Question

Find the area of a rhombus having each side equal to 15 cm and one of whose diagonals is 24 cm​

Answer 1

$\huge\bold\red{Answer}$

Let ABCD be the rhombus where diagonals intersect at O.

Then, AB = 15 cm and AC = 24 cm.

The diagonals of a rhombus bisect each other at right angles.

Therefore, Δ AOB is a right-angled triangle, right angled at O such that

OA = $\frac{1}{2}$ AC  = 12 cm and AB  = 15 cm.

By Pythagoras theorem, we have,

(AB)² = (OA)² + (OB)²

⇒ (15)² = (12)² + (OB)²

⇒ (OB)² = (15)² − (12)²

⇒ (OB)² = 225 − 144 = 81

⇒ (OB)² = (9)²

⇒ OB = 9 cm

∴ BD = 2 x OB = 2 x 9 cm = 18 cm

Hence,

Area of the rhombus ABCD = ($\frac{1}{2}$ × AC × BD)=($\frac{1}{2}$ × 24 × 18) = 216 cm²

$<font color="red">$$<b >$$<marquee>$ Hope it helps

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