Pythagoras' theorem (2D & 3D) – GCSE Maths exam questions with mark scheme

1Grade 4–5Calculator3 marks

The diagram shows a right-angled triangle.

Work out the length $x$. Give your answer correct to 1 decimal place. [3]

$7^2+11^2\;(=170)$M1
$\sqrt{170}$M1
13.0 cm (accept 13.03 to 13.04 seen)A1
Answer: 13.0 cm
Examiner tip: Square, add, then square root – check your answer is longer than both shorter sides.

2Grade 4–5Calculator3 marks

A right-angled triangle has hypotenuse 15.2 cm. One of the other sides is 9.6 cm.

Work out the length of the third side, $y$. Give your answer correct to 3 significant figures. [3]

$15.2^2-9.6^2\;(=138.88)$M1
$\sqrt{138.88}$M1
11.8 cm (accept 11.78…)A1
Answer: 11.8 cm
Examiner tip: When finding a shorter side, subtract the squares – adding them gives an answer longer than the hypotenuse, which is impossible.

3Grade 4–5Calculator3 marks

A ladder 6.5 m long leans against a vertical wall on horizontal ground. The foot of the ladder is 1.8 m from the bottom of the wall.

A window sill is 6.1 m above the ground. Does the top of the ladder reach the window sill? You must show your working. [3]

$6.5^2-1.8^2\;(=39.01)$
Alt: $6.1^2+1.8^2\;(=40.45)$ and $6.5^2\;(=42.25)$M1
height $=\sqrt{39.01}=6.24(58\ldots)$ m
Alt: $40.45$ and $42.25$ both correct (or ladder needed to reach sill $=\sqrt{40.45}=6.36\ldots$ m)A1
Yes, with correct comparison, e.g. $6.24\ldots > 6.1$ or $6.5 > 6.36\ldots$ or $42.25 > 40.45$ (dep on M1)C1
Answer: Yes – the top of the ladder is $\sqrt{39.01}=6.25$ m (3 s.f.) above the ground, which is higher than 6.1 m.
Examiner tip: Finish with a clear 'Yes' and the two values compared (e.g. 6.25 m > 6.1 m) – a value with no conclusion, or 'Yes' with no comparison, loses the final mark.

4Grade 4–5Non-calculator3 marks

Point A has coordinates $(2,-3)$. Point B has coordinates $(8,5)$.

Work out the length of the line segment AB. [3]

horizontal and vertical differences 6 and 8M1
$\sqrt{6^2+8^2}$M1
10 (units)A1
Answer: 10
Examiner tip: Subtract carefully with negatives: from $-3$ to $5$ is a change of 8, not 2.

5Grade 4–5Non-calculator3 marks

An isosceles triangle has sides 13 cm, 13 cm and 10 cm.

Work out the area of the triangle. [3]

splits into right-angled triangles with base 5: $13^2-5^2\;(=144)$M1
height $=\sqrt{144}=12$ and $\tfrac12\times10\times12$M1
60 cm$^2$A1
Answer: 60 cm$^2$
Examiner tip: The perpendicular height bisects the base – use 5 cm, not 10 cm, in Pythagoras.

6Grade 6–7HigherCalculator4 marks

The diagram shows a box in the shape of a cuboid ABCDEFGH. AB = 8 cm, BC = 5 cm and AE = 4 cm.

Gemma says a pencil of length 10.5 cm will fit inside the box, lying along the diagonal AG.

Is Gemma correct? You must show your working. [4]

$AC^2=8^2+5^2\;(=89)$ or $AC=\sqrt{89}=9.43\ldots$
Allow any face diagonal first, e.g. $AF^2=8^2+4^2\;(=80)$ or $AH^2=5^2+4^2\;(=41)$M1
$AG^2=89+4^2$ (or $80+5^2$ or $41+8^2$) or $\sqrt{8^2+5^2+4^2}$M1
AG = 10.2 cm or better (accept 10.24…, 10.25 or $\sqrt{105}$)
Alt: $AG^2 = 105$ and $10.5^2 = 110.25$A1
No, with correct comparison, e.g. 10.24… < 10.5 or 105 < 110.25 (dep on M2)C1
Answer: No – AG = 10.2 cm (to 3 s.f.), which is less than 10.5 cm.
Examiner tip: In 3D, use Pythagoras twice (or $\sqrt{a^2+b^2+c^2}$) – the base diagonal AC is not the longest length.

