Question : If the perimeter of a rhombus is 40 cm and one of its diagonals is 16 cm, what is the area (in cm2) of the rhombus?
Option 1: 72
Option 2: 48
Option 3: 96
Option 4: 192
Correct Answer: 96
Solution :
The perimeter of a rhombus is the sum of all its sides.
The side of rhombus $=\frac{40}{4} = 10\;\mathrm{cm}$
AB = BC = CD = DA = 10 cm
AC =16 cm
OA = OC = 8 cm
$\mathrm{OB} = \sqrt{(\mathrm{AB})^2 - (\mathrm{OA})^2} = \sqrt{(10)^2 - (8)^2} = \sqrt{36} = 6\;\mathrm{cm}$
⇒ BD = 6 × 2 = 12 cm
The area of a rhombus $=\frac{1}{2} \times (\mathrm{AC×BD})$
So, the area of the given rhombus $=\frac{1}{2} \times 16 \times 12 = 96\;\mathrm{cm^2}$
Hence, the correct answer is 96.
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