Question

PQRS is a rectangle inscribed in a quadrant of a circle of radius 13 cm. A is any point on PQ. If PS = 5 cm, then $ar\left(△RAS\right)$ is

Answer 1

In quadrant PLRM, rectangle PQRS is inscribed.
Radius of the circle = 13 cm

A is any point on PQ.
AR and AS are joined, PS = 5 cm
In right $△PRS,$ $P{R}^2=P{S}^2+S{R}^2\Rightarrow (13{)}^2=(5{)}^2+S{R}^2\Rightarrow 169=25+S{R}^2\Rightarrow S{R}^2=169-25=144=(12{)}^2\therefore SR=12cm\therefore AreaofrectanglePQRS=PS\times SR=5\times 12=60c{m}^2$
$\because$ Rectangle PQRS and $△RAS$ are on the same base SR and between the same parallels
$\therefore$ Area $△RAS=\frac{1}{2}AreaofrectanglePQRS$
$=\frac{1}{2}\times 60=30c{m}^2$