Chapter 7.3, Problem 38HP

To determine

To find: the value of YW.

Answer to Problem 38HP

  $YW=3.54$ .

Explanation of Solution

Given:

  $\overline{YW}$ is an altitude of a triangle of $\Delta XYZ$ .

  

Calculation:

According to the question,

  $\overline{YW}$ is an altitude of a triangle of $\Delta XYZ$ . Therefore, in $\Delta XYZ\text{ and }\Delta YZW$

  $\begin{array}{l}\angle XYZ=\angle YWZ\{\text{each 90}}\\ \angle Z=\angle Z\left\{\text{common}\right\}\\ \Delta XYZ\sim \Delta YZW\{byAA-similarity\}\end{array}$ .

Since, $\Delta XYZ$ is an isosceles triangle, therefore $\Delta YZW$ is an isosceles triangle.

Now use Pythagoras theorem.

  $\begin{array}{l}{\left(WY\right)}^2+{\left(WZ\right)}^2={\left(YZ\right)}^2\\ x^2+x^2=5^2\{\text{substitutes these value}}\\ \Rightarrow 2x^2=25\\ x^2=\frac{25}{2}\{\text{divide both sides by 2}}\\ \therefore x=\sqrt{\frac{25}{2}}\{\text{taking square root both sides}}\\ \Rightarrow x=\frac{5}{\sqrt{2}}=3.54\end{array}$

Hence, the value of $YW=3.54$ .