MCQ
In the given figure, $ABCD$ is a rectangle inscribed in a circle having length $8\ cm$ and breadth $6\ cm$. If $\pi=3.14$ then the area of the shaded region is:
- A$22\text{cm}^2$
- B$26\text{cm}^2$
- ✓$30.5\text{cm}^2$
- DNone of the above
✓
Answer
Correct option: C.
$30.5\text{cm}^2$
Join $AC$.
Now, $AC$ is the diameter of the circle.
We have:
$AC^2= AB^2+ BC^2$ [By pythagoras' theorem]
$\Rightarrow\text{AC}^2=\big\{(8)^2+(6)^2\big\}\text{cm}^2$
$\Rightarrow\text{AC}^2=(64+36)\text{cm}^2$
$\Rightarrow\text{AC}^2=100\text{cm}^2$
$\Rightarrow\text{AC}=10\text{cm}$
$\therefore$ Radius of the circle $=\frac{10}{2}\text{cm}$
$=5\text{cm}$
Now,
Area of the shaded region = Area of the circle with radius $5cm$ - Area of the rectangle $ABCD$
$=\big|(3.14\times5\times5)-(8\times6)\big|\text{cm}^2$
$=\Big|\Big(\frac{314}{100}\times25\Big)-48\Big|\text{cm}^2$
$=\Big(\frac{157}{2}-48\Big)\text{cm}^2$
$=\frac{61}{2}\text{cm}^2$
$=30.5\text{cm}^2$
Now, $AC$ is the diameter of the circle.
We have:
$AC^2= AB^2+ BC^2$ [By pythagoras' theorem]
$\Rightarrow\text{AC}^2=\big\{(8)^2+(6)^2\big\}\text{cm}^2$
$\Rightarrow\text{AC}^2=(64+36)\text{cm}^2$
$\Rightarrow\text{AC}^2=100\text{cm}^2$
$\Rightarrow\text{AC}=10\text{cm}$
$\therefore$ Radius of the circle $=\frac{10}{2}\text{cm}$
$=5\text{cm}$
Now,
Area of the shaded region = Area of the circle with radius $5cm$ - Area of the rectangle $ABCD$
$=\big|(3.14\times5\times5)-(8\times6)\big|\text{cm}^2$
$=\Big|\Big(\frac{314}{100}\times25\Big)-48\Big|\text{cm}^2$
$=\Big(\frac{157}{2}-48\Big)\text{cm}^2$
$=\frac{61}{2}\text{cm}^2$
$=30.5\text{cm}^2$
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