The height of a right circular cone is $63 ext{ cm}$ and the curved surface area is five times the area of its base. What is the volume of the cone in $ ext{cm}^3$? (Use $\pi = \frac{22}{7}$)

Question

The height of a right circular cone is $63 	ext{ cm}$ and the curved surface area is five times the area of its base. What is the volume of the cone in $	ext{cm}^3$? (Use $\pi = \frac{22}{7}$)

Accepted Answer

**Solution:**

1. Given:
   - Height of cone, $h = 63 	ext{ cm}$
   - Curved Surface Area (CSA) is five times the Base Area (BA)

2. Formulas:
   - Curved Surface Area, $	ext{CSA} = \pi r l$
   - Base Area, $	ext{BA} = \pi r^2$
   - Volume, $V = \frac{1}{3} \pi r^2 h$

3. Using the relation between CSA and BA:
   $$\pi r l = 5 	imes \pi r^2 \implies l = 5r$$

4. Using the Pythagorean theorem for the slant height $l$:
   $$l^2 = r^2 + h^2$$
   Substitute $l = 5r$ and $h = 63$:
   $$ (5r)^2 = r^2 + 63^2 $$
   $$ 25r^2 = r^2 + 3969 $$
   $$ 24r^2 = 3969 $$
   $$ r^2 = \frac{3969}{24} = 165.375 $$
   $$ r = \sqrt{165.375} \approx 12.86 	ext{ cm} $$

5. Calculate the volume:
   $$ V = \frac{1}{3} 	imes \frac{22}{7} 	imes 165.375 	imes 63 $$
   Simplify stepwise:
   $$ V = \frac{1}{3} 	imes \frac{22}{7} 	imes 165.375 	imes 63 $$
   $$ = \frac{1}{3} 	imes 22 	imes 165.375 	imes 9 $$
   $$ = 6 	imes 22 	imes 165.375 $$
   $$ = 10914.75 	ext{ cm}^3 $$

**Answer:** $10914.75 	ext{ cm}^3$

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