and the cost of filling follow the volume. For two similar solids of the same material, mass = density × volume with the same density, so mass ratio = volume ratio = k³.

The cost of painting is a surface area, so it uses k². Read which of the three — length, area or volume — the question is about before choosing a power.

Walkthrough 21 — from a length ratio to a volume Basic

Two similar cylinders have heights of 6 cm and 9 cm. The volume of the smaller cylinder is 96 cm³. Find the volume of the larger cylinder.

k = $\frac{9}{6}$ = $\frac{3}{2}$ (larger ÷ smaller)

A height is a length, so this ratio is k itself. Simplify it now: 3/2 cubes far more comfortably than 9/6 does.
$\frac{larger volume}{smaller volume} = k^{3} = {(\frac{3}{2})}^{3} = \frac{27}{8}$

Cube the fraction — top and bottom. 3³ = 27 and 2³ = 8. Quoting the rule as a line of working is worth a mark on its own.
Larger volume = 96 × $\frac{27}{8}$

Multiply the known volume by the volume factor. Because k > 1, the answer must be bigger — and by a lot more than the 1.5 the heights suggest.
= 12 × 27 = 324 cm³

96 ÷ 8 = 12, then × 27. Units are cm³. The cylinder radius was never mentioned and never needed: similarity scales everything at once.

Walkthrough 22 — the cube root, going back to a length Basic

Two similar solids have volumes of 128 cm³ and 250 cm³. The larger solid has a base radius of 7.5 cm. Find the base radius of the smaller solid.

$k^{3} = \frac{128}{250} = \frac{64}{125}$

The ratio of the volumes is k³, with the smaller volume on top because the smaller solid is the one asked about. Both sides divide by 2, giving 64 : 125, and both of those are perfect cubes: 64 = 4³, 125 = 5³.
k = $\sqrt[3]{\frac{64}{125}}$ = $\frac{4}{5}$

To come back from a volume ratio to a length ratio, take the cube root — of the top and of the bottom separately. This is the reverse direction, and it is the line the marks sit on.
Smaller radius = 7.5 × $\frac{4}{5}$

A radius is a length, so it scales by k, never by k³. Multiplying by 64/125 here would give 3.84 cm, and would be wrong.
= 6 cm

7.5 × 4 ÷ 5 = 6. Sense check: the volumes are roughly in the ratio 1 : 2, so the lengths should be in a much gentler ratio — and 6 : 7.5 is 4 : 5, as found.

Walkthrough 23 — same material, so mass follows the volume Intermediate

Two similar solid statues are made of the same material and have heights of 24 cm and 60 cm. The smaller statue has a mass of 3.2 kg. Find the mass of the larger statue.

Same material ⇒ same density, so $\frac{m_{2}}{m_{1}} = \frac{V_{2}}{V_{1}} = k^{3}$

Mass = density × volume. The phrase "the same material" is there to tell you the density cancels — so the masses are in exactly the same ratio as the volumes. Without that phrase the question could not be done.
k = $\frac{60}{24}$ = $\frac{5}{2}$

Heights are lengths, so this is k. Divide both by 12 before going anywhere near a cube.
$k^{3} = {(\frac{5}{2})}^{3} = \frac{125}{8}$

5³ = 125, 2³ = 8. The larger statue is only 2.5 times as tall but more than fifteen times as heavy — which is exactly why scale models of heavy things are so light.
Larger mass = 3.2 × $\frac{125}{8}$ = 0.4 × 125 = 50 kg

3.2 ÷ 8 = 0.4, then × 125. Keep the unit the question used — it gave kilograms, so the answer is in kilograms, with no conversion invented.

Where this comes from

The argument that gave k² in two dimensions runs once more in three. A volume is a length times a length times a length, so multiplying every length by k multiplies the volume by k³: double every edge of a cube and eight copies of the original fit inside it, two along, two across and two up.

Both powers live in the same picture. Doubling the edges of that cube multiplies its surface area by only 2² = 4 while the volume goes up by 2³ = 8, which is why a question has to say whether it is asking about a surface or a capacity before a power can be chosen.

Past-paper questions

G3 N2015 · Paper 1 · Q17 4 marks

Two bottles of water are geometrically similar.
The larger bottle holds 2 litres and the smaller bottle holds 1.25 litres.
The height of the larger bottle is 33.5 cm.