7Grade 6–7HigherCalculator4 marks

A pyramid has a square base of side 6 cm. The apex V is directly above the centre M of the base. Each sloping edge from V to a corner of the base is 9 cm.

Work out the perpendicular height VM of the pyramid. Give your answer correct to 3 significant figures. [4]

diagonal of base $\sqrt{6^2+6^2}=\sqrt{72}\;(=8.485\ldots)$
Alt: slant height of a face $\sqrt{9^2-3^2}=\sqrt{72}$M1
half the diagonal $\sqrt{18}\;(=4.242\ldots)$ (or $\sqrt{3^2+3^2}$ directly, which earns the first two M marks)
Alt: uses the triangle with sides VM, 3 and $\sqrt{72}$M1
$\sqrt{9^2-18}$ or $\sqrt{63}$
Alt: $\sqrt{72-3^2}$M1
7.94 cm (accept 7.937…)A1
Answer: 7.94 cm
Examiner tip: The right-angled triangle uses half the base diagonal, not half the base side.

8Grade 4–5Non-calculator3 marks

A builder wants to check that the corner of a room is a right angle.

She marks a point 1.2 m from the corner along one wall and a point 1.6 m from the corner along the other wall. She measures the straight-line distance between the two marks as 2.1 m.

Is the corner a right angle? You must show how you get your answer. [3]

$1.2^2 + 1.6^2\;(=1.44+2.56)$ (or an alternative method, e.g. $2.1^2 - 1.6^2$ or $2.1^2 - 1.2^2$)M1
$1.2^2+1.6^2 = 4$ (or $\sqrt{4}=2$); for the alternative, $1.85$ or $\sqrt{1.85} = 1.36\ldots$ (or $2.97$ or $1.72\ldots$)A1
No, with a correct comparison, e.g. $2 \ne 2.1$, or $4 \ne 4.41\;(=2.1^2)$, or $1.36 \ne 1.2$, or $1.72 \ne 1.6$ (dependent on M1)C1
Answer: No: for a right angle the marks would be $\sqrt{1.2^2+1.6^2}=2$ m apart, but they are 2.1 m apart ($4.41\ne4$).
Examiner tip: A bare 'no' scores nothing: write the two values you are comparing (2 and 2.1, or 4 and 4.41) next to your conclusion.

9Grade 1–3FoundationNon-calculator3 marks

The diagram shows a right-angled triangle with a square drawn on each of its sides.

Square A has area 81 cm². Square B has area 144 cm².

(a) Write down the area of square C. [1]

(b) Work out the perimeter of the triangle. [2]

(a) 225 (cm²)B1
(b) finds at least two of the side lengths 9, 12, 15 (e.g. $\sqrt{81}=9$ and $\sqrt{144}=12$)M1
(b) 36 cm (ft $9+12+\sqrt{\text{their (a)}}$ if their (a) is a square number)A1
Answer: (a) 225 cm² (b) 36 cm
Examiner tip: The areas are not the side lengths: square root each area before you add up the perimeter.

10Grade 4–5Calculator4 marks

The diagram shows a rectangular field ABCD. AB = 85 m and BC = 42 m.

Amir walks from A to C along the edges of the field, going via B.
Bella walks from A to C in a straight line across the field.

How much further does Amir walk than Bella? Give your answer correct to 1 decimal place. [4]

$85^2 + 42^2\;(=8989)$M1
$\sqrt{8989}\;(=94.81\ldots)$M1
$(85+42) - \text{their } 94.8\;$ i.e. $127 - 94.8$M1
32.2 m (accept 32.18 to 32.2)A1
Answer: 32.2 m
Examiner tip: Students often stop at the diagonal (94.8 m) – reread the question: it asks for the difference between the two routes.

11Grade 4–5Non-calculator4 marks

ABCD is a trapezium. Angle DAB = angle ADC = 90°.

AB = 18 cm, DC = 11 cm and AD = 24 cm.

Work out the perimeter of the trapezium. [4]

$18 - 11 = 7$ (horizontal distance for the triangle under BC)M1
$7^2 + 24^2\;(=49+576=625)$M1
$\sqrt{625} = 25$ (BC)M1
78 cmA1
Answer: 78 cm
Examiner tip: The slanted side is the hypotenuse of a triangle with base 18 − 11 = 7, not 18 or 11 – sketch that triangle first.

12Grade 6–7HigherNon-calculator6 marks

The diagram shows a right-angled triangle. All lengths are in centimetres.

