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(A) Let $ABCD$ be the rhombus-shaped field.Perimeter of the rhombus $= 400\, m$.Since all sides of a rhombus are equal,each side $= 400\, m / 4 = 100\

Vedclass · Accepted Answer

$4800$

Vedclass · Answer

(A) Let $ABCD$ be the rhombus-shaped field.Perimeter of the rhombus $= 400\, m$.Since all sides of a rhombus are equal,each side $= 400\, m / 4 = 100\, m$.Thus,$AB = BC = CD = DA = 100\, m$.Let the diagonal $BD = 160\, m$.The diagonal divides the rhombus into two congruent triangles,$	riangle ABD$ and $	riangle BCD$.For $	riangle ABD$,the sides are $a = 100\, m$,$b = 100\, m$,and $c = 160\, m$.The semi-perimeter $s$ is given by:$s = (100 + 100 + 160) / 2 = 360 / 2 = 180\, m$.Using Heron's formula,the area of $	riangle ABD$ is:Area $= \sqrt{s(s-a)(s-b)(s-c)}$Area $= \sqrt{180(180-100)(180-100)(180-160)}$Area $= \sqrt{180 	imes 80 	imes 80 	imes 20}$Area $= \sqrt{180 	imes 20 	imes 80^2}$Area $= \sqrt{3600 	imes 6400}$Area $= 60 	imes 80 = 4800\, m^2$.Since the land is divided into two equal parts,each of them will get an area of $4800\, m^2$.

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