Calculate the height of the smaller bottle. [2]
The ratio surface area of larger bottle : surface area of smaller bottle can be written in the form k : 1.
Find the value of k.
Give your answer to 2 decimal places. [2]

Worked solution

(a) Capacity is a volume, so it scales by the cube of the length scale factor. Writing h for the height of the smaller bottle,

${(\frac{h}{33.5})}^{3} = \frac{1.25}{2} = 0.625$

$h = 33.5 \times \sqrt[3]{0.625} = 28.642 \dots$ = 28.6 cm (3 s.f.)

(b) Surface area scales by the square of the same length scale factor, and the length scale factor larger : smaller is $\frac{33.5}{28.642 \dots}$ :

$k = {(\frac{33.5}{28.642 \dots})}^{2} = 1.36798 \dots$ k = 1.37 (2 d.p.)

The same answer in one line, straight from the capacities: $k = {(\frac{2}{1.25})}^{\frac{2}{3}} = {1.6}^{\frac{2}{3}} = 1.36798 \dots$

Why this works. One pair of solids seen through three powers — k, k², k³ — so a volume in means cube root out, and an area out means square. The ratio asked for is larger : smaller, so k comes out bigger than 1. Part (b) is an accuracy mark: keep the unrounded height in the calculator rather than 28.6.

G3 N2017 · Paper 1 · Q23(b) 3 marks

A company manufactures three sizes of the same brand of shampoo.
The bottles are all geometrically similar.
The height of the 250 ml bottle is 18.4 cm.
Calculate the height of the 75 ml bottle. [3]

Worked solution

Let h cm be the height of the 75 ml bottle. Capacities are volumes, so they are in the ratio of the cubes of the heights:

${(\frac{h}{18.4})}^{3} = \frac{75}{250} = \frac{3}{10}$

$\frac{h}{18.4} = \sqrt[3]{\frac{3}{10}} = 0.66943 \dots$

$h = 18.4 \times 0.66943 \dots = 12.317 \dots$ = 12.3 cm (3 s.f.)

Why this works. Three marks for one equation, so the cube is what is being paid for; the 400 ml bottle is a decoy, and only the two named capacities go into the ratio. Treating capacity as a length gives 18.4 × 75 ÷ 250 = 5.52 cm. Cube-rooting $\frac{3}{10}$ gives 0.669, so the smaller bottle is about two thirds as tall, which is what the picture shows.

G3 N2018 · Paper 1 · Q11 3 marks

The cargo volume of an aircraft is 184 m³.
A model of this aircraft is built to a scale of 1 : 20.
Find the cargo volume, in m³, of the model.
Give your answer in standard form. [3]

Worked solution

The scale 1 : 20 is a length ratio, so the volume ratio is

1³ : 20³ = 1 : 8000

cargo volume of the model $= \frac{184}{8000} = 0.023$ m³

= $2.3 \times 10^{−2}$ m³

Why this works. A scale printed as 1 : 20 describes lengths, so a volume needs that number cubed. The last instruction carries a mark of its own: standard form wants exactly one non-zero digit in front of the decimal point, so 0.023 becomes $2.3 \times 10^{−2}$ , not $23 \times 10^{−3}$ . Dividing cubic metres by a pure number leaves cubic metres.

G3 N2019 · Paper 1 · Q20 4 marks

The diagram shows two geometrically similar bottles.
The diameter of the base of the smaller bottle is 4 cm.
The diameter of the base of the larger bottle is 10 cm.

The height of the larger bottle is 30 cm.
Calculate the height of the smaller bottle. [2]
The smaller bottle has a capacity of 125 ml.
The manufacturer makes another geometrically similar bottle that holds 500 ml.
Calculate the diameter of the base of this bottle. [2]

Worked solution

(a) Both diameters are lengths, so every length of the smaller bottle is $\frac{4}{10}$ of the matching length of the larger one:

height of the smaller bottle $= 30 \times \frac{4}{10}$ = 12 cm

(b) Compare the new bottle with the 4 cm one. Capacities are volumes:

${(\frac{d}{4})}^{3} = \frac{500}{125} = 4$

$d = 4 \times \sqrt[3]{4} = 6.3496 \dots$ = 6.35 cm (3 s.f.)

