SSC CGL 2022 PYQ — Question 67

The correct answer is Option 4 i.e.${{600\pi-450\sqrt3cm^2}}$

Given:

Radius = 30 cm

Concept Used:

Pythagoras theorem, P² + B² = H² 

Explanation:

From the diagram above, OA = O'A = OB = O'B = OO' = 30 cm

Now, In triangle ACO, (AO)² = (AC)² + (OC)² 

AC² = (30)² - (15)²

AC² = 675

AC =  $15 \sqrt{3}$ cm

So, AB = $30 \sqrt{3}$ cm

Also, since OA = O'A = OO' = 30 cm, then triangle AOO' is an equilateral triangle.

So, angle AOO' = 60° and

angle AOB = 120°

Now, Area of AO'BA = Area of sector AO'B - Area of triangle AO'B

⇒ Area of AO'BA = ${120 \over 360} \times \pi(30)^2 - {1\over 2} \times 30 \sqrt{3} \times 15$

⇒ Area of AO'BA = $300\pi$ - $225 \sqrt{3}$

Now, Area of intersecting region = 2 × Area of AO'BA = 2 × (  $300\pi$ - $225 \sqrt{3}$) = $600\pi - 450 \sqrt{3}$ cm² 

Henve the required area of intersecting region is   $600\pi - 450 \sqrt{3}$ cm² .