Question

Rhombus has a one diagonal double the other if the area of the Rhombus is K the length of its side is what

Answer 1

Let the length of diagonal of rhombus be x and 2x,

Area of rhombus= $\frac{product Of Diagonals}{2}$

Area Of rhombus = $\frac{x * 2x}{2} = \frac{2x^2}{2}$



 K = $x^{2}$


$\sqrt{K} = x$

Length Of Diagonals :-

x = $\sqrt{K}$
2x = $2 \sqrt{K}$
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By Pythagoras Theorem,

$( \frac{ \sqrt{K} }{2}) ^{2} + ( \frac{2 \sqrt{K} }{2}) ^{2} = Side^2$


$( \frac{ \sqrt{K} }{2})^2$ + ( \sqrt{K})^2 = Side^2 [/tex]


$\frac{K}{4} + K =Side^2$


$\frac{K + 4K }{4}$ = $side^2$


$\frac{5 K }{4} = side^2$

$\sqrt{ \frac{5K}{4} } = side$

$\frac{ \sqrt{5K} }{2}$ =side


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Length Of side =$\frac{ \sqrt{5K} }{2}$
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i hope this will help you



-by ABHAY