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Question

ABCD is a field in the shape of a trapezium. AB||DC and ABC=90,DAB=60. 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.


Solution

Since AB||CD and ABC=90. Therefore BCD=90.

Also, BAD=60

CDA=18060=120

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.

=60360π(17.5)2+90360π(17.5)2+90360π(17.5)2+120360π(17.5)2
=16+14+14+13π(17.5)2
=962.5m2

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

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

In ΔALD, we have

tan 60=DLAL3=DL25So,DL=253m

Therefore area of trapezium ABCD =12 (AB+CD)DL

=12(75+50)×253

=2706.25m2

Hence,

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

= 2706.25 – 962.5 = 1743.75 m2

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