Pythagoras - Worksheets, Questions and Revision

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GCSE · Geometry and Measures

4.9 Pythagoras

EDEXCEL 1MA1 · Calculator allowed · about 65 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Revision notes: Pythagoras

Pythagoras' theorem relates the three sides of a RIGHT-ANGLED triangle: the square of the hypotenuse (the longest side, opposite the right angle) equals the sum of the squares of the other two sides, written a² + b² = c², where c is the hypotenuse.

Identify the hypotenuse: it is always the longest side, directly opposite the right angle.
To find the hypotenuse (when the two shorter sides are known), square both shorter sides, add the results together, then square root the total: c = √a² + b².
To find a shorter side (when the hypotenuse and one other side are known), square both known sides, SUBTRACT the smaller square from the larger, then square root the result: a = √c² - b².
Always check the answer is sensible: the hypotenuse must be longer than either of the other two sides.

Worked example. A right-angled triangle has shorter sides of length 5 cm and 12 cm. Work out the length of the hypotenuse.

Square both shorter sides: 5² = 25, and 12² = 144.
Add the squares: 25 + 144 = 169.
Square root the total to find the hypotenuse: √169 = 13.

Answer: 13 cm.

The diagram shows a right-angled triangle with the right angle marked. The sides are labelled p, q and r, where r is the side opposite the right angle.
Diagram: right-angled triangle, right angle shown at the bottom-left corner, side p along the base, side q up the vertical side, side r as the slanted side joining the two ends.

A) p
B) q
C) r

(Total for Question 1 is 1 mark)

A right-angled triangle has two shorter sides of length 9 cm and 12 cm.
Diagram: right-angled triangle with the right angle between the two shorter sides, marked 9 cm and 12 cm, hypotenuse unmarked labelled x cm.
Calculate the length of the hypotenuse.

Diagram NOT accurately drawn

(Total for Question 2 is 2 marks)

A right-angled triangle has a hypotenuse of length 17 cm and one shorter side of length 8 cm.
Diagram: right-angled triangle, hypotenuse marked 17 cm, one shorter side marked 8 cm, the other shorter side unmarked labelled y cm.
Work out the length of the third side.

Diagram NOT accurately drawn

(Total for Question 3 is 2 marks)

A right-angled triangle has two shorter sides of length 4.8 cm and 6.3 cm.
Diagram: right-angled triangle, shorter sides marked 4.8 cm and 6.3 cm meeting at the right angle, hypotenuse labelled h cm.
Calculate the length of the hypotenuse. Give your answer correct to 3 significant figures.

Diagram NOT accurately drawn

(Total for Question 4 is 3 marks)

A right-angled triangle has a hypotenuse of length 15.5 cm and one shorter side of length 9.3 cm.
Diagram: right-angled triangle, hypotenuse marked 15.5 cm, one shorter side marked 9.3 cm, the other shorter side unmarked labelled k cm.
Work out the length of the third side.

Diagram NOT accurately drawn

(Total for Question 5 is 3 marks)

A triangle has sides of length 7 cm, 24 cm and 25 cm.
Show that this triangle is a right-angled triangle.

(Total for Question 6 is 3 marks)

A triangle has sides of length 9 cm, 12 cm and 16 cm.
Determine, showing your working, whether this triangle is a right-angled triangle.

(Total for Question 7 is 3 marks)

A ladder of length 4.5 m leans against a vertical wall. The foot of the ladder is 1.5 m from the base of the wall.
Diagram: vertical wall on the right, ground shown horizontal, ladder drawn as the hypotenuse from a point 1.5 m from the wall on the ground up to a point on the wall, ladder length marked 4.5 m, height up the wall marked h m.
Calculate how far up the wall the ladder reaches. Give your answer correct to 3 significant figures.

Diagram NOT accurately drawn

(Total for Question 8 is 3 marks)

A rectangular television screen has width 80 cm and height 45 cm.
Diagram: rectangle representing a television screen, width marked 80 cm along the base, height marked 45 cm up the side, a diagonal line drawn from bottom-left to top-right labelled d cm.
Calculate the length of the diagonal of the screen, d. Give your answer correct to 3 significant figures.

Diagram NOT accurately drawn

(Total for Question 9 is 3 marks)

Point A has coordinates (2, 3) and point B has coordinates (7, 11).
Work out the distance AB. Give your answer correct to 3 significant figures.

(Total for Question 10 is 3 marks)

An isosceles triangle has two equal sides of length 10 cm and a base of length 12 cm. A line is drawn from the apex of the triangle perpendicular to the base, meeting the base at its midpoint.
Diagram: isosceles triangle with the two equal sides marked 10 cm, base marked 12 cm, and a dashed perpendicular line from the apex to the midpoint of the base labelled h cm.

Diagram NOT accurately drawn

(a)Explain why the perpendicular line splits the base into two lengths of 6 cm each.(1)

(b)Calculate the perpendicular height, h, of the triangle.(3)

(Total for Question 11 is 4 marks)

A rectangular field measures 120 m by 90 m. Instead of walking along two sides of the field to get from corner P to the opposite corner Q, Priya walks directly across the field along the diagonal.
Diagram: rectangle representing the field, length marked 120 m, width marked 90 m, corner P at bottom-left and corner Q at top-right, a diagonal line drawn from P to Q.
Work out how much shorter Priya's direct route is compared with walking along two sides of the field.

