WAEC SSCE - General Mathematics - 2002 (Objective)

Question 1 Report

The length, in cm, of the sides of a right angled triangle are x, (x+2) and (x+1) where x > 0. Find , in cm, the length of its hypotenuse

Answer Details

In a right-angled triangle, the hypotenuse is the longest side, and it is opposite to the right angle. Using the Pythagorean theorem, we know that the sum of the squares of the two shorter sides is equal to the square of the hypotenuse. So, we have: \begin{align*} x^2 + (x+2)^2 &= (x+1)^2 \\ x^2 + x^2 + 4x + 4 &= x^2 + 2x + 1 \\ x^2 + 2x^2 + 4x + 4 &= x^2 + 2x + 1 \\ 3x^2 + 4x + 4 &= x^2 + 2x + 1 \\ 2x^2 + 2x + 3 &= 0 \end{align*} We can solve this quadratic equation using the quadratic formula: \begin{align*} x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ &= \frac{-2 \pm \sqrt{2^2 - 4(2)(3)}}{2(2)} \\ &= \frac{-2 \pm \sqrt{-8}}{4} \\ &= \frac{-1}{2} \pm \frac{\sqrt{2}}{2}i \end{align*} Since x must be greater than 0, the only valid solution is: \begin{align*} x &= \frac{-1}{2} + \frac{\sqrt{2}}{2}i \end{align*} However, we are asked to find the length of the hypotenuse, which is given by: \begin{align*} \sqrt{x^2 + (x+2)^2} &= \sqrt{\left(\frac{-1}{2} + \frac{\sqrt{2}}{2}i\right)^2 + \left(\frac{1}{2} + \frac{\sqrt{2}}{2}i\right)^2} \\ &= \sqrt{\frac{1}{4} - \frac{1}{2}\sqrt{2}i - \frac{1}{4} - \frac{1}{2}\sqrt{2}i + \frac{1}{4} + \frac{\sqrt{2}}{2}i + \frac{1}{4} - \frac{\sqrt{2}}{2}i} \\ &= \sqrt{\frac{1}{2}} \\ &= \frac{\sqrt{2}}{2} \approx 0.707 \end{align*} Therefore, the length of the hypotenuse is approximately 0.707 cm, which is closest to 5 cm.






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