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ABCD is a field in the shape of a trapezium. AB||DC and ABC=90,DAB=60. Four sectors are formed withe centres A,B,C and D. The radius of each sector is 17.5 m. Find the area of remaining portion given that AB=75 m and CD=50 m.

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Solution

Let the central angles of 4 sectors be θ1,θ2,θ3,θ4.

Sum of area of sectors =θ1360×πr2+θ2360×πr2+θ3360×πr2+θ4360×πr2

=θ1+θ2+θ3+θ4360×π(17.5)2

But θ1+θ2+θ3+θ4=360 as they are vertex anges of a quadrilateral.
Hence,

Sum of area of sectors =360360×π×306.25

=962.5 m2

Since AB||CD and ABC= 90. Therefore

BCD= 90

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

AL=ABBL
=ABCD
=(7550)
=25 m

In ALD, we have

tan60=DLAL

3=DL25
DL=253 m

ar(ABCD)=12(AB+CD)×DL

=12(75+50)×253m2

=1562.5×1.732 m2
=2706.25 m2

Hence,
Area of the remaining portion =ar(ABCD) Sum of area of sectors

=2706.25 m2962.5 m2

=1743.75 m2.

1029894_1010016_ans_09ccfe53d9ec484580e548dba68e2c48.png

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