Chapter 11.7, Problem 20WE

(a)

To determine

To Find: The ratio of area of $TOSandQOP$ .

Answer to Problem 20WE

  $\frac{4}{9}$

Explanation of Solution

Given:

  

In triangles $TOSandQOP$

  $\begin{array}{l}\angle SOT=\angle POQ(v.o.a)\\ \angle OST=\angle OPQ(a.angle)\\ thus\\ \Delta SOT\sim \Delta POQ\\ i.e\\ SR=PQ=9\\ now\\ \frac{ar(TOS)}{ar(POQ)}=\frac{\frac{1}{2}\cdot 6\cdot h_{tos}}{\frac{1}{2}\cdot 9\cdot h_{qop}}\\ \frac{h_{tos}}{h_{poq}}=\frac{6}{9}=\frac{2}{3}\\ \frac{ar(TOS)}{ar(POQ)}=\frac{2}{3}\cdot \frac{2}{3}=\frac{4}{9}\end{array}$

(b)

To determine

To Find: The ratio of area of $TOSandTQR$ .

Answer to Problem 20WE

  $\frac{4}{25}$

Explanation of Solution

Given:

  

  $\begin{array}{l}\angle QTR=\angle OTS(common)\\ \angle QRT=\angle OST(samesideangle)\\ thus\\ \Delta SOT\sim \Delta TQR\\ now\\ \frac{ar(TOS)}{ar(TQR)}=\frac{6^2}{{15}^2}\\ \frac{ar(TOS)}{ar(TQR)}=\frac{36}{225}=\frac{4}{25}\end{array}$

Hence, $\frac{4}{25}$ .