The perimeter of a rhombus is 104 cm and one of its diagonal is 48 cm. find the area of the Rhombus.

Hello aspirant

I can help you with better information.

As perimer formula =2√(d1^2+d2^2)

From this d2 =73.1

From the above formula calculate d2 value

Now area of rhombus=½(d1+d2)

Finally area=60.55

Try to solve again.

Hope helpful to you

All the best

Thank you

Let S be the side of the rhombus.

Therefore perimeter of rhombus will be 4S.

Now 4S = 104.

Therefore, S = 26. One side of rhombus is 26 cms.

Now the diagonals in a rhombus form 4 congruent triangles. Also the diagonals of rhombus are perpendicular, so consider any one triangle and the angle between the diagonals is 90 degrees.

Apply Pythagoras Theorem in the triangle and you will get the following equation :-

(26)^2 = (24)^2 + b^2

[Where b is the half of the length of the other diagonal. This can be done because diagonals of rhombus bisect each other. ] {Since diagonals bisect each other we take 24 by (1/2)*48}

Therefore by solving above equation we get b as 10. Which implies length of other diagonal is 20 cms. (10*2)

Let A be the area of rhombus. Therefore the area of rhombus is

A =  1/2*(product of diagonals)

A = 1/2*(20*48)

Therefore the Area of Rhombus is 480 sq.cm.

In this way we can find area of rhombus.

If you still have any queries feel free to ask in the comment section down below.