Question

If a pool has to be constructed with a boundary of width $14m$ consisting of two straight sections $120m$ long joining semi-circular ends whose inner radius is $35m.$ Then, the area of the boundary is

• A. $7056m^2$
• B. $7000m^2$
• C. $7500m^2$
• D. $7050m^2$

Answer 1

The correct option is A

$7056m^2$

We have, $OC=O^{\prime }D=35m$
and $AC=BD=14m$
Also, $AB=CD=FE=GH=120m$
Now, $OA=O^{\prime }B=35+14=49m$

Area of the region (boundary) $=$ Area of rectangle $ABCD+$ Area of rectangle $EFGH+2$(area of semi-circle with radius $49m-$Area of semi-circle with radius $35m)$
$=120\times 14+120\times 14+2(\frac{1}{2}\times \frac{22}{7}\times {49}^2-\frac{1}{2}\times \frac{22}{7}\times {35}^2)=[1680+1680+\frac{22}{7}({49}^2-{35}^2\left)\right]=[3360+\frac{22}{7}(49+35\left)\right(49-35\left)\right]=3360+3696=7056m^2$