Question

One diagonal of a rhombus is half the other .If the length of the side of the rhombus is 10 cm, what is the area of the rhombus

Answer 1

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Answer 2

$In \: ABCD \: is \: a\: Rhombus .$

$AB = BC = CD = DA = 10 \: cm$

$AC \: and \: BD \: are \: two \: diagonals$

$Let \: AC = 4x \: cm$

$BD = \frac{AC}{2} = \frac{4x}{2} = 2x$

$In \: \trangle AOD , \: \angle {AOD} = 90\degree$

$\blue { ( By \: Pythagoras\: Theorem ) }$

$\pink { AD^{2} = OA^{2} + OD^{2} }$

$\implies 10^{2} = (2x)^{2} + x^{2}$

$\implies 100 = 4x^{2} + x^{2}$

$\implies 5x^{2} = 100$

$\implies x^{2} = \frac{100}{5}$

$\implies x^{2} = 20$

$\implies x = \sqrt{ 2^{2} \times 5 }$

$\implies x = 2\sqrt{5} \:cm$

$Now, BD = 2x \\= 2 \times 2\sqrt{5}\\ = 4\sqrt{5}\:cm$

$AC = 4x \\= 4 \times 2\sqrt{5} \\= 8\sqrt{5} \:cm$

$\red{Area \:of \: the \: Rhombus} \\= \frac{1}{2} \times AC \times BD \\= \frac{1}{2} \times 8\sqrt{5} \times 4\sqrt{5} \\= 16\times 5 \\= 80 \:cm^{2}$

Therefore.,

$\red{Area \:of \: the \: Rhombus} \green {=80 \:cm^{2} }$

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