33 Pythagoras’ theorem SKILL EXAM FACTS Use Pythagoras’ theorem to find one side of a right-angled triangle, given the lengths of the other two sides H M L KEY FACTS • In a right-angled triangle the side opposite the right angle (the longest side of the triangle) is called the hypotenuse of the triangle. c b Marks lost (%) Marks available • In the diagram the length of the hypotenuse is c. • Pythagoras’ theorem states that the area of the square on the hypotenuse is equal to the sum of the areas of the squares on the other two sides. a D • That is c2 = a2 + b2 • In triangle DEF, Pythagoras’ theorem gives DE2 = EF 2 + DF 2 F E • DE2 means that the length of the side DE is squared. Getting it right EXAM TIP D In triangle DEF 6.4 cm FE = 8.7 cm, F DF = 6.4 cm, 8.7 cm Angle DFE = 90°. Calculate the length of DE. Give your answer correct to 1 decimal place. DE2 = EF2 + DF2 DE 2 = 8.72 + 6.42 DE 2 = 75.69 + 40.96 96 AIMH_C33.indd 96 Diagram NOT E accurately drawn “Diagram NOT accurately drawn” means that taking measurements from the diagram will not give the correct answer. Identify the hypotenuse of the triangle (the side opposite the right-angle). Then write down Pythagoras’ theorem for the triangle. Substitute the given lengths. You would get 1 mark for this. Pythagoras’ theorem 13/6/07 09:00:29 WARNING DE2 = 116.65 DE = ! A common error is to fail to find the square root and give the answer as 116.65 116.65 = 10.80046… DE = 10.8 cm Remember to round your answer to 1 decimal place and write the units. PQR is a right-angled triangle. Angle PQR = 90°. QR = 15 cm. PR = 19 cm. Work out the length of PQ. Give your answer correct to 1 decimal place. P Q 19 cm Diagram NOT accurately drawn 15 cm WARNING R (1388 November 2005) PR2 192 361 361 − 225 PQ2 PQ = PQ2 + QR2 = PQ2 + 152 = PQ2 + 225 = PQ2 = 136 = 136 = 11.6619... PQ = 11.7 cm ! The side to be found is not opposite the right-angle so it is NOT the hypotenuse. A common error is to write incorrectly PQ2 = 152 + 192 This gives PQ = 24.2 which is not sensible, as PQ must be shorter than the hypotenuse, PR. Now try these In Questions 1–4, work out the lengths of the sides marked with letters. The diagrams are not accurately drawn. Give each answer correct to 1 decimal place. 1 7 cm a 12 cm 2 3 10.3 cm 9 cm 4 cm b 4.6 cm c 5.8 cm 4 d 13.6 cm Pythagoras’ theorem AIMH_C33.indd 97 97 13/6/07 09:00:32 P 5 PQR is a right-angled triangle. PR = 6 cm. QR = 4 cm Work out the length of PQ. Give your answer correct to 1 decimal place. Diagram NOT accurately drawn 6 cm Q 6 In triangle PQR QR = 9.3 cm. PQ = 5.7 cm. Angle PQR = 90°. Calculate the length of PR. Give your answer correct to 1 decimal place. 4 cm R (1387 June 2006) P Diagram NOT accurately drawn 5.7 cm Q R 9.3 cm (1388 November 2005) Diagram NOT accurately drawn 7 Work out the value of x. 7.5 cm x cm A 7.2 cm (4400 May 2006) 8 ABC is a triangle. AB = AC = 13 cm. BC = 10 cm. M is the midpoint of BC. Angle AMC = 90°. 13 cm 13 cm Diagram NOT accurately drawn B M 10 cm Work out the length of AM. 9 The diagram shows three cities. Norwich is 168 km due East of Leicester. York is 157 km due North of Leicester. Calculate the distance between Norwich and York. Give your answer correct to the nearest kilometre. C (4400 November 2006) York Diagram NOT accurately drawn 157 km Leicester Norwich 168 km N (1387 November 2006) 10 The diagram shows the positions of three telephone masts A, B and C. Mast C is 5 kilometres due East of Mast B. Mast A is due North of Mast B and 8 kilometres from Mast C. A Diagram NOT accurately drawn N 8 km Calculate the distance of A from B. Give your answer in kilometres, correct to 2 decimal places. B 98 Pythagoras’ theorem AIMH_C33.indd 98 5 km C (1385 June 1999) Pythagoras’ theorem 98 13/6/07 09:00:33

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