Find the total surface area of a right circular cone of radius 6 cm and height 12 cm. 2. The area of the total surface of a right circular cone is 64 m2 and its slant height is 5 times the radius of t

$$l = \sqrt{r^2 + h^2}$$ $$l = \sqrt{(6 \text{ cm})^2 + (12 \text{ cm})^2}$$ $$l = \sqrt{36 \text{ cm}^2 + 144 \text{ cm}^2}$$ $$l = \sqrt{180 \text{ cm}^2}$$ $$l = \sqrt{36 \times 5 \text{ cm}^2}$$ $$l = 6\sqrt{5} \text{ cm}$$

The formula for the total surface area of a cone is $TSA = \pi r (r + l)$.

$$TSA = \pi (6 \text{ cm}) (6 \text{ cm} + 6\sqrt{5} \text{ cm})$$ $$TSA = 6\pi (6 + 6\sqrt{5}) \text{ cm}^2$$ $$TSA = 36\pi (1 + \sqrt{5}) \text{ cm}^2$$

Using $\sqrt{5} \approx 2.236$:

$$TSA \approx 36\pi (1 + 2.236) \text{ cm}^2$$ $$TSA \approx 36\pi (3.236) \text{ cm}^2$$ $$TSA \approx 116.496\pi \text{ cm}^2$$ $$TSA \approx 116.496 \times 3.14159 \text{ cm}^2$$ $$TSA \approx 366.0 \text{ cm}^2$$

The total surface area is \boxed{\text{36\pi(1 + \sqrt{5}) cm}^2 \text{ or approximately 366.0 cm}^2}.

2. The area of the total surface of a right circular cone is $64 \text{ m}^2$ and its slant height is 5 times the radius of the base. Find the radius of the base.

$$TSA = \pi r (r + l)$$

Given $TSA = 64 \text{ m}^2$ and $l = 5r$.

$$64 \text{ m}^2 = \pi r (r + 5r)$$ $$64 \text{ m}^2 = \pi r (6r)$$ $$64 \text{ m}^2 = 6\pi r^2$$

$$r^2 = \frac{64}{6\pi} \text{ m}^2$$ $$r^2 = \frac{32}{3\pi} \text{ m}^2$$ $$r = \sqrt{\frac{32}{3\pi}} \text{ m}$$ $$r = \frac{\sqrt{32}}{\sqrt{3\pi}} \text{ m}$$ $$r = \frac{4\sqrt{2}}{\sqrt{3\pi}} \text{ m}$$

To rationalize the denominator:

$$r = \frac{4\sqrt{2}}{\sqrt{3\pi}} \times \frac{\sqrt{3\pi}}{\sqrt{3\pi}} \text{ m}$$ $$r = \frac{4\sqrt{6\pi}}{3\pi} \text{ m}$$

Using $\pi \approx 3.14159$:

$$r \approx \sqrt{\frac{32}{3 \times 3.14159}} \text{ m}$$ $$r \approx \sqrt{\frac{32}{9.42477}} \text{ m}$$ $$r \approx \sqrt{3.3953} \text{ m}$$ $$r \approx 1.8439 \text{ m}$$

The radius of the base is $\boxed{\text{}\frac{4\sqrt{6\pi}}{3\pi} \text{ m or approximately 1.84 m}}$.