Diagram NOT accurately drawn

(Total for Question 12 is 4 marks)

A rhombus has diagonals of length 16 cm and 30 cm. The diagonals of a rhombus bisect each other at right angles.
Diagram: rhombus with its two diagonals drawn, one diagonal marked 16 cm and the other marked 30 cm, crossing at right angles at the centre.

Diagram NOT accurately drawn

(a)State the lengths of the two half-diagonals formed where the diagonals cross.(1)

(b)Calculate the length of one side of the rhombus.(3)

(Total for Question 13 is 4 marks)

A right-angled isosceles triangle has two shorter sides, each of length 3 cm.
Diagram: right-angled triangle with the two shorter sides equal, each marked 3 cm, meeting at the right angle, hypotenuse unmarked labelled x cm.
Calculate the length of the hypotenuse, x. Give your answer as a surd in its simplest form.

Diagram NOT accurately drawn

(Total for Question 14 is 3 marks)

A cuboid has length 4 cm, width 4 cm and height 7 cm. The diagonal of the base of the cuboid is drawn, and then the space diagonal from one bottom corner to the opposite top corner is drawn.
Diagram: cuboid with length 4 cm, width 4 cm and height 7 cm, a diagonal drawn across the base labelled b cm, and a space diagonal drawn from a bottom corner to the diagonally opposite top corner labelled D cm.

Diagram NOT accurately drawn

(a)Calculate b², the square of the length of the base diagonal.(2)

(b)Calculate the length of the space diagonal, D.(3)

(Total for Question 15 is 5 marks)

A ship sails 24 km due north from port, then sails 18 km due east to reach a buoy.
Diagram: port marked at the start, a vertical arrow of length 24 km pointing north to a turning point, then a horizontal arrow of length 18 km pointing east to the buoy, and a dashed straight line drawn directly from the port to the buoy.

Diagram NOT accurately drawn

(a)Calculate the direct distance from the port to the buoy.(3)

(b)The ship then sails directly back to port along the straight line shown, instead of retracing its outward route. Calculate how much shorter this direct return journey is compared with retracing the original route (24 km then 18 km).(2)

(Total for Question 16 is 5 marks)

Mark scheme · 4.9 Pythagoras

Question 1

B1 r stated as the hypotenuse cao
Answer: C) r

Question 2

M1 √9² + 12² oe
A1 15 cm cao
Answer: 15 cm

Question 3

M1 √17² - 8² oe
A1 15 cm cao
Answer: 15 cm

Question 4

M1 4.8² + 6.3² oe (= 62.73)
M1 √their 62.73
A1 awrt 7.92 cm
Answer: 7.92 cm (3 s.f.)

Question 5

M1 15.5² - 9.3² oe (= 153.76)
M1 √their 153.76
A1 12.4 cm cao
Answer: 12.4 cm

Question 6

M1 7² + 24² (= 625)
M1 25² (= 625)
C1 correct conclusion: since 7² + 24² = 25², the triangle is right-angled cso
Answer: 7² + 24² = 25² (625 = 625), so the triangle is right-angled.

Question 7

M1 9² + 12² (= 225)
M1 16² (= 256)
C1 correct conclusion: since 225 does not equal 256, the triangle is not right-angled cso
Answer: Not a right-angled triangle, since 9² + 12² = 225 but 16² = 256, and 225 is not equal to 256.

Question 8

M1 4.5² - 1.5² oe (= 18)
M1 √their 18
A1 awrt 4.24 m
Answer: 4.24 m (3 s.f.)

Question 9

M1 80² + 45² oe (= 8425)
M1 √their 8425
A1 awrt 91.8 cm
Answer: 91.8 cm (3 s.f.)

Question 10

M1 (7-2)² + (11-3)² oe (= 89)
M1 √their 89
A1 awrt 9.43
Answer: 9.43 (3 s.f.)

Question 11

(a) B1 correct reason, e.g. the perpendicular from the apex of an isosceles triangle bisects the base (by symmetry), so each half is 12 / 2 = 6 cm oe
(a) Answer: The perpendicular from the apex of an isosceles triangle bisects the base by symmetry, so each half is 12 divided by 2 = 6 cm.
(b) M1 10² - 6² oe (= 64)
(b) M1 √their 64
(b) A1 8 cm cao
(b) Answer: 8 cm

Question 12

M1 √120² + 90² oe
A1 150 m (diagonal distance)
M1 120 + 90 (= 210 m, distance via two sides)
A1 60 m cao (210 - 150)
Answer: 60 m

Question 13

(a) B1 8 cm and 15 cm cao
(a) Answer: 8 cm and 15 cm
(b) M1 √8² + 15² oe
(b) A1 √289
(b) A1 17 cm cao
(b) Answer: 17 cm

Question 14

M1 3² + 3² oe (= 18)
M1 √18 = √9 x 2
A1 3sqrt(2) cm oe, simplest form
Answer: 3sqrt(2) cm

Question 15

(a) M1 4² + 4²
(a) A1 32 cao
(a) Answer: b² = 32
(b) M1 √their b² + 7² oe, i.e. √32 + 49
(b) M1 √81
(b) A1 9 cm cao
(b) Answer: 9 cm

Question 16

(a) M1 24² + 18² oe (= 900)
(a) M1 √their 900
(a) A1 30 km cao
(a) Answer: 30 km
(b) M1 24 + 18 oe (= 42 km)
(b) A1 12 km cao (42 - 30)
(b) Answer: 12 km

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Pythagoras - Worksheets, Questions and Revision

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Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

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