Question

In the given figure, two circles touch each other internally at point $P$, where $Q$ is the centre of larger circle. If $RS=17\text{cm},TU=9\text{cm}$ and $TV\perp RP$, then find the area of shaded region.

• A.
  $4090{\text{cm}}^2$
• B. $4096.91{\text{cm}}^2$
• C. $5000{\text{cm}}^2$
• D. $4050{\text{cm}}^2$

Answer 1

The correct option is B $4096.91{\text{cm}}^2$


Let the radii of two circles be ${}^{\prime }{R}^{\prime }$ and ${}^{\prime }{r}^{\prime }$
$PR=2R$
$PS=2r$
$PR-PS=RS$
$\Rightarrow 2R-2r=17$
$\Rightarrow 2(R-r)=17\Rightarrow R-r=\frac{17}{2}$
$\therefore R-r=8.5$
In $\Delta SQU$ and $\Delta UPQ,PR\perp TV$
$\Rightarrow \angle SQU=\angle UQP={90}^{\circ }$
Let $\angle QPU=\theta$
$\Rightarrow \angle PUQ={90}^{\circ }-\theta$
$\angle PUS={90}^{\circ }$ (triangle in a semi-circle is a right triangle)
$\Rightarrow \angle PUQ+\angle SUQ={90}^{\circ }$
$\Rightarrow {90}^{\circ }-\theta +\angle SUQ={90}^{\circ }$
$\therefore \angle SUQ=\theta =\angle QPU$
$\therefore \Delta SQU\sim \Delta UPQ$ (by AA similarity)
$\Delta SQU\sim \Delta UPQ$
$\frac{SU}{UP}=\frac{UQ}{PQ}=\frac{SQ}{UQ}$
$\frac{UQ}{PQ}=\frac{SQ}{UQ}$
$\frac{R-9}{R}=\frac{R-17}{R-9}$
$(R-9{)}^2=R(R-17)$
$R^2+81-18R=R^2=17R$
$-18R+17R=-81$
$-R=-81$
$\therefore R=81\text{cm}$
Substituting $R=81$ in (1), we get
$R-r=8.5$
$81-r=8.5$
$r=8.5-81$
$-r=-72.5$
$r=72.5$
$\therefore r=72.5\text{cm}$
Required shaded area:
$=\pi r^2-\pi r^2=\pi \left[\right(81{)}^2-\left(72.5{)}^2\right]$
$=3.14(6561-5256.25)$
$=3.14\times 1304.75$
$=4096.91{\text{cm}}^2$
Hence, required shaded area is $4096.91{\text{cm}}^2$