Perimeter of a rhombus is 2p units and the su | vedclass

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(C) Let the side of the rhombus be $a$. Perimeter $= 4a = 2p$,so $a = \frac{p}{2}$.Let the diagonals be $d_1$ and $d_2$. Given $d_1 + d_2 = m$.In a rh

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$\frac{1}{4}(m^{2}-p^{2})$ sq units

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(C) Let the side of the rhombus be $a$. Perimeter $= 4a = 2p$,so $a = \frac{p}{2}$.Let the diagonals be $d_1$ and $d_2$. Given $d_1 + d_2 = m$.In a rhombus,diagonals bisect each other at $90^{\circ}$. Let half-diagonals be $x = \frac{d_1}{2}$ and $y = \frac{d_2}{2}$.Then $2x + 2y = m$,so $x + y = \frac{m}{2}$.By Pythagoras theorem in the right-angled triangle formed by the sides and half-diagonals: $x^2 + y^2 = a^2 = (\frac{p}{2})^2 = \frac{p^2}{4}$.We know $(x + y)^2 = x^2 + y^2 + 2xy$.Substituting the values: $(\frac{m}{2})^2 = \frac{p^2}{4} + 2xy$.$\frac{m^2}{4} = \frac{p^2}{4} + 2xy \Rightarrow 2xy = \frac{m^2 - p^2}{4}$.The area of the rhombus is $\frac{1}{2} 	imes d_1 	imes d_2 = \frac{1}{2} 	imes (2x) 	imes (2y) = 2xy$.Therefore,Area $= \frac{1}{4}(m^2 - p^2)$ sq units.

Perimeter of a rhombus is $2p$ units and the sum of the lengths of its diagonals is $m$ units,then the area of the rhombus is

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