Question

$ABCD$ is a field in the shape of a trapezium. $AB||DC$ and $\angle ABC={90}^{\circ },\angle DAB={60}^{\circ }$. Four sectors are formed withe centres $A,B,C$ and $D.$ The radius of each sector is $17.5\text{m}.$ Find the area of remaining portion given that $AB=75\text{m}$ and $CD=50\text{m}.$

Answer 1

Let the central angles of $4$ sectors be ${\theta }_1,{\theta }_2,{\theta }_3,{\theta }_4$.

Sum of area of sectors $=\frac{{\theta }_1}{360}\times \pi r^2+\frac{{\theta }_2}{360}\times \pi r^2+\frac{{\theta }_3}{360}\times \pi r^2+\frac{{\theta }_4}{360}\times \pi r^2$

$=\frac{{\theta }_1+{\theta }_2+{\theta }_3+{\theta }_4}{360}\times \pi (17.5{)}^2$

But ${\theta }_1+{\theta }_2+{\theta }_3+{\theta }_4={360}^{\circ }$ as they are vertex anges of a quadrilateral.

Hence,

Sum of area of sectors $=\frac{{360}^{\circ }}{{360}^{\circ }}\times \pi \times 306.25$

$=962.5{\text{m}}^2$

Since $AB||CD$ and $\angle$ $ABC=$ ${90}^{\circ }$. Therefore

$\angle BCD=$ ${90}^{\circ }$

Let $DL$ be perpendicular drawn from $D$ on $AB.$ Then,

$AL=AB-BL$

$=AB-CD$

$=(75-50)$

$=25\text{m}$

In $△ALD,$ we have

$\tan {60}^{\circ }=\frac{DL}{AL}$

$\Rightarrow \sqrt{3}=\frac{DL}{25}$

$\Rightarrow DL=25\sqrt{3}\text{m}$

$ar\left(ABCD\right)=\frac{1}{2}(AB+CD)\times DL$

$=\frac{1}{2}(75+50)\times 25\sqrt{3}{m}^2$

$=1562.5\times 1.732{\text{m}}^2$

$=2706.25{\text{m}}^2$

Hence,

Area of the remaining portion $=ar\left(ABCD\right)-$ Sum of area of sectors

$=2706.25{\text{m}}^2-962.5{\text{m}}^2$

$=1743.75{\text{m}}^2$.