Chapter 8.2, Problem 11PSB

To determine

To calculate:The value of $\frac{x+y+z}{2}$

Answer to Problem 11PSB

The value of $\frac{x+y+z}{2}$ is 205

Explanation of Solution

Given information Triangles are similar, i.e., $\Delta KJM\sim \Delta OPR$

Formula used For similar triangles, corresponding angles are equalCalculation In given triangles, i.e., $\Delta KJM\sim \Delta OPR$

For finding unknown angle in $\Delta OPR,$ we have

Sum of the angles in a triangle is 180 $°$

  $\begin{array}{l}\text{ In }\Delta OPR,\\ \angle O+\angle P+\angle R=180°\\ \Rightarrow \angle O=180-60-40=80°\end{array}$

Now on comparing angles in both the given similar triangles, we have

  $\begin{array}{l}\therefore \angle JKM=80°\\ \angle JKM=\angle POR\\ \Rightarrow \left(\frac{1}{2}y\right)°=80°\\ \Rightarrow y=160\\ \\ \angle KJM=60°\\ \Rightarrow \left(x+10\right)°=60°\\ \Rightarrow x=50and\\ \\ \angle JMK=40°\\ \Rightarrow \left(0.3z-20\right)°=40°\\ \Rightarrow 0.3z=60\\ \Rightarrow z=200\\ \\ \therefore \frac{x+y+z}{2}=\frac{50+160+200}{2}\\ =\frac{410}{2}=205\end{array}$

  $\begin{array}{l}\angle JKM=\angle POR\\ In\Delta OPR,\\ \angle O+\angle P+\angle R=180\\ \Rightarrow \angle O=180-60-40=80°\\ \therefore \angle JKM=80°\\ \Rightarrow \left(\frac{1}{2}y\right)°=80°\\ \Rightarrow y=160\\ \angle KJM=60°\\ \Rightarrow \left(x+10\right)°=60°\\ \Rightarrow x=50and\\ \angle JMK=40°\\ \Rightarrow \left(0.3z-20\right)°=40°\\ \Rightarrow 0.3z=60\\ \Rightarrow z=200\\ \therefore \frac{x+y+z}{2}=\frac{50+160+200}{2}\\ =\frac{410}{2}=205\end{array}$