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(A) Let the sides of the parallelogram be $a = 60 \, m$ and $b = 40 \, m$,and the diagonal be $d = 80 \, m$.The area of a parallelogram can be calcula

Vedclass · Accepted Answer

$600 \sqrt{15} \, m^2$

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(A) Let the sides of the parallelogram be $a = 60 \, m$ and $b = 40 \, m$,and the diagonal be $d = 80 \, m$.The area of a parallelogram can be calculated as twice the area of the triangle formed by two sides and the diagonal.First,calculate the semi-perimeter $(s)$ of the triangle:$s = \frac{a + b + d}{2} = \frac{60 + 40 + 80}{2} = \frac{180}{2} = 90 \, m$.Using Heron's formula,the area of the triangle is:$	ext{Area} = \sqrt{s(s-a)(s-b)(s-d)}$$= \sqrt{90(90-60)(90-40)(90-80)}$$= \sqrt{90 	imes 30 	imes 50 	imes 10}$$= \sqrt{1350000} = 300 \sqrt{15} \, m^2$.The area of the parallelogram is twice the area of the triangle:$	ext{Area of parallelogram} = 2 	imes 300 \sqrt{15} = 600 \sqrt{15} \, m^2$.

The two adjacent sides of a parallelogram are $60 \, m$ and $40 \, m$ and one of the diagonals is $80 \, m$ long. The area of the parallelogram is

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