Question

Find the area of a quadrilateral $PQRS$ in which $\angle QPS=\angle SQR={90}^o,PQ=12cm,PS=9cm,QR=8cm$ and $SR=17cm$

Answer 1

Given : $\angle$$QPS=90$ $\angle$ $SQR=90$

$PQ=12cm$ , $PS=9cm$ , $QR=8cm$ and $SR=17cm$

By Pythagoras theorem in $\Delta SPQ$ we get :

$\Rightarrow$$SP$${}^2$$+PQ$${}^2$$=SQ$${}^2$

$\Rightarrow 9^2+{12}^2=S{Q}^2$

$\Rightarrow 81+144=S{Q}^2$

$\Rightarrow S{Q}^2=225$

$\Rightarrow SQ=\sqrt{225}$

$\Rightarrow SQ=15cm$

Area of the parallelogram = Area of triangle SPQ + area of triangle SQR.

$\Rightarrow$ Area of Triangle SPQ = $\frac{1}{2}\times base\times height$

$\Rightarrow \frac{1}{2}\times PQ\times SP$

$\Rightarrow \frac{1}{2}\times 12\times 9$

$\Rightarrow$ Area of triangle SPQ = $54c{m}^2$

$\Rightarrow$ Area of Triangle SQR = $\frac{1}{2}\times base\times height$

$\Rightarrow \frac{1}{2}\times QR\times SQ$

$\Rightarrow \frac{1}{2}\times 8\times 15$

$\Rightarrow$ Area of triangle SPQ = $60c{m}^2$

Total area = area of triangle SPQ + area of trinagle SQR

$\Rightarrow 54+60=114c{m}^2$

Area of quadrilateral PQRS is $114c{m}^2$