Answer

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Hint: If two tangents are drawn from a point to a circle, then the length of those two tangents from that particular point to the circle will be equal. In this question, $PQ,QR$ and $PR$ are the tangents to the circle.

In this question, we are given a right angle triangle having sides $PQ=24$cm, $QR=7$cm and $\angle

PQR={{90}^{\circ }}$. We can calculate the length of side $PR$ by using Pythagoras theorem.

Consider a triangle $ABC$ having $\angle ABC={{90}^{\circ }}$.

Using Pythagoras theorem the relation between the base, perpendicular and the hypotenuse of the

triangle is given by,

$AC=\sqrt{{{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}}.....................\left( 1 \right)$

Using Pythagoras theorem from equation $\left( 1 \right)$ in the triangle $PQR$, we get,

$PR=\sqrt{{{\left( PQ \right)}^{2}}+{{\left( QR \right)}^{2}}}$

In the question, it is given $PQ=24$cm, $QR=7$cm.

$\begin{align}

& \Rightarrow PR=\sqrt{{{\left( 24 \right)}^{2}}+{{\left( 7 \right)}^{2}}} \\

& \Rightarrow PR=\sqrt{576+49} \\

& \Rightarrow PR=\sqrt{625} \\

& \Rightarrow PR=25 \\

\end{align}$

Since the circle is inscribed in the triangle $PQR$, sides $PQ,QR,PR$ will act as a tangent to the

circle. Let us name the point of contact of circle with the tangent $PQ$ as $A$, tangent $QR$ as $B$

and tangent $PR$ as $C$. Let us join the point $A,B,C$ to the center of the circle $O$ . Also, let us

consider the radius of the circle equal to $x$.

Since $OA,OB,OC$ are the radius, their lengths are equal to $x$.

It can be seen from the figure that in quadrilateral $AOQB$, two adjacent sides ($OA,OB$) are equal

(since they are the radius of the circle) and all the angles are right angles. This means, $AOQB$ is a

square. So, $AQ=x$ and $BQ=x$.

There is a property of tangent which states that, if two tangents are drawn from a point to a circle,

then the length of those two tangents from that particular point to the circle will be equal. Using this

property in the above triangle,

$\Rightarrow PA=PC,QA=QB,RB=RC............\left( 2 \right)$

We have obtained $AQ=x$. Also $PQ=24$. Since $PAQ$ is a straight line, we can say,

$\begin{align}

& PA=PQ-QA \\

& \Rightarrow PA=24-x \\

\end{align}$

From $\left( 2 \right)$, we have $PA=PC$.

$\Rightarrow PC=24-x............\left( 3 \right)$

Also, we have obtained $BQ=x$. Also $QR=7$. Since $QBR$ is a straight line, we can say,

$\begin{align}

& RB=QR-QB \\

& \Rightarrow RB=7-x \\

\end{align}$

From $\left( 2 \right)$, we have $RB=RC$.

\[\Rightarrow RC=7-x............\left( 4 \right)\]

Since $PCR$ is a straight line, we can write,

$PR=PC+RC$

Using Pythagoras theorem, we obtained $PR=25$cm. Also, from equation $\left( 1 \right)$ and

equation $\left( 2 \right)$, we have $PC=24-x$ and \[RC=7-x\]. Substituting in the above equation,

we get,

$\begin{align}

& 25=24-x+7-x \\

& \Rightarrow 2x=6 \\

& \Rightarrow x=3 \\

\end{align}$

Hence, the radius of the inscribed circle is $3$.

Note: There is an alternative method to find the radius of the inscribed circle. The radius of the

inscribed circle of a triangle is given by the formula, $r=\dfrac{K}{s}$ where $K$ is the area of the

triangle and $s$ is the half of the perimeter of the triangle.

In this question, we are given a right angle triangle having sides $PQ=24$cm, $QR=7$cm and $\angle

PQR={{90}^{\circ }}$. We can calculate the length of side $PR$ by using Pythagoras theorem.

Consider a triangle $ABC$ having $\angle ABC={{90}^{\circ }}$.

Using Pythagoras theorem the relation between the base, perpendicular and the hypotenuse of the

triangle is given by,

$AC=\sqrt{{{\left( AB \right)}^{2}}+{{\left( BC \right)}^{2}}}.....................\left( 1 \right)$

Using Pythagoras theorem from equation $\left( 1 \right)$ in the triangle $PQR$, we get,

$PR=\sqrt{{{\left( PQ \right)}^{2}}+{{\left( QR \right)}^{2}}}$

In the question, it is given $PQ=24$cm, $QR=7$cm.

$\begin{align}

& \Rightarrow PR=\sqrt{{{\left( 24 \right)}^{2}}+{{\left( 7 \right)}^{2}}} \\

& \Rightarrow PR=\sqrt{576+49} \\

& \Rightarrow PR=\sqrt{625} \\

& \Rightarrow PR=25 \\

\end{align}$

Since the circle is inscribed in the triangle $PQR$, sides $PQ,QR,PR$ will act as a tangent to the

circle. Let us name the point of contact of circle with the tangent $PQ$ as $A$, tangent $QR$ as $B$

and tangent $PR$ as $C$. Let us join the point $A,B,C$ to the center of the circle $O$ . Also, let us

consider the radius of the circle equal to $x$.

Since $OA,OB,OC$ are the radius, their lengths are equal to $x$.

It can be seen from the figure that in quadrilateral $AOQB$, two adjacent sides ($OA,OB$) are equal

(since they are the radius of the circle) and all the angles are right angles. This means, $AOQB$ is a

square. So, $AQ=x$ and $BQ=x$.

There is a property of tangent which states that, if two tangents are drawn from a point to a circle,

then the length of those two tangents from that particular point to the circle will be equal. Using this

property in the above triangle,

$\Rightarrow PA=PC,QA=QB,RB=RC............\left( 2 \right)$

We have obtained $AQ=x$. Also $PQ=24$. Since $PAQ$ is a straight line, we can say,

$\begin{align}

& PA=PQ-QA \\

& \Rightarrow PA=24-x \\

\end{align}$

From $\left( 2 \right)$, we have $PA=PC$.

$\Rightarrow PC=24-x............\left( 3 \right)$

Also, we have obtained $BQ=x$. Also $QR=7$. Since $QBR$ is a straight line, we can say,

$\begin{align}

& RB=QR-QB \\

& \Rightarrow RB=7-x \\

\end{align}$

From $\left( 2 \right)$, we have $RB=RC$.

\[\Rightarrow RC=7-x............\left( 4 \right)\]

Since $PCR$ is a straight line, we can write,

$PR=PC+RC$

Using Pythagoras theorem, we obtained $PR=25$cm. Also, from equation $\left( 1 \right)$ and

equation $\left( 2 \right)$, we have $PC=24-x$ and \[RC=7-x\]. Substituting in the above equation,

we get,

$\begin{align}

& 25=24-x+7-x \\

& \Rightarrow 2x=6 \\

& \Rightarrow x=3 \\

\end{align}$

Hence, the radius of the inscribed circle is $3$.

Note: There is an alternative method to find the radius of the inscribed circle. The radius of the

inscribed circle of a triangle is given by the formula, $r=\dfrac{K}{s}$ where $K$ is the area of the

triangle and $s$ is the half of the perimeter of the triangle.

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