Question

ABCD Is a field in the shape of a trapezium AB IS PARLLEL to CD and angle ABC =120angle DAB =90 anglr BCD =60 angle CDA =90 four sector s are formed with centres A,B , C , D The radius of each sector is 17.5m find the area of remaining portion of the trapezium when AB=50m and CD = 75 m use root 3=1.732

Answer 1

Here is the answer. It is same as your questions answere.

Answer 2

Answer:

Area of remaining portion = 1744.138 $m^{2}$

Step-by-step explanation:

Radius of each sector (R) = 17.5 m

Area of sector = πr^{2}$\frac{\theta}{360}$

Area of sector I = 160.352 $m^{2}$

Area of sector II = 320.704 $m^{2}$

Area of sector III = 240.528 $m^{2}$

Area of sector IV = 240.528 $m^{2}$

Area of trapezium = $\frac{h(a+b)}{2}$

a = 75m b = 50 m

For h, construct a perpendicular AE, now DE = 50m and EC = 25m

By trignometric ratios, tan(60°) = $\frac{BE}{EC}$

                                               h    = 25√3 = 43.3m (using root 3=1.732)

Area of trapezium =  $\frac{43.3(75+50)}{2}$

                               = 2706.25 $m^{2}$

Area of remaining portion = Aea of trapezium - Area of all 4 sectors

                                           = 2706.25-( 160.352+320.704+240.528+240.528)

                                           = 1744.138 $m^{2}$