Why this works. The wording names the power each time: "diameter" and "height" are lengths (k), "capacity" is a volume (k³). In (b) the 500 ml bottle pairs with the 125 ml one on the 4 cm base, the only other bottle whose capacity is given. Then 500 ÷ 125 = 4, so the diameter is $4 \sqrt[3]{4}$ cm — about 1.6 times as wide.

G3 N2021 · Paper 1 · Q18(b) 3 marks

Another cruise ship has a length of 330 m.
The capacity of the fuel tanks is 1.228×106 litres.
A model of the ship is made to a scale of 1 : 60.
1. Find the length of the model. [1]
2. Find the capacity of the fuel tanks of the model. [2]

Worked solution

(b)(i) Length is a length, so divide by the scale factor once:

length of the model $= \frac{330}{60}$ = 5.5 m

(b)(ii) Capacity is a volume, so the ratio is the cube of the scale factor: 1³ : 60³ = 1 : 216 000.

capacity of the model's fuel tanks $= \frac{1.228 \times 10^{6}}{216 000} = 5.6851 \dots$ = 5.69 litres (3 s.f.)

Why this works. Part (i) divides by 60 once; part (ii) is a capacity, so it needs 60³ = 216 000, and writing that line down is the method mark. The capacity is already in litres and stays in litres, because dividing by a pure number leaves the unit alone. Round once, at the end: 5.68518… is 5.69 to three significant figures.

G3 N2022 · Paper 2 · Q5(b) 3 marks

A, B and C are similar cones.
The ratio volume of cone A : volume of cone B : volume of cone C = 1 : 3 : 8.

Find the ratio height of cone A : height of cone C. [1]
Find the surface area of cone B as a percentage of the surface area of cone C. [2]

Worked solution

(a) Heights are lengths, so take the cube root of each volume:

height of A : height of C = $\sqrt[3]{1}$ : $\sqrt[3]{8}$ = 1 : 2

(b) height of B : height of C = $\sqrt[3]{3}$ : $\sqrt[3]{8}$ = $\sqrt[3]{3}$ : 2

Surface areas scale by the square of that:

surface area of B : surface area of C = ${(\sqrt[3]{3})}^{2}$ : $2^{2}$

required percentage $= \frac{{(\sqrt[3]{3})}^{2}}{4} \times 100 % = \frac{2.08008 \dots}{4} \times 100 %$

= 52.0 % (3 s.f.)

Why this works. No lengths are given, which is the point: cube root down from volumes to lengths, then square back up to areas, so 3 : 8 becomes $3^{\frac{2}{3}}$ : $8^{\frac{2}{3}}$ , which is 2.08008… : 4. One hop instead gives 37.5%, and $\sqrt{3 \div 8}$ gives 61.2%. Three significant figures makes the answer 52.0%, not 52%.

Cones, cups and the part that is filled

The most common Paper 2 use of k³ is a cone-shaped container with liquid in it. The trick is to notice that the liquid is itself a cone, similar to the container, so a single scale factor links the two.

Fig. 7.17

An inverted conical cup of depth H, filled to a depth h. The liquid is a cone of the same shape as the cup, so the scale factor is h ÷ H. Every length — the depth, the surface radius, the slant — is in that ratio; the surface areas are in the ratio of its square, and the volumes in the ratio of its cube.

Walkthrough 24 — how full is the cup? Advanced

An inverted conical container has a depth of 15 cm. It contains water to a depth of 9 cm. Find (i) the fraction of the container's capacity that is filled, and (ii) the ratio of the area of the water surface to the area of the top of the container.

The water forms a cone similar to the container, so $k = \frac{9}{15} = \frac{3}{5}$

Say why the two are similar: the water's surface is parallel to the top of the container, so the water cone and the whole cone have the same shape. The two depths are matching lengths, so their ratio is k.
$\frac{volume of water}{capacity of container} = k^{3} = {(\frac{3}{5})}^{3}$

"Capacity" is a volume, so the power is 3. Nothing about the radius of the cone is needed — and nothing about π either, because it cancels on both sides of a ratio.
= $\frac{27}{125}$

3³ = 27, 5³ = 125. So the cup is a little over a fifth full even though the water reaches three-fifths of the way up — the answer the eye never guesses, and the reason the question is asked.
$\frac{water surface}{top of container} = k^{2} = {(\frac{3}{5})}^{2} = \frac{9}{25}$ i.e. 9 : 25

Both surfaces are circles, and they are areas, so the power is 2 — the same k, a different power. The question asked for a ratio, so give it as 9 : 25 rather than as a fraction.

