Question

If the area of a rhombus-shaped box is $6\sqrt{7}{cm}^2$ and its one diagonal is $6cm$. Find the side of the rhombus.

Answer 1

Let the side of the rhombus = a

Area of the triangle containing the diagonal = $\frac{1}{2}$ times area of the rhombus

One of the diagonal = 6 cm

Semi perimeter = $\frac{6+a+a}{2}=a+3$ (1 mark)

Now the area of the triangle containing the diagonal =$\sqrt{s(s-a)(s-b)(s-c)}$

$3\sqrt{7}$ = $\sqrt{(a+3)(a+3-a)(a+3-a)(a+3-6)}$
= $\sqrt{(a+3)(a-3)\left(9\right)}$
= $\sqrt{(a^2-9)\left(9\right)}$ (1 mark)
We take square on both sides and get,

63 = $(a^2-9)\times 9$
$\Rightarrow (a^2-9)$ = 7
$\Rightarrow a^2$ = 7+9 = 16

$\Rightarrow$ a= 4cm ( 1 mark)