If the area of the rhombus is ^@336 \space cm^2^@ and one of its | Math Question and Answer

Answer:

$100 \space cm$

Step by Step Explanation:

Let $ABCD$ be the given rhombus with $AC = 14 \space cm$ and BD = $x \space cm$ .

We know, $\begin{aligned} \text { Area of rhombus } & = \dfrac { 1 } { 2 } \times \text { Product of its diagonals } \\ \implies & 336 = \dfrac { 1 } { 2 } \times AC \times BD \\ \implies & 336 = \dfrac { 1 } { 2 } \times 14 \times x \\ \implies & \dfrac { 2 \times 336 } { 14 } = x \\ \implies & 48 = x \\ \end{aligned}$ Thus, the length of diagonal $BD = 48 \space cm.$
Let the diagonals $AC$ and $BD$ bisect at a point $O$ .
We know that the diagonals of a rhombus bisect each other at right angles.
So, $AO = \dfrac { 1 } { 2 } AC$ and $BO = \dfrac { 1 } { 2 } BD.$ $\therefore AO = \dfrac { 1 } { 2 } \times 14 = 7 \space cm \text { and } BO = \dfrac { 1 } { 2 } \times 48 = 24 \space cm.$ Also, $\angle AOB = 90^\circ.$
Using Pythagous' theorem in right $\triangle AOB$ , we have $\begin{aligned} & AB^2 = AO^2 + BO^2 \\ \implies & AB^2 = (7)^2 + (24)^2 \\ \implies & AB^2 = 49 + 576 \\ \implies & AB^2 = 625 \\ \implies & AB = 25 \space cm \\ \end{aligned}$
Now, $\begin{aligned} \text{ Perimeter of rhombus } & = 4 \times \text { Length of side } \\ & = 4 \times AB \\ & = 4 \times 25 \\ & = 100 \space cm \end{aligned}$ Thus, the perimeter of the rhombus is $100 \space cm.$