Congruence & similarity (incl. area/volume scale factors) – GCSE Maths exam questions with mark scheme

1Grade 4–5Non-calculator4 marks

Triangle ABC is similar to triangle PQR.
AB = 6 cm, BC = 8 cm, PQ = 9 cm and PR = 15 cm.
Angle ABC = angle PQR.

(a) Work out the length QR, marked $x$. [2]

(b) Work out the length AC. [2]

(a) scale factor $9 \div 6 = 1.5$ (or $\tfrac{3}{2}$), or $8 \times \tfrac{9}{6}$, or $\sqrt{15^2-9^2}$ using the right angle at Q shown on the diagramM1
(a) $12$ cmA1
(b) $15 \div 1.5$ (or $15 \times \tfrac{6}{9}$), or $\sqrt{6^2+8^2}$ using the right angle at B shown on the diagramM1
(b) $10$ cmA1
Answer: (a) 12 cm (b) 10 cm
Examiner tip: Pair up corresponding sides first (AB with PQ, BC with QR) — using 6 and 15 together gives a wrong scale factor.

2Grade 1–3FoundationNon-calculator2 marks

In triangle ABC, AB = 5 cm, AC = 7 cm and angle BAC $= 40°$.
In triangle PQR, PQ = 5 cm, PR = 7 cm and angle QPR $= 40°$.

Are the two triangles congruent?
Give a reason for your answer. [2]

YesB1
SAS — two sides and the included angle are equal (accept 'side, angle, side')B1
Answer: Yes — SAS (two sides and the angle between them are equal)
Examiner tip: The angle must be BETWEEN the two sides for SAS — check which vertex the angle is at before you write the condition.

3Grade 6–7HigherNon-calculator4 marks

ADB and AEC are straight lines. DE is parallel to BC.
AD = 4 cm, DB = 6 cm, DE = 5 cm and AC = 9 cm.

(a) Work out the length of BC. [2]

(b) Work out the length of EC. [2]

(a) scale factor $\dfrac{10}{4}$ (= 2.5) or $5 \times \dfrac{10}{4}$M1
(a) $12.5$ cmA1
(b) $AE = 9 \div 2.5 \;(= 3.6)$ or $9 \times \dfrac{4}{10}$, or $EC = 9 \times \dfrac{6}{10}$ directlyM1
(b) $5.4$ cmA1
Answer: (a) 12.5 cm (b) 5.4 cm
Examiner tip: The scale factor uses AB = 4 + 6 = 10, not DB = 6 — compare the whole side of the big triangle with the small one.

4Grade 6–7HigherCalculator3 marks

Shapes A and B are mathematically similar.
The area of shape A is 20 cm². The area of shape B is 45 cm².
The length of the base of shape A is 6 cm.

Work out the length of the base of shape B. [3]

area scale factor $45 \div 20 = 2.25$M1
length scale factor $\sqrt{2.25} = 1.5$M1
$9$ cmA1
Answer: 9 cm
Examiner tip: The ratio of areas is the length scale factor SQUARED — multiplying 6 by 2.25 is the most common wrong answer (13.5).

5Grade 8–9HigherCalculator4 marks

Two bottles, P and Q, are mathematically similar.
The surface area of P is 150 cm² and the surface area of Q is 384 cm².
The volume of Q is 1250 cm³.

Work out the volume of P. Give your answer correct to 3 significant figures. [4]

$384 \div 150 = 2.56$ (or $150 \div 384$)M1
length scale factor $\sqrt{2.56} = 1.6$ (or $0.625$)M1
volume scale factor $1.6^3 = 4.096$ and $1250 \div 4.096$ (or $1250 \times 0.625^3$)M1
$305$ cm³ (from $305.17...$)A1
Answer: 305 cm³
Examiner tip: Go area → length (square root) → volume (cube); never go straight from the area ratio to volume.

6Grade 8–9HigherCalculator4 marks

A solid cone has base radius 9 cm and vertical height 24 cm.
A small cone of vertical height 8 cm is cut off the top, parallel to the base, leaving a frustum.

