Multiple choice

A sector containing an angle of 120o is cut off from a circle of radius 21 cm. and folded in to a cone. Find the curved surface area of the cone. 21 सेमी. त्रिज्या के 120o कोण का एक त्रिज्यखण्ड काटा गया और एक शंकु के रूप में मोड़ा गया। तो शंकु का तिर्यक पृष्ठ क्षेत्रफल ज्ञात कीजिए।

  1. 462 cm2./सेमी2

  2. 426 cm2./सेमी2

  3. 246 cm2./सेमी2

  4. 642 cm2./सेमी2
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A Correct answer
Explanation

When a sector of angle 120° is folded into a cone, the radius of the sector becomes the slant height (l = 21 cm), and the arc length becomes the circumference of the base. Arc length = (120°/360°) × 2π × 21 = 14π cm. This equals 2πr (base circumference), so r = 7 cm. Curved surface area = πrl = π × 7 × 21 = 462 cm².

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