| In the given figure, ABCD is a rectangle of dimensions\[\text{21 cm}\times \text{14 cm}\]. A semicircle is drawn with BC as diameter. |
| The area and the perimeter of the shaded region in the figure, are: (CBSE 2017) |
 |
A) \[217\,c{{m}^{2}},\,\,78\,cm\]
B) \[215\,c{{m}^{2}},\,\,82\,cm\]
C) \[219\,c{{m}^{2}},\,76\,cm\]
D) None of these
Correct Answer: A
Solution :
| [a] Given, length of the rectangle |
| \[ABCD(l)=21cm\] |
| and breadth of the rectangle |
| \[ABCD(b)=14cm\] |
| From figure, |
| Diameter of semicircle = breadth of rectangle = 14 cm |
| \[\therefore \] Radius of semicircle \[(r)=\frac{14}{2}=7cm\] |
| (i) Now, area of rectangle |
| \[ABCD=l\times b=21\times 4\] |
| \[=294\,c{{m}^{2}}\] |
| and area of semicircle |
| \[=\frac{\pi {{r}^{2}}}{2}=\frac{22\times {{(7)}^{2}}}{7\times 2}=77c{{m}^{2}}\] |
| So. Area of shaded region |
| \[=\text{Area of rectangle ABCD}-\text{Area of semicircle}\]\[=294-77=217c{{m}^{2}}\] |
| (ii) Perimeter of shaded region |
| \[=2l+b+\] circumference of semicircle |
| \[=2\times 17+14+\frac{1}{2}\times 2\pi r\] |
| \[=34+14+\frac{22}{7}\times 7=78cm\] |
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