Work out the volume of the frustum. Give your answer correct to 3 significant figures.
[Volume of a cone $= \tfrac{1}{3}\pi r^2 h$] [4]

volume of large cone $\tfrac{1}{3}\pi \times 9^2 \times 24 \;(= 648\pi = 2035.75...)$M1
radius of small cone $= 9 \times \tfrac{8}{24} = 3$, or volume scale factor $\left(\tfrac{1}{3}\right)^3$, giving small cone $= 24\pi \;(=75.39...)$M1
$648\pi - 24\pi$ (large cone − small cone), or $648\pi \times \tfrac{26}{27}$M1
$1960$ cm³ (accept answers in the range $1960$ to $1961$, e.g. $1960.35...$; $624\pi$ on its own gains M3 but not A1, since 3 s.f. was asked for)A1
Answer: 1960 cm³ (= $624\pi$)
Examiner tip: The small cone is similar to the big cone — its radius is 3 cm, not 9 − something. Use the height ratio 8 : 24.

7Grade 6–7HigherNon-calculator3 marks

ABCD is a parallelogram. BD is a diagonal.

Prove that triangle ABD is congruent to triangle CDB. [3]

one correct pair of equal sides or angles with a correct reason, e.g. AB = CD (opposite sides of a parallelogram are equal), or angle ABD = angle CDB (alternate angles, AB parallel to DC)B1
all three pairs needed for one condition, each with a correct reason, e.g. AB = CD and AD = CB (opposite sides of a parallelogram) and BD is common to both triangles; or angle ABD = angle CDB and angle ADB = angle CBD (alternate angles) with BD commonB1
fully correct proof with the matching condition stated: SSS, SAS (AB = CD, angle ABD = angle CDB, BD common) or ASA (two pairs of alternate angles and BD common). Do not accept AAA or SSA/ASSC1
Answer: AB = CD and AD = CB (opposite sides of a parallelogram); BD is common ⇒ congruent by SSS
Examiner tip: State the congruence condition (SSS, SAS, ASA, RHS) at the end AND a reason for each equal pair — 'it's a parallelogram' on its own is not enough.

8Grade 1–3FoundationNon-calculator3 marks

Six shapes, A to F, are drawn on a centimetre grid.

(a) Write down the letter of the shape that is congruent to shape A. [1]

(b) Write down the letter of the shape that is similar to shape A but not congruent to shape A. [1]

(a) BB1
(b) CB1
(c) 3 (accept ×3; do not accept 1 : 3 or $\tfrac{1}{3}$)B1
Answer: (a) B (b) C (c) 3
Examiner tip: Congruent shapes can be turned round or flipped over — B is A rotated, so check side lengths, not which way the shape faces.

9Grade 4–5Non-calculator4 marks

Here is information about three pairs of triangles.

Pair 1 Both triangles have a right angle. Both have a hypotenuse of 13 cm and one other side of 5 cm.

Pair 2 Both triangles have angles of 50°, 60° and 70°.

Pair 3 Both triangles have angles of 35° and 80°, and in both triangles the side between these two angles is 6 cm.

(a) For Pair 1, write down the condition of congruence that shows the two triangles are congruent. [1]

(b) For Pair 3, write down the condition of congruence that shows the two triangles are congruent. [1]

(c) Leon says, “The triangles in Pair 2 must be congruent because all their angles are the same.”
Is Leon correct? Give a reason for your answer. [2]

(a) RHS (accept right angle, hypotenuse, side)B1
(b) ASA (accept AAS; accept angle, side, angle)B1
(c) No (Leon is not correct) — may be implied by a correct reasonB1
(c) reason, e.g. the triangles could be different sizes / no side lengths are given / they are only similar (AAA is not a condition of congruence)C1
Answer: (a) RHS (b) ASA (c) No — equal angles show the triangles are similar, but one could be an enlargement of the other
Examiner tip: AAA proves similarity, not congruence — every congruence condition (SSS, SAS, ASA, RHS) needs at least one pair of equal sides.

