Question

if diagonal of a rhombus are 24cm and 32 cm then perimeter of that rhombus

Answer 1

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Answer 2

The perimeter of the given rhombus is 80 cm.

Step-by-step explanation:

We have drawn a rhombus with nomenclature.

ABCD is a rhombus in which AC and BD are diagonals that intersect each other at point O.

Given,

Length of AC = 24 cm.

Length of BD = 32 cm.

We have to find out the perimeter of the rhombus.

Solution,

Since we know that the diagonals of a rhombus perpendicularly bisects each other.

$length\ of\ AO =\frac{length\ of\ AC}{2}=\frac{24}{2}=12\ cm$

Similarly,

$length\ of\ BO =\frac{length\ of\ BD}{2}=\frac{32}{2}=16\ cm$

Now, In right angled ΔABO, we can calculate the length of AB by using Pythagoras Theorem.

"The square of the third side is equal to the sum of the squares of the other two sides".

$AB^2=AO^2+OB^2$

On putting the values, we get;

$AB^2=(12)^2+(16)^2\\\\AB^2=144+256=400$

On taking square root on both side we get;

$\sqrt {AB^2}=\sqrt{400}\\\\AB=20\ cm$

We know that all the four sides of a rhombus are equal in length and perimeter is equal to sum of all sides.

$Perimeter=4\times AB=4\times20=80\ cm$

Hence The perimeter of the given rhombus is 80 cm.