Past-paper questions

G3 N2016 · Paper 1 · Q7 2 marks

The diagram shows a cone of height 30 cm.

The volume of the liquid in the cone is half the volume of the cone.
Calculate the depth, h centimetres, of the liquid. [2]

Worked solution

The liquid has the same point and the same slant as the container, so the liquid is itself a cone similar to the whole cone, with heights h and 30. Volumes scale by the cube of the height ratio:

${(\frac{h}{30})}^{3} = \frac{1}{2}$

$h^{3} = \frac{1}{2} \times 30^{3} = 13 500$

$h = \sqrt[3]{13 500} = 23.811 \dots$ h = 23.8 cm (3 s.f.)

Why this works. The cone volume formula is never needed — the radius is not given and would cancel anyway. A partly filled cone standing on its point holds a smaller cone similar to itself, so volume ratio = (length ratio)³ settles it in two lines. The cube root of $\frac{1}{2}$ is 0.7937, so the liquid reaches almost four fifths of the way up, as the diagram shows.

G3 N2023 · Paper 1 · Q18 4 marks

Diagram A shows a sealed cone containing some water.
Diagram B shows the cross-section of the cone and water.
The radius of the base of the cone is 10 cm and the height of the cone is 20 cm.
The radius of the surface of the water is r cm, the depth of the water is d cm and the surface of the water is h cm below the top of the cone.

Show that the depth of the water, d, is 20 − 2r. [1]
The volume of water in the cone is equal to the volume of the empty space in the cone.
Calculate d. [3]

Worked solution

(a) The empty space at the top is a cone similar to the whole cone (the water surface is parallel to the base), with radius r and height h:

$\frac{r}{10} = \frac{h}{20}$ ⇒ $h = 2 r$

The depth of the water is what is left of the 20 cm: d = 20 − h = 20 − 2r (shown)

(b) Water and empty space together make the whole cone, and they are equal, so the empty cone is half the whole cone:

$\frac{1}{3} π r^{2} (2 r) = \frac{1}{2} \times \frac{1}{3} π 10^{2} (20)$

$2 r^{3} = 1000$ , so $r^{3} = 500$ and $r = \sqrt[3]{500} = 7.9370 \dots$

$d = 20 - 2 (7.9370 \dots) = 4.1259 \dots$ d = 4.13 cm (3 s.f.)

Alternatively, the empty cone is half the whole one, so its height is $20 \div \sqrt[3]{2} = 15.874 \dots$ cm and d = 20 − 15.874… = 4.13 cm — the same N2016/P1/Q7 idea, upside down.

Why this works. The water is a frustum with no similar partner, so the question works with the empty space, a cone similar to the container. Part (a) is one mark for saying that as a ratio, which turns two unknowns into one; in (b) "equal volumes" makes the empty cone half, so the $\frac{1}{3} π$ cancels. The question asks for d, so carry 7.9370… into the subtraction.

G3 N2024 · Paper 1 · Q26(b) 4 marks

A second container is also made from a cylinder and cone, each with diameter 20 cm.
The height of the cylinder is 25 cm and the height of the cone is 90 cm.
The container is half full of water, as shown.
Calculate the depth, d cm, of the empty space. [4]

Worked solution

The diameter is 20 cm, so both radii are 10 cm.

total volume $= π 10^{2} (25) + \frac{1}{3} π 10^{2} (90) = 2500 π + 3000 π = 5500 π$ cm³

The container is half full, so the empty space has volume $\frac{1}{2} \times 5500 π = 2750 π$ cm³.

That empty space is a cone of height d, similar to the 90 cm cone, so its radius R satisfies

$\frac{R}{10} = \frac{d}{90}$ ⇒ $R = \frac{d}{9}$

$\frac{1}{3} π {(\frac{d}{9})}^{2} d = 2750 π$

$\frac{d^{3}}{243} = 2750$ ⇒ $d^{3} = 668 250$

$d = \sqrt[3]{668 250} = 87.427 \dots$ d = 87.4 cm (3 s.f.)

