Question 1: [2 marks]
ABC is a right-angled triangle with
AB = 6 cm
AC = 8 cm
Calculate the length of the missing side BC.
Answer type: Simple text answer
ANSWER: 10 cm
WORKING:
6^2+8^2=36+64=100
c = \sqrt{100} = 10 cm
Question 2: [3 marks]
EFG is a right-angled triangle with
FG = 6 cm
EG = 15 cm
Calculate the length of the missing side EG.
Give your answer correct to 1 decimal place.
Answer type: Simple text answer
ANSWER: 16.2 cm
WORKING:
15^2+6^2=225+36=261
\sqrt{261}=16.2 cm (1 dp)
Question 3: [3 marks]
ABC is a right-angled triangle with
AB = 7.8 cm
AC = 9.2 cm
Calculate the length of the missing side BC.
Give your answer correct to 1 decimal place.
Answer type: Simple text answer
ANSWER: 4.9 cm
WORKING:
9.2^2-7.8^2=84.64-60.84=23.8
\sqrt{23.8} = 4.9 cm (1 dp)
Question 4:
PQR is a right-angled triangle with
PQ = 5.9 cm
PR = 6.7 cm
Question 4(a): [3 marks]
Calculate the length of the missing side QR.
Give your answer correct to 1 decimal place.
Answer type: Simple text answer
ANSWER: 8.9 cm
WORKING:
5.9^2+6.7^2=34.81+44.89=79.70
\sqrt{79.70} = 8.9 cm (1 dp)
Question 4(b): [1 mark]
Calculate the perimeter of the triangle PQR.
Give your answer correct to 1 decimal place.
Answer type: Simple text answer
ANSWER: 21.5 cm
WORKING:
5.9+6.7+8.9=21.5 cm
Question 5: [4 marks]
XYZ is a right-angled triangle with
XY = 3.7 cm
YZ = 4.9 cm
Calculate the area of the triangle.
Give your answer correct to 1 decimal place.
Answer type: Simple text answer
ANSWER: 5.9 cm^2
WORKING:
4.9^2 - 3.7^2 = 10.32
\sqrt{10.32} = 3.2 cm
\text{Area} = \dfrac{1}{2} \times 3.7 \times 3.2 = 5.9 cm^2 (1 dp)
Question 6: [3 marks]
A builder places a 3.6 m ladder against a vertical wall as shown.
To be safe to use, the base of this ladder must be 1.5 m away from the wall.
The ladder is also placed on flat horizontal ground.
How far up the wall does the ladder reach?
Give your answer correct to 1 decimal place.
Answer type: Simple text answer
ANSWER: 3.3 cm
WORKING:
3.6^2 - 1.5^2 = 12.96 - 2.25 = 10.71
\sqrt{10.71} = 3.3 cm (1 dp)
Question 7:
ABC is an isosceles triangle.
M is the midpoint of the vertices A and C.
The distance from A to B is 9 cm.
The distance from A to C is 14 cm.
Question 7(a): [3 marks]
Work out the length of B to M.
Give your answer to 1 decimal place.
Answer type: Simple text answer
ANSWER: 5.7 cm
WORKING:
9^2 - 7^2 = 81 - 49 = 32
\sqrt{32} = 5.7 cm
Question 7(b): [2 marks]
Find the area of the triangle.
Give your answer to 2 significant figures.
Answer type: Simple text answer
ANSWER: 40
WORKING:
\dfrac{1}{2} \times 14 \times 5.7 = 40 cm^2 (2 sf)
Question 8: [4 marks]
Two points have the coordinates (2, 3) and (7, 9).
Calculate the distance between the two points.
Give your answer to 1 decimal place.
Answer type: Simple text answer
ANSWER: 7.8 cm
WORKING:
\text{Change in } x = 7 - 2 = 5
\text{Change in } y = 6 - 3 = 3
5^2+6^2 = 25+36 = 61
\sqrt{61} = 7.8 cm (1 dp)
Question 9: [2 marks]
The diagram below shows a triangle.
MC is perpendicular to AB.
AB=c
AM=x
MB=c-x
By applying Pythagoras’ Theorem to triangles AMC and CMB,
choose the correct expression from below.
Answer type: Multiple choice type 2
A: c^2 - 2cx = a^2 - b^2
B: c^2 - 2cx = b^2 - a^2
C: c^2 - 2hx = a^2 - b^2
D: c^2 - x^2 = a^2 - b^2
ANSWER: A: c^2 - 2cx = a^2 - b^2
WORKING:
AMC: \, x^2 + h^2 = b^2
CMB: \, (c-x)^2 + h^2 = a^2
Subtract second equation from first, to eliminate h from the expression.
(c-x)^2 - x^2 = a^2 - b^2
c^2 - 2cx = a^2 - b^2