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 17.5m. Find the

i). Total area of the four Sectors.

ii). Area of remaining portion given that AB=75 m and CD =50m.

Answer 1

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

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

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

i) 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.

$=\frac{60}{360}\pi (17.5{)}^2+\frac{90}{360}\pi (17.5{)}^2+\frac{90}{360}\pi (17.5{)}^2+\frac{120}{360}\pi (17.5{)}^2$
$=\frac{1}{6}+\frac{1}{4}+\frac{1}{4}+\frac{1}{3}\pi (17.5{)}^2$
$=962.5m^2$

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

AL= AB-BL=AB-CD =(75-50)m = 25m

In $\Delta ALD$, we have

$tan{60}^{\circ }=\frac{DL}{AL}\Rightarrow \sqrt{3}=\frac{DL}{25}So,DL=25\sqrt{3}m$

Therefore area of trapezium ABCD =$\frac{1}{2}$ (AB+CD)DL

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

$=2706.25m^2$

Hence,

Area of the remaining portion = Area of trapezium ABCD - Area of 4 sectors

= 2706.25 – 962.5 = 1743.75 $m^2$