Check: 87.4 is less than 90, so the empty space really does sit inside the cone, as the picture shows.

Why this works. "Half full" is about the whole container, so the total volume comes first, cylinder plus cone; leaving it as $5500 π$ keeps the numbers clean, because the $π$ cancels on the next line. The empty part is the piece that is similar to something, so work with that. Its radius is $\frac{d}{9}$ , not 10, because the cone it is cut from is 90 cm tall.

Cutting a cone: the frustum

Slice the top off a cone with a cut parallel to the base and two things are true at once: the piece removed is a small cone similar to the whole one, and what is left — the frustum — is the difference of the two.

Fig. 7.18

A cone of height H with a cut parallel to the base, h from the vertex O. The small cone and the whole cone are similar with scale factor h ÷ H, so their radii are in that ratio too. The frustum is whole cone − small cone; it is not similar to either of them.

Keep the answer in terms of π until the last line, because the rounding instruction applies once, at the end.

Walkthrough 25 — the volume of a frustum Advanced

A cone has height 20 cm and base radius 15 cm. The top of the cone is removed by a cut parallel to the base, 8 cm from the vertex. Find (i) the radius of the circular cut and (ii) the volume of the remaining frustum, in terms of π.

The small cone is similar to the whole cone, with $k = \frac{8}{20} = \frac{2}{5}$

A cut parallel to the base leaves a cone of exactly the same shape, so the two are similar and the heights give k. Simplify 8/20 to 2/5 straight away.
Radius of the cut = k × 15 = $\frac{2}{5}$ × 15 = 6 cm

A radius is a length, so it scales by k — the first power, not the second or third. Part (i) is done.
Volume of the whole cone = ⅓π(15)²(20) = ⅓π(4500) = 1500π cm³

Now an actual volume is wanted, so reach for the formula sheet: volume of a cone = ⅓πr²h. Keep the π symbolic; the question asked for the answer in terms of π, and carrying it costs nothing.
Volume of the small cone = ⅓π(6)²(8) = ⅓π(288) = 96π cm³

The same formula with the small cone's own radius and height. Check it against the ratio rule: 1500π × (2/5)³ = 1500π × 8/125 = 96π ✓ — two independent routes to the same number.
Volume of the frustum = 1500π − 96π = 1404π cm³ (= 4410 cm³ to 3 s.f.)

Subtract: a frustum is the whole cone minus the small cone and is similar to neither, so no power of k gives it directly. π was asked for, so 1404π is the answer and the decimal is only a check.

More detail

This is the one place in the chapter where the printed formula page earns its keep. Volume of a cone = ⅓πr²h is printed on the paper, as is volume of a sphere = ⁴⁄₃πr³, so when a similar-solids question asks for an actual volume rather than a ratio, the formula can be looked up rather than remembered. What the page does not hold is k² or k³, and a frustum question usually needs both halves.

Check yourself

Two similar cones have base radii 4 cm and 10 cm. The volume of the smaller is 48 cm³. Find the volume of the larger.
Answer
k = 10/4 = 5/2, so k³ = 125/8 and the volume is 48 × 125/8 = 750 cm³.
Two similar solids have volumes 54 cm³ and 128 cm³. The height of the smaller is 7.5 cm. Find the height of the larger.
Answer
k³ = 128/54 = 64/27, so k = ∛(64/27) = 4/3 and the height is 7.5 × 4/3 = 10 cm. Cube root, because the given ratio was a volume one.
Two similar solid ornaments made of the same material have heights 15 cm and 25 cm. The smaller has a mass of 1.08 kg. Find the mass of the larger.
Answer
Same material, so mass ratio = volume ratio. k = 25/15 = 5/3 and k³ = 125/27, so the mass is 1.08 × 125/27 = 5 kg.
The total surface areas of two similar jugs are 100 cm² and 225 cm². The smaller jug holds 400 ml. Find the capacity of the larger jug.
Answer
Surface areas give k² = 225/100, so k = 15/10 = 3/2 — a square root. Capacity is a volume, so the ratio is k³ = 27/8 and the larger jug holds 400 × 27/8 = 1350 ml. Two different powers of the same k, in one question.
An inverted conical cup of depth 12 cm holds water to a depth of 8 cm. What fraction of its capacity is filled?
Answer
k = 8/12 = 2/3, so the fraction is (2/3)³ = 8/27 — not 2/3, and not 4/9.
A model car is made to a scale of 1 : 24. The real car's boot has a capacity of 415 litres. Find the capacity of the model's boot, in cm³, correct to 3 significant figures.
Answer
415 litres = 415 000 cm³, and k³ = 1/24³ = 1/13 824, so the model's boot holds 415 000 ÷ 13 824 = 30.02… = 30.0 cm³ (to 3 s.f.). The scale of a model is a length scale, so it has to be cubed before it touches a capacity.

