Suppose we have a pavement all around the garden bed of which \[1\,m\] as shown below. The area of the pavement is: |
A) \[(21+15\pi )\,{{m}^{2}}\]
B) \[(42+15\pi )\,{{m}^{2}}\]
C) \[(42-15\pi )\,m\]
D) \[(84+30\pi )\,{{m}^{2}}\]
Correct Answer: B
Solution :
We have \[FG=1+14+1=16m\] |
Area of the pavement = Area of rectangle EFGH \[+2\times \] (Area of semi-circle with diameter \[16m\]) - (Area of rectangle \[ABCD+2\times \] (Area of semi-circle with diameter \[14m\])] |
\[=EF\times FG+2\times \frac{1}{2}\pi {{\left( \frac{16}{2} \right)}^{2}}-\left[ AB\times BC+2\times \frac{1}{2}\pi {{\left( \frac{14}{2} \right)}^{2}} \right]\]\[=21\times 16+\pi \times {{(8)}^{2}}-[21\times 14+\pi +{{(7)}^{2}}]\] |
\[=336+64\pi -294-49\pi =(42+15\pi ){{m}^{2}}\] |
So, option [b] is correct. |
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