Question

ABCD is a rhombus, O is the intersecting point of diagonal OD = 4 cm and OA = 3 cm, find the perimeter of ABCD.

Answer 1

In AOD triangle angle AOD=90degree, so AO²+OD²=AD², So AD=✓(OA²+OD²)=✓(3²+4²)=5
. Now perimeter=4×side=4×5=20

Answer 2

Given,

ABCD is a rhombus.

O is the intersecting point of diagonals.

OD = 4 cm, OA = 3 cm.

To find,

The perimeter of ABCD.

Solution,

The perimeter of the rhombus, ABCD, will be 20 cm.

We can easily solve this problem by following the given steps.

Now, we know that the diagonals of a rhombus intersect each other at 90°.

Here, the diagonals AC and BD are intersecting at O.

Therefore, ∆ AOD is a right-angled triangle.

Using the Pythagoras theorem in ∆AOD,

AD² = OA² + OD²

AD² = (3)² + (4)²

AD² = 9+16

AD² = 25

AD = √25

AD = 5 cm

We know that the perimeter of the rhombus is 4×side because all the sides of the rhombus are equal.

The perimeter of the rhombus, ABCD = 4×side

The perimeter of the rhombus, ABCD = 4 × AD

The perimeter of the rhombus, ABCD = 4 × 5

The perimeter of the rhombus, ABCD = 20 cm

Hence, the perimeter of the rhombus, ABCD, is 20 cm.