Question 1 of 16intermediate🔧 ApplyNumerical5 marks

The length of the two equal sides of an isosceles right triangle is given. Find the length of the hypotenuse. Find bounds on the length of the hypotenuse such that they have at least one digit after the decimal point.

(i)

Equal sides = 3

Answer

Hypotenuse = 32\sqrt{2} ≈\approx 4.24 units. Bounds: 4.2 < 32\sqrt{2} < 4.3 (since 4.2² = 17.64 and 4.3² = 18.49, while (32\sqrt{2})² = 18)

(ii)

Equal sides = 4

Answer

Hypotenuse = 42\sqrt{2} ≈\approx 5.66 units. Bounds: 5.6 < 42\sqrt{2} < 5.7 (since 5.6² = 31.36 and 5.7² = 32.49, while (42\sqrt{2})² = 32)

(iii)

Equal sides = 6

Answer

Hypotenuse = 62\sqrt{2} ≈\approx 8.49 units. Bounds: 8.4 < 62\sqrt{2} < 8.5 (since 8.4² = 70.56 and 8.5² = 72.25, while (62\sqrt{2})² = 72)

(iv)

Equal sides = 8

Answer

Hypotenuse = 82\sqrt{2} ≈\approx 11.31 units. Bounds: 11.3 < 82\sqrt{2} < 11.4 (since 11.3² = 127.69 and 11.4² = 129.96, while (82\sqrt{2})² = 128)

(v)

Equal sides = 9

Answer

Hypotenuse = 92\sqrt{2} ≈\approx 12.73 units. Bounds: 12.7 < 92\sqrt{2} < 12.8 (since 12.7² = 161.29 and 12.8² = 163.84, while (92\sqrt{2})² = 162)

Explanation

(i) Using the formula c² = 2a², we get c² = 2×92 \times 9 = 18. So c = 18\sqrt{18} = 32\sqrt{2}. To find bounds: 4² = 16 < 18 < 20.25 = 4.5². Checking 4.2² = 17.64 and 4.3² = 18.49, we get 4.2 < 18\sqrt{18} < 4.3. (ii) Using c² = 2a², we get c² = 2×162 \times 16 = 32. So c = 32\sqrt{32} = 42\sqrt{2}. Checking: 5.6² = 31.36 < 32 < 32.49 = 5.7². (iii) Using c² = 2a², we get c² = 2×362 \times 36 = 72. So c = 72\sqrt{72} = 62\sqrt{2}. Checking: 8.4² = 70.56 < 72 < 72.25 = 8.5². (iv) Using c² = 2a², we get c² = 2×642 \times 64 = 128. So c = 128\sqrt{128} = 82\sqrt{2}. Checking: 11.3² = 127.69 < 128 < 129.96 = 11.4². (v) Using c² = 2a², we get c² = 2×812 \times 81 = 162. So c = 162\sqrt{162} = 92\sqrt{2}. Checking: 12.7² = 161.29 < 162 < 163.84 = 12.8².

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