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HIGHER

4.9H Pythagoras: Higher Tier Practice

EDEXCEL Edexcel 1MA1 · Calculator allowed · about 75 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Work out the length of the hypotenuse of a right-angled triangle with legs 8 cm and 15 cm.

(Total for Question 1 is 1 mark)

A square field has diagonal length 26 m. Work out the side length of the square.

(Total for Question 2 is 2 marks)

Show that a triangle with sides 7 cm, 24 cm and 25 cm is a right-angled triangle. State which side is the hypotenuse.

(Total for Question 3 is 3 marks)

The ends of a horizontal metal rod are 3 m apart. A support is placed so that the two halves of the rod form a right triangle with the ground: one half makes a vertical support of 1.2 m from the ground to the rod's midpoint. Calculate the length of one half of the rod. (Diagram: right triangle with vertical 1.2 m and horizontal half-base 1.5 m.)

(Total for Question 4 is 4 marks)

Points A and B have coordinates A(2, -1) and B(11, 5). Work out the distance AB. Give your answer to 1 decimal place.

(Total for Question 5 is 4 marks)

A circle has centre O and a chord AB of length 14 cm. The perpendicular from O to AB meets AB at M and OM = 6 cm. Find the radius of the circle.

(Total for Question 6 is 5 marks)

An engineer fits a diagonal brace across a rectangular panel 2.4 m by 1.6 m. The brace runs from one corner to the opposite corner. The engineer cuts the brace from a straight length of metal so it overhangs by 5 cm at each end. Calculate the required length of metal to order, correct to the nearest centimetre.

(Total for Question 7 is 6 marks)

A ladder of length 5.5 m leans against a vertical wall. The foot of the ladder is 2.1 m from the base of the wall. Work out the height up the wall the ladder reaches. Give your answer to 2 decimal places.

(Total for Question 8 is 6 marks)

Give a proof that for any positive real number x, the triangle with sides 1, x and √1 + x² is right-angled. Your proof should show which angle is right and use algebraic manipulation.

(Total for Question 9 is 5 marks)

Mark scheme · 4.9H Pythagoras: Higher Tier Practice

Question 1

B1 17 cm cao
Answer: 17 cm

Question 2

M1 use 2 x side² = 26² or side² = 26² / 2 oe
A1 side = 26 / √2 = 13*√2 cao
Answer: 13*√2 m

Question 3

M1 identify largest side 25 and calculate 7² + 24² = 49 + 576 = 625 oe
A1 compare with 25² = 625 and conclude equality oe
B1 state: triangle is right-angled with hypotenuse 25 cm cao
Answer: 25 cm

Question 4

M1 identify half-base = 3/2 = 1.5 m and use Pythagoras: length² = 1.5² + 1.2² oe
M1 calculate 1.5² + 1.2² = 2.25 + 1.44 = 3.69
A1 length = √3.69 = 1.92... m oe
A1 give final answer 1.92 m to 2 decimal places awrt
Answer: 1.92 m

Question 5

M1 use distance formula: sqrt[(11-2)² + (5-(-1))²] oe
M1 compute differences: 9² + 6² = 81 + 36 = 117
A1 √117 seen
A1 distance = 10.8 to 1 dp (10.816... -> 10.8) cao
Answer: 10.8

Question 6

M1 half-chord AM = 14/2 = 7 cm and right triangle OMA used
M1 use Pythagoras: radius² = OM² + AM² oe
M1 compute radius² = 6² + 7² = 36 + 49 = 85
A1 radius = √85 oe
A1 give decimal to 3 s.f. 9.22 cm (√85=9.2195...) awrt
Answer: 9.22 cm

Question 7

M1 use diagonal = √2.4² + 1.6² oe
M1 compute 2.4² + 1.6² = 5.76 + 2.56 = 8.32
M1 diagonal = √8.32 = 2.884... m
M1 add overhangs: total length = diagonal + 0.05 + 0.05 = diagonal + 0.10 m
A1 compute total = 2.884... + 0.10 = 2.984... m
A1 convert to cm and round to nearest cm: 2.984... m = 298.4... cm -> 298 cm cao
Answer: 298 cm

Question 8

M1 use Pythagoras: height² = 5.5² - 2.1² oe
M1 compute 5.5² - 2.1² = 30.25 - 4.41 = 25.84
M1 take square root: height = √25.84 oe
A1 height = 5.083... m
A1 round to 2 dp: 5.08 m cao
A1 state final answer 5.08 m awrt
Answer: 5.08 m

Question 9

M1 square the two shorter sides: 1² + x² = 1 + x² oe
M1 compare with square of longest side: (√1 + x²)² = 1 + x² oe
M1 establish equality 1 + x² = 1 + x² and conclude Pythagoras holds
A1 state triangle is right-angled with hypotenuse √1 + x² cao
A1 explicitly state the right angle is between sides of length 1 and x (i.e. the angle opposite √1 + x²)
Answer: The triangle is right-angled with hypotenuse √1 + x²

Pythagoras: Higher Tier Practice - Worksheets, Questions and Revision