Pythagoras’ theorem – Year 9 maths test answers to download as PDF

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1. Answer to Question 1 (4 marks)

   1. The unknown is the hypotenuse, so square and add: \(7^{2}+24^{2}=49+576=625\), then \(\sqrt{625}=\)25 cm. (1 mark)
   2. Square and add: \(6^{2}+9^{2}=36+81=117\). Then \(\sqrt{117}=10.81\ldots\), which is 10.8 cm. (1 mark)
   3. The hypotenuse is known, so square and subtract: \(15^{2}-9^{2}=225-81=144\), and \(\sqrt{144}=\)12 cm. (1 mark)
   4. Square and subtract: \(8^{2}-5^{2}=64-25=39\), and \(\sqrt{39}=6.24\ldots\), which is 6.2 cm. (1 mark)

   Parts (a) and (c) are Pythagorean triples (7, 24, 25 and 9, 12, 15), so their answers are exact. A quick sense check: the hypotenuse is always the longest side.

2. Answer to Question 2 (4 marks)

   1. Compare the sum of the squares of the two shorter sides with the square of the longest side: \(8^{2}+15^{2}=64+225=289\), and \(17^{2}=289\) (1 mark). The two values are equal, so by the converse of Pythagoras' theorem the triangle is right-angled, with the right angle opposite the 17 cm side (1 mark).
   2. \(5^{2}+6^{2}=25+36=61\), but \(8^{2}=64\) (1 mark). The values are different, so the triangle is not right-angled (1 mark).

   In (b), 61 is less than 64, which means the longest side is too long for a right angle: the angle opposite it is obtuse. If the sum of the squares had been greater than 64, the largest angle would be acute instead.

3. Answer to Question 3 (4 marks)

   1. Draw a right-angled triangle with AB as its hypotenuse. The horizontal difference is \(8-2=6\) and the vertical difference is \(9-1=8\) (1 mark). Then \(AB=\sqrt{6^{2}+8^{2}}=\sqrt{36+64}=\sqrt{100}=\)10 units (1 mark).
   2. Horizontal difference \(5-(-3)=8\), vertical difference \(4-(-2)=6\) (1 mark). Then \(CD=\sqrt{8^{2}+6^{2}}=\sqrt{64+36}=\)10 units (1 mark).

   The sign of a difference does not matter because it is squared: \((-6)^{2}=36\) as well. Both segments are 6, 8, 10 triangles, a multiple of the 3, 4, 5 triple. Subtracting negative coordinates is where errors creep in, so write the brackets.

4. Answer to Question 4 (4 marks)

   1. The ladder is the hypotenuse, the wall height is one short side and the distance from the wall is the other. So \(d=\sqrt{6.5^{2}-6^{2}}=\sqrt{42.25-36}\) (1 mark) \(=\sqrt{6.25}=\)2.5 m (1 mark).
   2. The ladder is still 6.5 m long and still the hypotenuse. Subtract the square of the new distance, \(3.9^{2}=15.21\): \(h=\sqrt{42.25-15.21}=\sqrt{27.04}\) (1 mark) \(=\)5.2 m (1 mark).

   Moving the foot out from 2.5 m to 3.9 m lowers the top by 0.8 m. Both triangles come from well-known triples: 2.5, 6, 6.5 is the 5, 12, 13 triple halved, and 3.9, 5.2, 6.5 is the 3, 4, 5 triple scaled by 1.3.

5. Answer to Question 5 (4 marks)

   1. The diagonal is the hypotenuse of a right-angled triangle with sides 60 m and 80 m: \(\sqrt{60^{2}+80^{2}}=\sqrt{3600+6400}=\sqrt{10000}=\)100 m. (1 mark)
   2. Dan walks along two sides, \(60+80=140\) m, while Eve walks 100 m. Dan walks \(140-100=\)40 m further. (1 mark)
   3. Time = distance ÷ speed. Eve: \(100\div1.5=66.7\) s, which is 67 s (1 mark). Dan: \(140\div1.5=93.3\) s, which is 93 s (1 mark).

   Eve arrives about 27 seconds before Dan. The diagonal is always shorter than going round two sides, a fact known as the triangle inequality.

Total : 20 marks