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.
Open in App
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 =360∘360∘×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=AB−BL
=AB−CD
=(75−50)
=25 m
In △ALD, we have
tan60∘=DLAL
⇒√3=DL25
⇒DL=25√3 m
ar(ABCD)=12(AB+CD)×DL
=12(75+50)×25√3m2
=1562.5×1.732 m2
=2706.25 m2
Hence,
Area of the remaining portion =ar(ABCD)− Sum of area of sectors