Question

The area of rhombus is 96cm square . If its one diagonal is 12 cm , find the length of its side.

Answer 1

Let the diagonals of the Rhombus are d1 and d2

respectively.

Then we know that area of Rhombus is

$\frac{1}{2} \times d_1 \times d_2$

Here, d1 is equals to 12cm and area of rhombus is equals to 96 sq. cm . And have to find value of d2 .

As,

$\frac{1}{2} \times 12 \times d_2 = 96 \\ \implies6d_2 = 96 \\ \implies \: d_2 = \frac{96}{6} \\ \implies d_2 = 16cm$

So,

length of another diagonal will be 16 cm.

As we know that

side of a triangle using its diagonal is

$side = \dfrac{ \sqrt{ {d_1 }^{2} + {d_2}^{2} } }{2} \\ \\ = \dfrac{ \sqrt{ {12}^{2} + {16}^{2} } }{2} \\ \\ = \dfrac{ \sqrt{144 + 256} }{2} \\ \\ = \dfrac{ \sqrt{400} }{2} \\ \\ = \dfrac{20}{2} \\ \\ = 10cm$

So, side of the Rhombus will be 10 cm.

Answer 2

Solution:-

• given:-

1)The area of rhombus is 96cm square .

2) one diagonal is 12 cm.

we know all side of rhombus are equal.

let, ac = 1st diagonal = 12 cm and

bd = 2cd diagonal ,

A = Area = 96cm.

=> Area of rhombus = [ac×bd]/2

=> A = [ac×bd]/2

=> 96 = [ 12 × bd ]/2

=> 96×2 = 12 × bd

=> 192 = 12 × bd

=> bd = 192/12

=> bd = 16 cm

•we know,

ac = ao + oc and bd = bo + od

• diagonals of a rhombus bisect each other at an angle of 90° or right angle.

means ao = oc = 12/2 = 6 cm &

bo = od = 16/2 = 8 cm.

in Δ aod

by Pythagoras theorem

=> (ad)² = (ao)² + (od) ²

=> (ad)² = (6)² + (8) ²

=> (ad)² = 36 + 64

=> (ad)² = 100

=> ad = √100

=> ad = 10 cm

Hence all side of rhombus are

10 cm.

i hope it helps you .