Question : If the perimeter of a rhombus is 80 cm and one of its diagonals is 24 cm, then what is the area (in cm2) of the rhombus?
Option 1: 218
Option 2: 192
Option 3: 384
Option 4: 768
Correct Answer: 384
Solution :
Given: Perimeter of rhombus = 80 cm
Let AB, BD, DC, and CA be the sides, BC($d_1$) and AD($d_2$) be the diagonals of a rhombus meeting at O.
Side AB = $\frac{80}{4}$ = 20 cm
Diagonal ($d_1$) BC = 24 cm
OB = $\frac{1}{2}×24 = 12$ cm
We know that the diagonals of a rhombus bisect each other at 90$^\circ$.
Using the Pythagoras theorem,
$AB^2 = OA^2 + OB^2$
⇒ $20^2 = x^2 + 12^2$
⇒ $x^2 = 400–144$
⇒ $x^2 = 256$
⇒ $x = 16$
So, AD ($d_2$) = 16 × 2 = 32 cm
Now, area of the rhombus = $\frac{1}{2}×d_1×d_2$
= $\frac{1}{2} ×24×32$ = $384$ cm$^2$
Hence, the correct answer is 384.
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