Question

Each side of a rhombus is $10$ cm long and one of its diagonals measures $16$ cm. Find the length of the other diagonal and hence find the area of the rhombus.

Answer 1

Diagonals of a rhombus bisect each other at right angles.

Let $ABCD$ be the rhombus, Diagonal $AC=16$ cm and side $AB=10$ cm.

In right $△AOB$, $AB=10$ cm, $AO=8$ cm

By Pythagoras theorem, ${AB}^2={AO}^2+{BO}^2$

$\Rightarrow {10}^2=8^2+{BO}^2$

$\Rightarrow {BO}^2=100-64=36$

$\therefore BO=6$cms

Diagonal $BD=2\times 6=12$cm

Area of Rhombus$=\frac{1}{2}\times \left(productofthediagonals\right)$

$=\frac{1}{2}\times 16\times 12=8\times 12=96$sq.cm

Hence length of other diagonal $=12$ cm;

And Area of rhombus $=96$sq.cm