Question

In the fig. $ABCD$ is a trapezium with $AB||DC$ and $\angle BCD={60}^{\circ }$.If $BFEC$ is a sector of a circle with the center $C$ and $AB=BC=7cm$ and $DE=4cm$ then find the area of the shaded region (use $\pi =22/7$ and $\sqrt{3}=1.732$ )

Answer 1

Given: $CE=CB=7cm$

$\therefore$ $CD=CE+ED=(7+4)cm=11cm$

In $\Delta$ CLB, we have

$sin{60}^{\circ }=\frac{BL}{BC}\Rightarrow \frac{\sqrt{3}}{2}=\frac{BL}{7}\Rightarrow BL=\frac{7\sqrt{3}}{2}cm$

$\therefore$ Area of trapezium = $\frac{1}{2}(AB+CD)\times BL=\frac{1}{2}(7+11)\times \frac{7\sqrt{3}}{2}c{m}^2=\frac{63\sqrt{3}}{2}c{m}^2$

Area of sector BFEC = $\frac{{60}^{\circ }}{{360}^{\circ }}\times \frac{22}{7}\times 7^{2}c{m}^2=\frac{77}{3}c{m}^2$.......(Area of sector$=\frac{\theta }{360}\times \pi r^2$)

$\therefore$ Required area = $(\frac{63\sqrt{3}}{2}-\frac{77}{3})c{m}^2=(54.558-25.666)c{m}^2=28.89c{m}^2$