Question

The perimeter of a rhombus is $96cm$ and obtuse angle of it is ${120}^{\circ }$,the lengths of its smallest diagonal is ____.

Answer 1

Let ABCD be the rhombus with perimeter = $96$
Let $\angle ABC={120}^{\circ }$

Then, $\angle CDA=\angle ABC=120$ (Opposite angles)
and, $\angle C=\angle A=x$

Sum of angles of rhombus = 360
$\angle A+\angle B+\angle C+\angle D=360$
$2x=360-240$
$x={60}^{\circ }$
Now, all sides of rhombus are equal. hence, length of the side = $\frac{96}{4}$= $24$ cm
In $△ABC$
$\cos B=\frac{A{B}^2+B{C}^2-A{C}^2}{2\times AB\times BC}$
$\cos 120=\frac{{24}^2+{24}^2-A{C}^2}{2\times 24\times 24}$
$\frac{-1}{2}=\frac{{24}^2+{24}^2-A{C}^2}{2\times 24\times 24}$
$-{24}^2=2\times {24}^2-A{C}^2$
$AC=24\sqrt{3}$
$AC=41.56$ cm
In $△ABD$
$\angle A={60}^{\circ }$
$\cos A=\frac{A{B}^2+A{D}^2-B{D}^2}{2\times AB\times AD}$
$\cos 60=\frac{{24}^2+{24}^2-B{D}^2}{2\times 24\times 24}$
$\frac{1}{2}=\frac{2\times {24}^2-B{D}^2}{2\times 24\times 24}$
${24}^2=2\times {24}^2-B{D}^2$
$BD=24$ cm