Question

A rhombus has sides 10 cm long and an angle of 60°. Find the diagonals of the rhombus.

Answer 1

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Answer 2

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

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

$\angle A = 60\degree$

$i ) In \: \triangle {AOD}$

$\angle {DAO} = 30 \degree$

$sin \: 30\degree = \frac{OD}{AD}$

$\implies \frac{1}{2} = \frac{OD}{10}$

$\implies \frac{10}{2} = OD$

$\implies OD = 5 \:cm$

$\therefore Diagonal \:BD = 2 \times OD$

$= 2 \times 5$

$= 10 \: cm$

$ii) In \: \triangle AOD$

/* By Phythagorean theorem */

$AO^{2} + OD^{2} = AD^{2}$

$\implies AO^{2} = AD^{2} - OD^{2}$

$= 10^{2} - 5^{2}$

$= 100 - 25$

$= 75$

$AO = \sqrt{ 75 }$

$\implies AO = 5\sqrt{3}$

$Diagonal \: (AC ) = 2 \times AO$

$= 2 \times 5\sqrt{3}$

$= 10\sqrt{3}$

$= 10 \times 1.732$

$= 17.32 \:cm$

Therefore.,

$\green { Diagonal \: AC = 17.32\:cm }$

$\green { Diagonal \: BD = 5\:cm }$

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