Chapter 11.3, Problem 5MRE

To determine

To Complete: In a ${45}^{°}-{45}^{°}-{90}^{°}$ triangle has leg $20cm$ , then the length of the altitude of the hypotenuse is…..

Answer to Problem 5MRE

  $h=10\sqrt{2}cm$

Explanation of Solution

Given: In a ${45}^{°}-{45}^{°}-{90}^{°}$ triangle has leg $20cm$ , then the length of the altitude of the hypotenuse is…..

  

Using Pythagoras theorem finding hypotenuse.

  $\begin{array}{l}A{C}^2={20}^2+{20}^2\\ A{C}^2=400+400\\ A{C}^2=800\\ AC=20\sqrt{2}cm\end{array}$

Now area of triangle for finding the altitude

  $\begin{array}{l}A=\frac{1}{2}bh\\ A=\frac{1}{2}(20)(20)\\ A=\frac{1}{2}(AC)(h)\mathrm{....}(i)\\ A=\frac{1}{2}\left(20\sqrt{2}\right)(h)\\ A=10\sqrt{2}h\mathrm{...}(ii)\\ from(i)and(ii)\\ h=10\sqrt{2}cm\end{array}$

Hence, $h=10\sqrt{2}cm$