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Foundation program for Bank

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Maths / Mensuration / Perimeter of Rhombus
(A Brief Glimpse of ABHYAS Content - Have aLook !!!!)

## Perimeter of Rhombus

### Perimeter of A Rhombus:

First, all four sides of a rhombus are equal, meaning that if we find one side, we can simply multiply the side by four to find the perimeter.

Second, the diagonals of a rhombus are perpendicular bisectors of each other, thus giving us four right triangles and splitting each diagonal in half. Using Pythagorean Theorem on any one of them will give us the length of our sides.

Illustration: If the area of rhombus be 24 $cm^{2}$  and one of its diagonals be 4 cm, find the perimeter of rhombus.

Solution :Let  ABCD be a rhombus such that its one diagonals AC = 4 cm. Support the diagonals AC and BD intersect at O. We have, Area of rhombus ABCD = 24 $cm^{2}$

$\dpi{120} \fn_jvn \large \Rightarrow$ $\Rightarrow \frac{1}{2}\times 4\times BD= 24$

Hence BD = 12cm. Thus,  we have AC = 4 cm and  BD = 12 cm

$\dpi{120} \fn_jvn \large \therefore$$OA=\frac{1}{2}AC=2 cm$      and $OB=\frac{1}{2}BD=6 cm$

Since the diagonals of a rhombus bisect each other at right angle. Therefore, $\dpi{120} \fn_jvn \large \Delta$ OAB  is right triangle, right angled at O. Using Pythagoras theorem in $\dpi{120} \fn_jvn \large \Delta$ AOB, we have

$AB^{2}$ = $OA^{2}$  +  $OB^{2}$

$\dpi{120} \fn_jvn \large \Rightarrow$ $AB^{2}$  =  $2^{2}$  +  $6^{2}$ = 4 + 36   = 40

$\Rightarrow AB =\sqrt{40}cm =2\sqrt{10}cm$

Hence, Perimeter of the rhombus ABCD = $4\times 2\sqrt{10}$ cm =$8\sqrt{10} cm$

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