Past-paper questions

G3 N2018 · Paper 2 · Q8(d) 2 marks

The diagram shows a pyramid ABCDE.
The base of the pyramid is a square of side 15 cm.
E is vertically above the centre of the square base.
The vertical height of the pyramid is 20 cm.
A smaller, similar pyramid is removed from the top of the original pyramid.
The diagram below shows the solid remaining.

The vertical height of the solid remaining is 4 cm.
Calculate the area of the top, PQRS, of the remaining solid. [2]

Worked solution

PQRS is the base of the small pyramid that was cut off, and that pyramid is similar to the whole one. Its vertical height is

20 − 4 = 16 cm

so the length scale factor from the original pyramid to the removed one is

$k = \frac{16}{20} = \frac{4}{5}$

side of PQRS $= 15 \times \frac{4}{5} = 12$ cm

area of PQRS = 12 × 12 = 144 cm²

(Or in one step from the area ratio: $15 \times 15 \times {(\frac{4}{5})}^{2} = 225 \times \frac{16}{25} = 144$ .)

Why this works. Both marks turn on which height goes with which pyramid: the 4 cm belongs to the frustum, similar to nothing, while PQRS belongs to the pyramid that was removed, of height 20 − 4 = 16 cm. So the first line is the subtraction: k is $\frac{16}{20}$ , not $\frac{4}{20}$ . Then scale the side by k and square, or the area by k².

What to carry into the exam

Two similar figures, one scale factor. Nearly everything in this chapter is a consequence of choosing the right power of it — and of naming the reason for every line you write.

The one idea: `k`, then `k`², then `k`³ ✕ Not in G2

The chain, forwards. If every length of the second figure is k times the matching length of the first, then ✕ Not in G2

\frac{length}{length} = k \frac{area}{area} = k^{2} \frac{volume}{volume} = k^{3}

The chain, backwards — the half that needs a root:

k = \sqrt{\frac{area}{area}} k = \sqrt[3]{\frac{volume}{volume}}

If the question gives you…	…to get a length	…to get an area	…to get a volume
a length ratio `k` sides, heights, radii, slant heights, perimeters, map scales, model scales	× `k`	× `k`²	× `k`³
an area ratio areas, surface areas, cost of painting	take the square root first	use it as it stands	square root, then cube
a volume ratio volumes, capacities, masses of the same material, cost of filling	take the cube root first	cube root, then square	use it as it stands

Worked in numbers, forwards. Two similar solids whose lengths are in the ratio 2 : 3 have surface areas in the ratio 4 : 9 and volumes in the ratio 8 : 27. ✕ Not in G2

And backwards. Volumes in the ratio 8 : 27 means lengths in the ratio ∛8 : ∛27 = 2 : 3, and therefore areas 4 : 9. Every question in 7.4 and 7.5 is one of those two journeys.

The four congruence tests ✕ Not in G2

Test	Needs	Watch for
SSS	3 pairs of equal sides	a common side counts — write it down as "(common side)"
SAS	2 pairs of equal sides and the included angle	the angle must be between the two sides you quoted
AAS	1 pair of equal sides and 2 pairs of equal angles	ASA is the same test under a different name — the third angle is free
RHS	a right angle, equal hypotenuses and one other pair of equal sides	the hypotenuse is the side opposite the right angle — say the word in your reason
SSA	is not a test	two sides and an angle that is not between them fit two different triangles; RHS is the only exception
AAA	is not a test	equal angles fix the shape only — that is similarity

