A field is in the shape of a parallelogram. I | vedclass

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(B) Let the sides of the parallelogram be $a = 65 \, m$,$b = 119 \, m$,and the diagonal be $d = 156 \, m$.The area of the parallelogram is twice the a

Vedclass · Accepted Answer

$71400$

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(B) Let the sides of the parallelogram be $a = 65 \, m$,$b = 119 \, m$,and the diagonal be $d = 156 \, m$.The area of the parallelogram is twice the area of the triangle formed by these three sides.First,calculate the semi-perimeter $s$ of the triangle:$s = \frac{a + b + d}{2} = \frac{65 + 119 + 156}{2} = \frac{340}{2} = 170 \, m$.Using Heron's formula,the area of the triangle is:$	ext{Area} = \sqrt{s(s-a)(s-b)(s-d)}$$= \sqrt{170(170-65)(170-119)(170-156)}$$= \sqrt{170 	imes 105 	imes 51 	imes 14}$$= \sqrt{12751200} = 3570 \, m^2$.Area of the parallelogram $= 2 	imes 3570 = 7140 \, m^2$.Cost of gravelling $= 	ext{Area} 	imes 	ext{Rate} = 7140 	imes 10 = ₹ 71400$.

$A$ field is in the shape of a parallelogram. Its adjacent sides and one diagonal are $65 \, m$,$119 \, m$,and $156 \, m$ respectively. Find the cost of gravelling it at the rate of $₹ 10$ per $m^2$ (in $Rs.$).

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