Question

In the figure, find the area of $△PQR$ and the height $PS$.

Answer 1

As area of right angled triangle$=\frac{1}{2}$(Base)(Height)

$\Rightarrow ar\left(\Delta PQR\right)=ar\left(QTR\right)-ar\left(PQT\right)=\frac{1}{2}\left(QT\right)\left(TR\right)-\frac{1}{2}\left(QT\right)\left(PT\right)$

$=\frac{1}{2}\left(QT\right)(TR-PT)=\frac{1}{2}\left(QT\right)\left(PR\right)=\frac{1}{2}\left(12\right)\left(15\right)=90c{m}^2$

$\Rightarrow \frac{1}{2}\left(PS\right)\left(QR\right)=90$

$\Rightarrow PS\times 20=180$

$\Rightarrow PS=9$cm.