10Grade 4–5Calculator4 marks

Kemi has a photograph that is 12.5 cm wide and 17.5 cm tall.

(a) She orders a print that is mathematically similar to the photograph.
The print is 30 cm wide.
Work out the height of the print. [2]

(b) Kemi has a frame that is 40 cm wide and 55 cm tall.
She says, “I can order a print similar to my photograph that fills this frame exactly.”
Is Kemi correct? You must show your working. [2]

(a) $30 \div 12.5$ (= 2.4) or $17.5 \div 12.5$ (= 1.4) or $17.5 \times \dfrac{30}{12.5}$M1
(a) 42 (cm)A1
(b) a correct comparison method, e.g. $40 \div 12.5$ (= 3.2) and $17.5 \times 3.2$ (= 56), or $55 \div 40$ (= 1.375) and $17.5 \div 12.5$ (= 1.4), or $55 \div 17.5$ (= 3.14…) and $40 \div 12.5$ (= 3.2)M1
(b) No, with correct supporting values, e.g. height would be 56 cm not 55 cm, or 1.375 ≠ 1.4, or 3.2 ≠ 3.14…C1
Answer: (a) 42 cm (b) No — a 40 cm wide print would be 56 cm tall (ratios 1.4 and 1.375 differ)
Examiner tip: Similar means every length is multiplied by the same scale factor — adding the same amount (40 − 12.5 = 27.5) to each side is the classic wrong method.

11Grade 6–7HigherNon-calculator5 marks

AE and BD are straight lines that cross at C.
AB is parallel to DE.
AB = 8 cm, DE = 12 cm, AC = $x$ cm and CE = $(x + 3)$ cm.

(a) Give reasons why triangle ABC is similar to triangle EDC. [2]

(b) Work out the value of $x$. [3]

(a) angle ACB = angle ECD because vertically opposite angles are equalB1
(a) angle BAC = angle DEC (or angle ABC = angle EDC) because alternate angles are equal (AB parallel to DE); so all three angles match. Do not accept 'Z angles'B1
(b) a correct ratio statement, e.g. $\dfrac{x+3}{x} = \dfrac{12}{8}$ or scale factor 1.5 used with $x$ and $x + 3$, e.g. $1.5x = x + 3$M1
(b) correct linear equation without fractions, e.g. $12x = 8(x + 3)$ or $12x = 8x + 24$ or $0.5x = 3$M1
(b) $x = 6$A1
Answer: (a) vertically opposite angles at C equal; alternate angles (AB ∥ DE) equal (b) $x = 6$
Examiner tip: Match corresponding sides: CE in the larger triangle matches AC in the smaller one, so $\dfrac{x+3}{x} = \dfrac{12}{8}$, not $\dfrac{x}{x+3}$.

12Grade 6–7HigherNon-calculator5 marks

Two solid statues are mathematically similar and are made from the same metal.
The small statue is 15 cm tall and has a mass of 540 g.
The large statue is 25 cm tall.

(a) Work out the mass of the large statue. [3]

(b) The base of the large statue has an area of 75 cm².
Work out the area of the base of the small statue. [2]

(a) length scale factor $\dfrac{25}{15}$ or $\dfrac{5}{3}$ (or $\dfrac{3}{5}$), or ratio 3 : 5M1
(a) volume (mass) scale factor $\left(\dfrac{5}{3}\right)^3 = \dfrac{125}{27}$, or e.g. $540 \div 27 \times 125$M1
(a) 2500 (g) (accept 2.5 kg)A1
(b) $75 \times \left(\dfrac{3}{5}\right)^2$ or $75 \div \dfrac{25}{9}$ or $75 \times \dfrac{9}{25}$M1
(b) 27 (cm²)A1
Answer: (a) 2500 g (b) 27 cm²
Examiner tip: Mass depends on volume, so cube the length scale factor; area squares it — using $\tfrac{5}{3}$ on its own gives the very common wrong answer 900 g.