The three similarity tests ✕ Not in G2

Test	Needs	Mirrors
AA	2 pairs of equal angles — and that is all	AAS Congruence Test
SSS Similarity	all 3 ratios of matching sides equal	SSS Congruence Test
SAS Similarity	2 ratios of matching sides equal and the angle between them equal	SAS Congruence Test

Where similar triangles hide. Nested: a line parallel to one side of a triangle — corresponding angles plus a common angle, so AA. Butterfly: two parallel segments joined by crossing lines — vertically opposite angles plus alternate angles, so AA again. ✕ Not in G2

In an exam the two triangles overlap rather than sit side by side. In the nested case the matching side of AP is the whole AB, never the piece PB.

Bisectors and scale drawings

Object	What it gives you
Perpendicular bisector of `AB`	⊥ to `AB`, through the midpoint; every point on it satisfies `PA` = `PB`; it splits the picture into two congruent triangles
Angle bisector of ∠`BAC`	halves the angle; every point on it is the same perpendicular distance from `AB` and from `AC`
Scale 1 : `n`	drawing → real: × `n`. real → drawing: ÷ `n`. Areas use `n`². Both sides of the ratio must be in the same unit before `n` means anything
Scale factor `k` of an enlargement	every length × `k`; every angle unchanged — which is why enlargement produces a similar figure

The habits that earn the marks

1. Name the correspondence before you use it. △ABC ≡ △XZY means A ↔ X, B ↔ Z, C ↔ Y. Write the pairing out; the six equal sides and angles then read straight off, and the alphabetical-looking answer is usually the wrong one.
2. Every line of a proof carries its reason in brackets. "(given)", "(common side)", "(vert. opp. ∠s)", "(corr. ∠s, PQ // BC)", "(opp. sides of a parallelogram)". Three facts, three reasons, then the conclusion.
3. Name the test. A congruence or similarity statement with no test after it is an unfinished answer.
4. In the nested figure, add the pieces first. The matching side is the whole one: AB = AP + PB. Part-to-part ratios must be turned into part-to-whole ones before they touch a similarity.
5. Keep the same triangle on top all the way through an equation of ratios. Flipping one fraction turns k into 1/k.
6. Convert units before writing a ratio — 1 m = 100 cm, 1 km = 100 000 cm. A ratio 1 : n means nothing until both sides are in the same unit.
7. Answer in the units asked for: cm for a length, cm² for an area, and a ratio in its simplest form when a ratio was requested.

… and four more for the ratio questions ✕ Not in G2

8. Decide which quantity the question is about — a length, an area or a volume — before choosing a power. Mass, capacity and the cost of filling are volumes; surface and the cost of painting are areas; perimeters, radii, slant heights and model scales are lengths.
9. Coming back down, take the root. Square root from an area ratio, cube root from a volume ratio, and only then multiply a length by it.
10. A frustum, a trapezium and an "L-shape left over" are never similar to what they came from — find them by subtracting.
11. Squared and cubed units convert by squared and cubed factors — 1 m² = 10 000 cm², 1 m³ = 1 000 000 cm³, 1 litre = 1000 cm³.

Syllabus at a glance

The five G3 Mathematics (K310) outcomes this chapter covers, quoted from the Secondary Three/Four section of the syllabus — and, beside them, the four G2 outcomes for the same topic, which are not the same list.

G3 Mathematics (K310, first examination 2027) — these notes are written to it. Topic G2. Congruence and similarity, Secondary Three/Four.

LO	What you must be able to do	Covered in
2.6	scale drawings	7.1 Scale drawings, 7.1 Area on a plan
2.7	properties and construction of perpendicular bisectors of line segments and angle bisectors	7.1 Perpendicular bisector, 7.1 Angle bisector, 7.1 Which bisector?
2.8	determining whether two triangles are: congruent similar ✕ Not in G2	7.2 Congruence tests, 7.2 SSA, 7.3 Similarity tests
2.9	ratio of areas of similar plane figures ✕ Not in G2	7.4 Areas
2.10	ratio of volumes of similar solids ✕ Not in G2	7.5 Volumes, 7.5 Cones and cups, 7.5 Frustum

Wording is quoted verbatim from the Secondary Three/Four section of the G3 Mathematics syllabus, and matches the G2 row of the K310 content table. Four things this chapter uses are not in that Sec 3/4 list because they are Secondary One/Two carry-over, listed earlier in the same G2 topic: 2.1 congruent figures, 2.2 similar figures, 2.3 properties of similar triangles and polygons ("corresponding angles are equal; corresponding sides are proportional"), and 2.4 enlargement and reduction of a plane figure. They are recapped in 7.2, 7.3 and 7.1. Map scales for distance and area (7.1) are N2.4, also Secondary Two. Pythagoras' Theorem, used once in Walkthrough 10, is G4 4.1, and the volume of a cone, used in Walkthrough 25, is G5 — and is printed on the formula sheet.

