The Baudhayana-Pythagoras Theorem — Question 2

If a is the length of two equal sides of an isosceles right triangle, then hypotenuse = \(\sqrt{a^2+a^2}=\sqrt{2 a^2}=a \sqrt{2}\) (i) a = 3 Hypotenuse = a√2 = 3√2 = \(\sqrt{3 \times 3 \times 2}\) = √18 √18 lies between √16 and √25 √16 < √18 < √25 ⇒ 4 < 3√2 < 5 In one decimal point 4.1 2 = 16.81, 4.2 2 = 17.64, 4.3 2 = 18.49 So, 4.2 < 3√2 < 4.3 (ii) a = 4 Hypotenuse = a√2 = 4√2 = √32 √25 < √32 < √36 ⇒ 5 < 4√2 < 6 In one decimal point 5.1 2 = 26.01, 5.2 2 = 27.04, 5.3 2 = 28.09, 5.4 2 = 29.16, 5.5 2 = 30.25, 5.6 2 = 31.36, 5.7 2 = 32.49 So, 5.6 < 4√2 < 5.7 (iii) a = 6 Hypotenuse = a√2 = 6√2 = \(\sqrt{6 \times 6 \times 2}\) = √72 √64 < √72 < √81 ⇒ 8 < 6√2 < 9 In one decimal point 8.1 2 = 65.61, 8.2 2 = 67.24, 8.3 2 = 68.89, 8.4 2 = 70.56, 8.5 2 = 72.25 So, 8.4 < 6√2 < 8.5. (iv) a = 8 Hypotenuse = a√2 = 8√2 = √128 √121 < √128 < √144 ⇒ 11 < 8√2 < 12 In one decimal point 11.1 2 = 123.21, 11.2 2 = 125.44, 11.3 2 = 127.69, 11.4 2 = 129.96 So, 11.3 < 8√2 < 11.4. (v) a = 9 Hypotenuse = a√2 = 9√2 = √162 √144 < √162 < √169 ⇒ 12 < 9√2 < 13 In one decimal point 12.1 2 = 146.41, 12.2 2 = 148.84, 12.3 2 = 151.29, 12.4 2 = 153.76, 12.5 2 = 156.25, 12.6 2 = 158.76, 12.7 2 = 161.29, 12.8 2 = 163.84 So, 12.7 < 9√2 < 12.8