13Grade 8–9HigherCalculator5 marks

A drinking glass is in the shape of an inverted cone.
When it is full, the water is 12 cm deep and the glass holds 360 ml.

The water in a partly filled glass always forms a cone that is mathematically similar to the full cone.

(a) Water is poured into the empty glass until it is 9 cm deep.
Work out the volume of water in the glass. [2]

(b) Sam says, “If the glass is half full, the water will be 6 cm deep.”
Show that Sam is wrong and work out the correct depth when the glass is half full.
Give your answer correct to 3 significant figures. [3]

(a) $360 \times \left(\dfrac{9}{12}\right)^3$ or $360 \times 0.75^3$ or $360 \times \dfrac{27}{64}$M1
(a) 152 ml (accept 151.875 or 151.9 or 151 to 152)A1
(b) shows Sam is wrong, e.g. at 6 cm deep volume $= 360 \times \left(\dfrac{6}{12}\right)^3 = 45$ ml (not 180 ml), or states volume scale factor is cube of length scale factor so halving depth gives $\tfrac{1}{8}$ of the volumeC1
(b) $\left(\dfrac{d}{12}\right)^3 = \dfrac{1}{2}$ or length scale factor $\sqrt[3]{0.5}$ (= 0.7937…) or $d = 12 \times \sqrt[3]{0.5}$ or $d^3 = 864$M1
(b) 9.52 cm (accept 9.524…)A1
Answer: (a) 152 ml (151.875 ml) (b) at 6 cm the glass holds only 45 ml; half full (180 ml) is at 9.52 cm
Examiner tip: Half the volume is NOT half the depth — the volume scale factor is the cube of the length scale factor, so you need a cube root.

14Grade 6–7HigherNon-calculator4 marks

Triangle P and triangle Q are drawn on the grid.

(a) Describe fully the single transformation that maps triangle P onto triangle Q. [3]

(b) Triangle P has an area of 8 square units.
Without counting squares, explain how you can find the area of triangle Q, and write down its area. [1]

(a) enlargementB1
(a) scale factor $-\dfrac{1}{2}$ (accept −0.5)B1
(a) centre $(-1, 0)$B1
(b) 2 (square units) with a reason, e.g. area scale factor is $\left(-\dfrac{1}{2}\right)^2 = \dfrac{1}{4}$, so $8 \times \dfrac{1}{4} = 2$B1
Answer: (a) Enlargement, scale factor $-\tfrac{1}{2}$, centre $(-1, 0)$ (b) 2 square units ($8 \times \tfrac{1}{4}$)
Examiner tip: A shape that is smaller AND upside down on the other side of the centre has a negative fractional scale factor — and the area scale factor is the square, which is positive.

15Grade 8–9HigherNon-calculator5 marks

ABCD is a square.
E is a point on BC and F is a point on CD such that BE = CF.
AE and BF meet at G.

(a) Prove that triangle ABE is congruent to triangle BCF. [3]

(b) Hence prove that AE is perpendicular to BF. [2]

(a) AB = BC (sides of a square are equal) and BE = CF (given)B1
(a) angle ABE = angle BCF = 90° (angles of a square)B1
(a) congruent by SAS, with all three pairs correctly stated and reasons given. Do not accept SSA or RHS unless the hypotenuse pair AE = BF has been justifiedC1
(b) angle BAE = angle CBF (corresponding angles of congruent triangles), i.e. angle BAG = angle CBFM1
(b) complete argument, e.g. angle ABG + angle CBF = angle ABC = 90°, so angle BAG + angle ABG = 90°, so angle AGB = 180° − 90° = 90° (angles in a triangle), hence AE ⊥ BFC1
Answer: (a) AB = BC, angle ABE = angle BCF = 90°, BE = CF ⇒ congruent (SAS) (b) angle BAG = angle CBF, so angle BAG + angle ABG = 90° and angle AGB = 90°
Examiner tip: In a congruence proof every pair needs a reason AND the matching condition at the end — and the 90° must be the angle BETWEEN the two pairs of sides for SAS.

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