There are different ways of doing it, one method is as follows…..
Rhombus ABCD, AB = 10 cm, So BC = 10 cm, as all sides of rhombus are equal.
Diagonal AC = 12 cm
So, area of triangle ABC by Heron's formula
= √{ s(s - a)(s - b)(s - c)}, s is semi perimeter & a, b, c are sides of the triangle
= s = 32/2 = 16
Area = √{16(16 – 10)(16 – 10)(16 – 12)
⇒ area = √(16 × 6 × 6 × 4)
⇒ area = √(22 × 22 × 22 × 32 × 22)
⇒ area = 2 × 2 × 2 × 3 × 2 = 48 cm2
⇒ area (Rhombus ABCD) = 48 × 2 = 96 cm2
Alternate Method
Given:
The side of a rhombus is 10 cm and the smaller diagonal is 12 cm
Concept use:
• All sides of a rhombus are equal
• Diagonals of a rhombus bisect each other at a right angle (90°)
• Area of a rhombus = 1/2 × (product of two diagonals)
Calculation:
Let ABCD be the rhombus, where AC and BD be the diagonals intersecting at point O
Here, AC = 12, then OC = 6 and BC = 10
Now, ΔBOC is a right-angle triangle where right angle at O
So, BO = √(102 - 62) = 8
So, BD = 8 × 2 = 16 cm
Now, area of the rhombus is 1/2 × 16 × 12 = 96 cm2
∴ The area of the rhombus is 96 cm2