Question

Find the area of the rhombus having each side equal to 10 cm and one of its diagonals equal to 12 cm.

Answer 1

ABCD is a rhombus in which AB = BC = CD = DA = 10 cm
AC = 12 cm

AO = OC = 6 cm [Diagonals of a rhombus bisect each other.]

$Therefore,AO=6cm$
$In\Delta AOD$,
$A{D}^2$ =$A{O}^2$ + $O{D}^2$ [Pythgoras theorem]
⇒ ${10}^2$ = $6^2$ +$O{D}^2$
⇒ $100=36+O{D}^2$
⇒ $64=O{D}^2$
⇒ $OD=8$
$Therefore,BD=2OD$
$=2\times 8$
$=16cm$
Area of rhombus = $\frac{1}{2}\left(theproductofdiagonals\right)$
$=\frac{1}{2}\times 16\times 12$
$=96c{m}^2$