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 with centers A, B, C, and D. The radius of each sector is 17.5 m. Find the total area of the four sectors

Answer 1

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

$\angle$BAD = ${60}^{\circ }$

$\therefore \angle CDA={180}^{\circ }-{60}^{\circ }={120}^{\circ }$

We have,
Total area of the four sectors

$=$ Area of sector at A + Area of sector at B + Area of sector at C + Area of sector at D.

$=\frac{60}{360}\times \pi \times (17.5{)}^2+\frac{90}{360}\times \pi \times (17.5{)}^2+\frac{120}{360}\times \pi \times (17.5{)}^2$......$(Areaofsector=\frac{\theta }{360}\times \pi r^2)$

$=\{(\frac{1}{6}+\frac{1}{4}+\frac{1}{4}+\frac{1}{3})\times \pi \times (17.5{)}^2\}{m}^2$

$=\pi \times {\left(\frac{35}{2}\right)}^2{m}^2=\frac{22}{7}\times \frac{35}{2}\times \frac{35}{2}{m}^2=962.5m^2$