Question

A Rhombus of side of $10cm$ has two angles of ${60}^o$ each. Find the length of diagonals and also find its area.

• A. The length of diagonals $AC=20\sqrt{3}cm$ and $BD=10cm$
  Area of Rhombus $=35\sqrt{3}sq.cm$
• B. The length of diagonals $AC=10\sqrt{3}cm$ and $BD=10cm$
  Area of Rhombus $=50\sqrt{3}sq.cm$
• C. The length of diagonals $AC=16\sqrt{3}cm$ and $BD=10cm$
  Area of Rhombus $=20\sqrt{3}sq.cm$
• D. The length of diagonals $AC=19\sqrt{3}cm$ and $BD=10cm$
  Area of Rhombus $=30\sqrt{3}sq.cm$

Answer 1

The correct option is B The length of diagonals $AC=10\sqrt{3}cm$ and $BD=10cm$

Area of Rhombus $=50\sqrt{3}sq.cm$
Let $ABCD$ be a rhombus of side $10cm$ and $\angle BAD=\angle BCD={60}^o$. Diagonals of parallelogram bisect each other.
So, $AO=OC$ and $BO=OD$
In right triangle $AOB$
$\sin {30}^o=\frac{OB}{AB}$
$\Rightarrow$ $\frac{1}{2}=\frac{OB}{10}$
$\Rightarrow$ $OB=5cm$
$\therefore$ $BD=2\left(OB\right)$
$\Rightarrow$ $BD=2\left(5\right)$
$\Rightarrow$ $BD=10cm$
$\cos {30}^o=\frac{OA}{AB}$
$\Rightarrow$ $\frac{\sqrt{3}}{2}=\frac{OA}{10}$
$\Rightarrow$ $OA=5\sqrt{3}$
$\therefore$ $AC=2\left(OA\right)$
$\Rightarrow$ $AC=2\left(5\sqrt{3}\right)$
$\Rightarrow$ $AC=10\sqrt{3}cm$
So, the length of diagonals $AC=10\sqrt{3}cm$ and $BD=10cm$
Area of Rhombus $=\frac{1}{2}\times AC\times BD$
$=\frac{1}{2}\times 10\sqrt{3}\times 10$
$=50\sqrt{3}$