(a) Show that $x^2 - 4x - 32 = 0$ [3]

(b) Work out the perimeter of the triangle. [3]

(a) $x^2 + (x+7)^2 = (x+9)^2$M1
(a) correct expansion of both brackets: $x^2 + 14x + 49$ and $x^2 + 18x + 81$M1
(a) $2x^2 + 14x + 49 = x^2 + 18x + 81$ simplified to $x^2 - 4x - 32 = 0$ with no errorsA1
(b) $(x-8)(x+4)\;(=0)$ or correct substitution into the quadratic formulaM1
(b) $x = 8$ (ignore $x=-4$ unless chosen as the answer)A1
(b) 40 cm (ft $3x+16$ for their positive $x$)A1
Answer: (a) shown (b) 40 cm (sides 8, 15, 17)
Examiner tip: $(x+7)^2$ is not $x^2+49$: write the bracket out twice and expand to get the $14x$ term.

13Grade 6–7HigherCalculator4 marks

A closed tin is in the shape of a cylinder. The radius of the tin is 3.5 cm and its height is 12.4 cm.

Theo has a thin straight rod of length 14.5 cm. He says the rod will fit inside the closed tin.

Is Theo correct? You must show your working. [4]

uses the diameter 7 in Pythagoras: $7^2 + 12.4^2\;(=202.76)$M1
$\sqrt{202.76}$M1
14.2 cm (accept 14.23 to 14.24)A1
No, with correct comparison, e.g. $14.2 < 14.5$, or $202.76 < 210.25\;(=14.5^2)$ (dependent on M2 or on $7^2+12.4^2$ compared with $14.5^2$)C1
Answer: No – the longest rod that fits is $\sqrt{7^2+12.4^2}=14.2$ cm (3 s.f.), which is less than 14.5 cm.
Examiner tip: The longest rod goes from one edge of the base to the opposite edge of the top, so use the diameter (7 cm), not the radius.

14Grade 8–9HigherCalculator5 marks

The diagram shows a ramp in the shape of a triangular prism.

ABCD is a horizontal rectangle with AB = 5.2 m and BC = 2.4 m.
BCEF is a vertical rectangle with BF = 1.8 m.

(a) Work out the length of AE. Give your answer correct to 3 significant figures. [3]

(b) Work out the size of the angle between AE and the plane ABCD. Give your answer correct to 1 decimal place. [2]

(a) $5.2^2 + 2.4^2\;(=32.8)$ for $AC^2$, or $5.2^2+1.8^2\;(=30.28)$ for $AF^2$M1
(a) $\sqrt{5.2^2 + 2.4^2 + 1.8^2}$ or $\sqrt{32.8 + 1.8^2}$ or $\sqrt{30.28 + 2.4^2}$ $(=\sqrt{36.04})$M1
(a) 6.00 m (accept 6.003)A1
(b) $\sin\theta = \dfrac{1.8}{\text{their } AE}$ or $\tan\theta = \dfrac{1.8}{\sqrt{32.8}}$ or $\cos\theta = \dfrac{\sqrt{32.8}}{\text{their } AE}$M1
(b) 17.4° (accept 17.4 to 17.5)A1
Answer: (a) 6.00 m (b) 17.4°
Examiner tip: The angle with the base is at A in triangle ACE – the side on the base is AC (the diagonal), not AB.

15Grade 8–9HigherNon-calculator5 marks

ABCDEFGH is a cube with sides of length 6 cm.
M is the midpoint of the edge GH.

(a) Show that AM = 9 cm. [3]

(b) Work out the length of CM. Give your answer in the form $a\sqrt{b}$, where $a$ and $b$ are integers. [2]

(a) a correct 2D step, e.g. $AH^2 = 6^2+6^2 = 72$, or $AN^2 = 6^2 + 3^2 = 45$ where N is the midpoint of DCM1
(a) $AM^2 = 72 + 3^2$ or $45 + 6^2$ or $3^2+6^2+6^2$M1
(a) $\sqrt{81} = 9$ with all working shownA1
(b) $6^2 + 3^2\;(=45)$ (triangle CGM, right angle at G)M1
(b) $3\sqrt{5}$ (cm)A1
Answer: (a) $AM=\sqrt{6^2+6^2+3^2}=\sqrt{81}=9$ cm (b) $3\sqrt{5}$ cm
Examiner tip: In a 'show that' you must reach $\sqrt{81}$ before writing 9 – just writing 9 cm earns nothing.

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