It is proposed to add to a square lawn with each side 58 m two circular ends the centre of each circle being the point of intersection of the diagonals of the square the area of the whole lawn is

Foundation

Mathematics Foundation

Area of Triangle

Area Related to Circle

Mensuration

Question

It is proposed to add to a square lawn with each side 58 m two circular ends the centre of each circle being the point of intersection of the diagonals of the square the area of the whole lawn is Question Image

415

m^{2}

4200

m^{2}

4319.5

m^{2}

4215.5

m^{2}

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Let ABCD be a square with side $58$ metres
Let its diagonals intersect at O

Then $O A = O B = O C = O D$ and

$∠ A O D = ∠ B O C = 9 0^{\circ}$

With radius $= \frac{1}{2} A C arcs have been drawn namely A E D a n d C B F$

$∴ Radius of each arc = \frac{1}{2} \times d i a g o n a l$

$= \frac{1}{2} \sqrt{2} \times 58$

$= 1.41 \times 29 = 40.89 m e t r e s .$

$58 \times 58 + 2 \times [\frac{22}{7} \times {(40.89)}^{2} \times \frac{90}{360} - \frac{1}{2} \times {(40.89)}^{2} 9 0^{\circ}]$

$33.64 + \frac{11 \times {(40.89)}^{2}}{7} - {(40.89)}^{2}$

$= (3364 + 2627.4 - 1671.9)$ sq metres

$= 4319.5$ sq metres