Question

# 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)2But θ1+θ2+θ3+θ4=360∘ as they are vertex anges of a quadrilateral.Hence,Sum of area of sectors =360∘360∘×π×306.25 =962.5 m2Since AB||CD and ∠ ABC= 90∘. Therefore ∠BCD= 90∘Let DL be perpendicular drawn from D on AB. Then,AL=AB−BL =AB−CD =(75−50) =25 mIn △ALD, we havetan60∘=DLAL⇒√3=DL25⇒DL=25√3 mar(ABCD)=12(AB+CD)×DL =12(75+50)×25√3m2 =1562.5×1.732 m2 =2706.25 m2Hence,Area of the remaining portion =ar(ABCD)− Sum of area of sectors =2706.25 m2−962.5 m2 =1743.75 m2.

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