Question

A right circular cone in such that the angle of at its vertex is 90º and its base radius is 49 cm. Then find the curved surface area of the cone?

Let slant height be x and ⊥ will bisect vertex angle.

∴ Δ ABC is right angle Δ

∴ x² + x² = (49 × 2)²

x² = 4802

Curved surface area of cone = `square`   ...(Formula)

= `22/7 xx 49 xx  square`

= `22/7 xx  square xx 49sqrt(2)`

= `square` cm²

Answer 1

Let slant height be x and ⊥ will bisect vertex angle.

∴ Δ ABC is right angle Δ

∴ x² + x² = (49 × 2)²

x² = 4802

Curved surface area of cone = \[\boxed{πrl}\]   ...(Formula)

= \[\frac{22}{7} \times 49 \times \boxed{49\sqrt{2}}\]

= \[\frac{22}{7} \times \boxed{49} \times 49\sqrt{2}\]

= \[\boxed{7546\sqrt{2}}\] cm²