Question

If a pool has to be constructed with a boundary of width 14 m consisting of two straight sections 120 m long joining semi-circular ends whose inner radius is 35 m. 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, OB =O'C =35 m
and AB =CD =14 m
Also, AD =FG =120 m
Now, OA =O'D =(35+14) =49 m

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