Question

The perimeter of a rhombus is equal to $48$, and the sum of the lengths of the diagonals is equal to $26$. Find the area of the rhombus.

Answer 1

Given that,

The perimeter of a rhombus is $48$ and the sum of the lengths of the diagonals is $26$.

To find out,

The area of the rhombus.

We know that, all four sides in a rhombus are equal.

Let $x$ be the length of each side.

Also, perimeter of a rhombus with side $a=4a$

Hence, $4x=48$

$\Rightarrow x=\frac{48}{4}$

$=12$

Now, let $d_1,d_2$ be the diagonals

The diagonals of a rhombus are perpendicular bisectors of each other.

Hence, there will be four identical right-angled triangles due to the two diagonals and four sides.

Applying the Pythagoras theorem in one of those triangles, we will get:

${\left(\frac{d_1}{2}\right)}^2+{\left(\frac{d_2}{2}\right)}^2=x^2$

$\frac{(d_1{)}^2}{4}+\frac{(d_2{)}^2}{4}=x^2$

${d_1}^2+{d_2}^2=4x^2$

$x=12$

$\therefore {d_1}^2+{d_2}^2=4(12{)}^2$

$=576$

Also, given $d_1+d_2=26$

We know that, $(a+b{)}^2=a^2+2ab+b^2$

Hence, $ab=\frac{(a+b{)}^2-(a^2+b^2)}{2}$

So, $d_1{d}_2=\frac{(d_1+d_2{)}^2-({d_1}^2+{d_2}^2)}{2}$

$\Rightarrow d_1{d}_2=\frac{(26{)}^2-(576)}{2}[d_1+d_2=26and{d_1}^2+{d_2}^2=576]$

$\Rightarrow d_1{d}_2=\frac{676-576}{2}$

$\Rightarrow d_1{d}_2=\frac{100}{2}$

$\therefore d_1{d}_2=50$

Now, we know that, area of a rhombus with diagonals $d_1$ and $d_2=\frac{1}{2}{d}_1{d}_2$

Hence, area of the given rhombus $=\frac{1}{2}\times 50[d_1{d}_2=50]$

$=25\text{Square units}$

Hence, the area of the rhombus is $25\text{Square units}$.