Question

1. The length of each side of a rhombus is 10 cm. If
one of its diagonals is 12 cm, find its area.
(answer = 96 cm square) ​

Answer 1

Answer:

96sq.cm

Step-by-step explanation:

Diagonals of a rhombus bisect each other at right angles.

Let ABCD be the rhombus, Diagonal AC=16 cm and side AB=10 cm.

In right △AOB, AB=10 cm, AO=8 cm

By Pythagoras theorem, AB  

2

=AO  

2

+BO  

2

 

⇒10  

2

=8  

2

+BO  

2

 

⇒BO  

2

=100−64=36

∴BO=6cms

Diagonal BD=2×6=12cm

Area of Rhombus=  

2

1

​

×(productofthediagonals)

=  

2

1

​

×16×12=8×12=96sq.cm

Hence length of other diagonal = 12 cm;  

And Area of rhombus = 96sq.cm

Answer 2

We know that the diagonals of a rhombus bisect at right angles

We get AO = OC = ½ AC

We know that AC = 16 cm

AO = OC = ½ (16) = 8 cm

Consider △ AOB

Using the Pythagoras Theorem

AB^2 = AO^2 + OB^2

By substituting values

10^2 = 8^2 + OB^2

On further calculation OB^2 = 100 – 64

By subtraction OB^2 = 36

So we get OB = √36

OB = 6 cm

We know that the length of other diagonal

BD = 2 × OB

By substituting the value BD = 2 × 6 BD = 12 cm

Area of rhombus = 1/2 * d1* d1

So, 12*8

=96cm^2