Question

$PQRS$ is a rectangle inscribed in a quadrant of a circle of radius $13\mathrm{cm}$. $A$ is any point on $PQ$. If $PS=5\mathrm{cm}$, then $ar\left(PAS\right)=30{\mathrm{cm}}^2$. Write True or False and justify your answer.

Answer 1

Determine whether the given statement is true or false.

It is given that $PQRS$ is a rectangle inscribed in a quadrant of a circle of radius $PS=5\mathrm{cm}$.

It is given that $PS=5\mathrm{cm}$ and the radius is $13\mathrm{cm}$.

From the diagram, observe that the radius $QS=13\mathrm{cm}$.

So, apply Pythagoras theorem in$∆PQS$.

$\begin{array}{rcl}Q{S}^2& =& P{Q}^2+P{S}^2\\ & \Rightarrow & {13}^2=P{Q}^2+5^2\\ & \Rightarrow & P{Q}^2=169-25\\ & \Rightarrow & P{Q}^2=144\\ & \Rightarrow & PQ=12\end{array}$

It is known that the area of a triangle is $A=base\times height$.

From the diagram, conclude that the base is $PS$ and height is $PQ$.

Therefore, the area of the triangle $∆PQS$ is as follows:

$\begin{array}{rcl}A& =& \frac{1}{2}\times PS\times PQ\\ & \Rightarrow & A=\frac{1}{2}\times 5\times 12\left[\therefore \mathrm{PQ}=12,\mathrm{PS}=5\right]\\ & \Rightarrow & A=30\end{array}$

Thereofore, the area of $∆PQS$ is $30{\mathrm{cm}}^2$.

Since $A$ is any point on $PQ$, the area of $∆PAS$ is $30{\mathrm{cm}}^2$.

Hence, the statement is true.