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.

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Solution

## Since AB||CD and ∠ABC=90∘. Therefore ∠BCD=90∘. Also, ∠BAD=60∘ ∠CDA=180∘−60∘=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∘=DLAL⇒√3=DL25So,DL=25√3m Therefore area of trapezium ABCD =12 (AB+CD)DL =12(75+50)×25√3 =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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