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)
Answers
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
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