Question

find the area of Rhombus whose perimeter is 80 m and one of whose diagonal is 24 m

Answer 1

hope it will help u ..if I m wrong then comment..

Answer 2

Answer:

The area of the rhombus is 384 sq. m.

Step-by-step explanation:

Given : Rhombus whose perimeter is 80 m and one of whose diagonal is 24m.

To find : The area of rhombus?

Solution :

Let p and q are the diagonals of rhombus.

Let a be the side of the rhombus.

Perimeter of rhombus $P=4\times side=4a$

Side of the rhombus $a=\frac{\sqrt{p^2+q^2}}{2}$

Area of rhombus is $A=\frac{1}{4}p\times \sqrt{P^2-4p^2}$

Where P is the perimeter P=80

And p is the diagonal = 24 m

Substitute the value in the formula,

$A=\frac{1}{4}p\times \sqrt{P^2-4p^2}$

$A=\frac{1}{4}(24)\times \sqrt{(80)^2-4(24)^2}$

$A=6\times \sqrt{6400-2304}$

$A=6\times \sqrt{4096}$

$A=6\times 64$

$A=384 m^2$

Therefore, The area of the rhombus is 384 sq. m.