G2 Mathematics, Secondary Three/Four — the same topic G2. Congruence and similarity, but a shorter list.

LO	What a G2 student must be able to do	Covered in
2.4	enlargement and reduction of a plane figure	7.1 Enlargement
2.5	scale drawings	7.1 Scale drawings
2.6	properties and construction of perpendicular bisectors of line segments and angle bisectors	7.1 Bisectors
2.7	solving simple problems involving congruence and similarity	7.2 Congruence, 7.3 Similarity, 7.3 Finding a length

Read the two lists side by side and the differences are exact. G3's 2.6 and 2.7 are G2's 2.5 and 2.6, word for word — scale drawings and the two bisectors are examined at both levels, so nothing in 7.1 is hidden from anyone. G2's 2.4, enlargement and reduction, is a Secondary Two item for G3 (2.4 there too), which is why 7.1 Enlargement carries an ⓘ Recap note rather than a band stamp. And G2's 2.7, "solving simple problems involving congruence and similarity", is word for word G3's Secondary Two item 2.5 — so understanding what congruent and similar mean, and using a congruence or similarity you are given to find a missing side or angle, belongs to both courses.

What G2 never meets is the other three. There is no G2 outcome anywhere that asks a student to determine whether two triangles are congruent or similar, so the four congruence tests, the three similarity tests and the SSA warning are G3-only. Neither is there a G2 outcome for the ratio of areas of similar plane figures or the ratio of volumes of similar solids, so the whole of 7.4 and 7.5 is G3-only as well. Those three differences are the only ones, and they are the ones the level buttons act on.

One more thing worth knowing about the G2 list. Some outcomes in it are underlined, and the underline is not decoration: the G2 scheme of papers says that Paper 2 Section B — two questions of which a candidate answers only one, worth 7 or 8 marks — "will be based on the underlined content", with one question from the Geometry and Measurement strand and one from Statistics and Probability. In the Geometry strand that underlining falls on properties of circles, the sine and cosine rules, problems in two and three dimensions, and the radian mensuration items. None of G2's 2.4, 2.5, 2.6 or 2.7 is underlined — so everything a G2 student does have in this chapter is core Section A content that every candidate must answer, not an optional extra.

📐 Congruence & Similarity — Revision Notes

The chapter in one table

The perpendicular bisector of a line segment

The angle bisector

Which bisector does the question want?

Past-paper questions

Enlargement and reduction ⓘ Recap

Scale drawings, and the 1 : n form

Past-paper questions

Area on a plan ⓘ Recap

Check yourself

Past-paper questions

What congruent means, and why the order of the letters matters

The four congruence tests ✕ Not in G2

Past-paper questions

SSA is not a congruence test ✕ Not in G2

Check yourself

Similar figures, and the scale factor k

The three similarity tests ✕ Not in G2

The two configurations the paper keeps using

Finding an unknown length

Check yourself

Past-paper questions

Areas scale by k²

Check yourself

Past-paper questions

Volumes scale by k³

Past-paper questions

Cones, cups and the part that is filled

Past-paper questions

Cutting a cone: the frustum

Check yourself

Past-paper questions

What to carry into the exam

The one idea: k, then k², then k³ ✕ Not in G2

The four congruence tests ✕ Not in G2

The three similarity tests ✕ Not in G2

Bisectors and scale drawings

The habits that earn the marks

… and four more for the ratio questions ✕ Not in G2

Syllabus at a glance

Scale drawings, and the 1 : `n` form

Similar figures, and the scale factor `k`

Areas scale by `k`²

Volumes scale by `k`³

The one idea: `k`, then `k`², then `k`³ ✕ Not in G2