Question

if this side of rhombus is 10 cm and one diagonal is 16 cm
then find the area of Rhombus

Answer 1

I hope it will helps you.

Answer 2

$In \: ABCD \: Rhombus \: AB = 10\:cm \: and \\diagonal (AC) = 16 \:cm$

$AO = \frac{AC}{2} \\= \frac{16}{2} \\= 8 \:cm$

$\blue {( In \: a \: Rhombus \: diagonals }\\blue { bisects \: perpendicularly )}$

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

$\underline { \orange { By \: Pythagoras \:Theorem :}}$

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

$\implies 10^{2} = 8^{2} + OD^{2}$

$\implies 100 = 64 + OD^{2}$

$\implies 100 - 64 = OD^{2}$

$\implies 36 = OD^{2}$

$\implies 6^{2} = OD^{2}$

$\implies OD = 6 \:cm$

$Now, Diagonal (BD )= 2\times OD \\= 2 \times 6 \\= 12 \:cm$

$Area \: of \: a \: Rhombus = \frac{1}{2} d_{1} d_{2}$

$= \frac{1}{2} \times AC \times BD \\= \frac{1}{2} \times 16 \times 12 \\= 8 \times 12 \\= 96 \:cm^{2}$

Therefore.,

$\red {Area \: of \: a \: Rhombus}\green {=96 \:cm^{2} }$

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