Question

It is proposed to add to a square lawn measuring 58 m on a side, two circular ends.The center of each circle being the point of intersection of the diagonals of the square.Find the area of the whole lawn.

Answer:

The area of the lawn is 4322.74 square meters.

Let ABCD be the square lawn and AED and BFC be the circular ends of the lawn.
AB = BC = CD = DA = 58 meters as ABCD is a square.
By applying Pythagoras Theorem in triangle BCD, we get:
(BD)² = (BC)² + (DC)² which implies that (BD)² = (58)² + (58)² which further implies that (BD)² = 2*(58)² and so we get BD equal to 58√2
As ABCD is a square, BD = 2CO = 2BO.
Therefore, OA = OC = BO = CO = BC/2 = 29√2 = Radius of the circular ends AED and BFC
angle AOD = angle BOC = 90° as diagonals of a square are perpendicular bisectors of each other.
Area of segment AED:
= Area of sector OAD - Area of triangle OAD
= πr²*θ/360 - OA*OD/2
= π*29*29*2*90/360 - 29*29*2/2
= 841(2π/4 - 1)
= 841(π/2 - 1)
= 841(π-2)/2 square meters
Similarly, we get area of segment BFC as 841(π-2)/2 square meters

Area of the entire lawn:
= Area of square lawn + Area of the two circular ends
= 58*58 + 2[841(π-2)/2]
= 29*29*4 + 841(π-2)
= 841(4+π-2)
= 841(2+π)
= 841(2+3.14)
= 841(5.14)
= 4322.74 square meters.


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