Question

The lengths of one side of a rhombus and one of the two diagonals are 6 cm each. Find the area of the rhombus (in $cm^2$).

• A. $27√3$
• B. $18$
• C. $9√3$
• D. $18√3$

Answer 1

Answer:

$18√3$

Diagonals of a Rhombus are perpendicular bisector

Let ABCD be a rhombus and AC = 6 cm with midpoint O and Side AB = 6 cm

So, in ΔAOB,

⇒ AO² + OB² = AB²

⇒ (3)² + OB² = 6²

⇒ 9 + OB² = 36

⇒ OB² = 27

⇒ OB = 3√3 cm

⇒ BD = 2 $\times$ OB = 6√3 cm

⇒ Area of Rhombus = $\frac{1}{2}\times$(Product of diagonal of Rhombus) $=\frac{1}{2}[d1\times{d2}$]

⇒ $\frac{1}{2}\times{(6\times{6\sqrt{3}})}$ = $8\sqrt{3}$ cm²