IPMAT Indore 2024 - QA (MCQ) Q19 Explanation

Explanation:

(c) Note: ∆AOP will be an equilateral ∆ because ∠POA = ∠PAO = 60°

(∴ AO = PO = radius)

Area of circle = π(k/2)^2 = π⋅k2/4

Area of semi-circle = π/8 k2

Area of one shaded segment = area of sector AOP – are of equilateral ∆AOP

= (60°/360)π(k/2)^2 - √3/4 (k/2)^2

= (k/2)^2 {π/6 - √3/4} = k^2/4{2π - 3√3/12}

∴ Area of 2–shaded segments =2⋅ {k^2/4 (2π - 3√3/12)} = k^24{2π - 3√3}

Area of the portion of the triangle lying inside the circle = Area of semi-circle - area of 2 shaded segments

= k^2/8 = k^2 (2π−3√3 )

= k^2/24 [3π −2π+3√3 ]

= (k^2/24) (3√3 + π)

Hence option (c). Ans.