Question

Abcd is a field in the shape of a trapezium ad parallel to bc angle abc=90

Answer 1

YOUR QUESTION IS INCOMPLETE..

I THINK THIS IS UR 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.

HERE IS UR ANSWER:

Radius of each sector (R) = 17.5 m

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

360

θ

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

2

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

2

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

2

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

2

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

2

h(a+b)

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}

EC

BE

h = 25√3 = 43.3m (using root 3=1.732)

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

2

43.3(75+50)

= 2706.25 m^{2}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}m

2