In a right-angled triangle with shorter sides a and b and hypotenuse c:
a² + b² = c²
You can rearrange this to find any missing side.
Question 1
Find the length of the hypotenuse.
cm
Hint
c² = a² + b². Substitute a = 6 and b = 8.
Full Solution
c² = 6² + 8² = 36 + 64 = 100
c = √100 = 10 cm
Question 2
Find the missing side x.
cm
Hint
Rearrange: x² = c² − b². The hypotenuse is always the longest side.
Full Solution
x² = 17² − 15² = 289 − 225 = 64
x = √64 = 8 cm
Question 3
An isosceles triangle has two equal sides of 13 cm and a base of 10 cm. Find its height.
cm
Hint
The height splits the base in half, making two right-angled triangles with base 5 and hypotenuse 13.
Full Solution
The height splits the triangle into two right-angled triangles.
h² + 5² = 13²
h² = 169 − 25 = 144
h = √144 = 12 cm
Question 4
A 10 m ladder leans against a wall. The foot of the ladder is 6 m from the base of the wall. How high up the wall does the ladder reach?
m
Hint
The ladder is the hypotenuse. The ground distance and wall height are the two shorter sides.
Full Solution
h² + 6² = 10²
h² = 100 − 36 = 64
h = √64 = 8 m
Question 5
Find the length of the diagonal of this rectangle.
cm
Hint
The diagonal of a rectangle splits it into two right-angled triangles.
Full Solution
d² = 7² + 24² = 49 + 576 = 625
d = √625 = 25 cm
Question 6
Find the distance between points A(1, 2) and B(7, 10).
units
Hint
Find the horizontal distance (7 − 1) and vertical distance (10 − 2), then use Pythagoras.
Full Solution
Horizontal distance = 7 − 1 = 6
Vertical distance = 10 − 2 = 8
d² = 6² + 8² = 36 + 64 = 100
d = √100 = 10 units
3D Pythagoras
Use Pythagoras' theorem in 3D shapes to find unknown lengths. Remember: in a cuboid with sides a, b, and c, the space diagonal d satisfies:
d² = a² + b² + c²
You can also break it into two steps: first find the face diagonal, then use it with the remaining side.
Question 1
Find the length of the space diagonal AG in this cuboid.
cm
Hint
Step 1: Find diagonal AB on the base using a² + b².
Step 2: Use AB and the height to find AG.
Full Solution
Step 1: Find the face diagonal on the base (AB to G projected down).
AB² = 12² + 6² = 144 + 36 = 180
AB = √180
Step 2: Use this with the height.
AG² = AB² + 8² = 180 + 64 = 244
AG = √244 = 15.6 cm (1 d.p.)
Question 2
A room is 5 m long, 4 m wide and 3 m high. A spider sits in one corner of the ceiling. What is the straight-line distance to the opposite corner on the floor?
m
Hint
Use d² = length² + width² + height² directly.
Full Solution
d² = 5² + 4² + 3²
d² = 25 + 16 + 9 = 50
d = √50 = 7.1 m (1 d.p.)
Question 3
A triangular prism has the cross-section shown. Find the length PQ, the distance from one end of the top edge to the opposite bottom corner.
The triangle cross-section has a base of 10 cm and height of 8 cm. The prism is 15 cm long.
cm
Hint
P is at the top of the back triangle. Q is at the bottom-left of the front triangle. Think about what horizontal and vertical distances separate them.
Full Solution
P is at the apex of the back face. Q is at the bottom-left of the front face.
Horizontal distance along the base: half the base = 5 cm (apex is centred above the base for the height measurement)
Depth of prism: 15 cm
Vertical drop: 8 cm
PQ² = 5² + 15² + 8²
PQ² = 25 + 225 + 64 = 314
PQ = √314 = 17.7 cm (1 d.p.)
Question 4
A cube has side length 9 cm. Find the length of the space diagonal.
cm
Hint
For a cube with side s, the space diagonal = s√3.
Full Solution
d² = 9² + 9² + 9² = 81 + 81 + 81 = 243
d = √243 = 9√3 = 15.6 cm (1 d.p.)
Question 5 (Challenge)
A cone has a base radius of 5 cm and a slant height of 13 cm. A straight tunnel is drilled from a point on the base circumference, through the inside of the cone, to the tip of the cone on the opposite side. What is the length of this tunnel?
The tunnel goes from a point on the base edge to the apex. This is the slant height!
cm
Hint
Think carefully: what IS the distance from a point on the base circumference to the apex?
Full Solution
The distance from any point on the base circumference to the apex is the slant height.
The tunnel length = 13 cm
(This question tests whether you recognise when 3D Pythagoras is NOT needed!)
Question 6
A rectangular box is 20 cm long, 15 cm wide and 12 cm tall. Will a straight rod of length 28 cm fit inside the box? Show your working.
Hint
The longest thing that fits inside a cuboid is the space diagonal.
Full Solution
Space diagonal² = 20² + 15² + 12²
= 400 + 225 + 144 = 769
Space diagonal = √769 = 27.7 cm (1 d.p.)
27.7 < 28, so the space diagonal (27.7 cm) is shorter than the rod (28 cm).