Question : Two identical circles each of radius 30 cm intersect each other such that the circumference of each one passes through the centre of the other. What is the area of the intersecting region?
Option 1: $400 \pi-250 \sqrt{3} \mathrm{~cm}^2$
Option 2: $300 \pi-150 \sqrt{3} \mathrm{~cm}^2$
Option 3: $500 \pi-350 \sqrt{3} \mathrm{~cm}^2$
Option 4: $600 \pi-450 \sqrt{3} \mathrm{~cm}^2$
Correct Answer: $600 \pi-450 \sqrt{3} \mathrm{~cm}^2$
Solution :
OA = O’A = OB = OB’ = OO’ = 30 cm
In triangle AOO’,
OA = O’A = OO’ = 30 cm ( As all three sides of AOO' are equal, so it is an equilateral triangle)
⇒ $∠AOO’ = 60°$
⇒ $∠AOB = 120°$
In triangle ACO,
$AC^2 = OA^2 - OC^2 =30^2-15^2= 900 - 225$
⇒ $AC^2 = 675$
⇒ $AC = 15\sqrt3$
So, $AB = 30\sqrt3$
Area of AO’BA = Area of sector AO’BO $-$ Area of triangle AOB =
$=\frac{\pi(30)^2}{3}-\frac{1}{2}\times 30\sqrt3\times 15$
$=\frac{900π}{3}- 225\sqrt3$
$=300π - 225\sqrt3$
$\therefore$ Area of intersecting region = 2(Area of AO’BA)
$=600π - 450\sqrt3$ cm
2
Hence, the correct answer is $600π - 450\sqrt3$ cm
2
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