Find the volume of the figure below;

The figure is a composite shape consisting of a cylinder and a cone.
For the cylinder:

• Radius $r = 7$ cm
• Height $h_{cyl} = 10$ cm

For the cone:

• Radius $r = 7$ cm (same as the cylinder's base)
• Slant height $l = 13$ cm

We use the Pythagorean theorem, where the radius, height, and slant height form a right-angled triangle:

$$h_{cone}^2 + r^2 = l^2$$ $$h_{cone}^2 + (7 \text{ cm})^2 = (13 \text{ cm})^2$$ $$h_{cone}^2 + 49 \text{ cm}^2 = 169 \text{ cm}^2$$ $$h_{cone}^2 = 169 \text{ cm}^2 - 49 \text{ cm}^2$$ $$h_{cone}^2 = 120 \text{ cm}^2$$ $$h_{cone} = \sqrt{120} \text{ cm} = 2\sqrt{30} \text{ cm}$$

The formula for the volume of a cylinder is $V_{cyl} = \pi r^2 h_{cyl}$.

$$V_{cyl} = \pi (7 \text{ cm})^2 (10 \text{ cm})$$ $$V_{cyl} = \pi (49 \text{ cm}^2) (10 \text{ cm})$$ $$V_{cyl} = 490\pi \text{ cm}^3$$

The formula for the volume of a cone is $V_{cone} = \frac{1}{3} \pi r^2 h_{cone}$.

$$V_{cone} = \frac{1}{3} \pi (7 \text{ cm})^2 (2\sqrt{30} \text{ cm})$$ $$V_{cone} = \frac{1}{3} \pi (49 \text{ cm}^2) (2\sqrt{30} \text{ cm})$$ $$V_{cone} = \frac{98\sqrt{30}}{3}\pi \text{ cm}^3$$

The total volume is the sum of the volume of the cylinder and the volume of the cone.

$$V_{total} = V_{cyl} + V_{cone}$$ $$V_{total} = 490\pi \text{ cm}^3 + \frac{98\sqrt{30}}{3}\pi \text{ cm}^3$$ $$V_{total} = \pi \left(490 + \frac{98\sqrt{30}}{3}\right) \text{ cm}^3$$

Now, we calculate the numerical value using $\pi \approx 3.14159$ and $\sqrt{30} \approx 5.477225$:

$$V_{total} \approx 3.14159 \left(490 + \frac{98 \times 5.477225}{3}\right) \text{ cm}^3$$ $$V_{total} \approx 3.14159 \left(490 + \frac{536.76805}{3}\right) \text{ cm}^3$$ $$V_{total} \approx 3.14159 (490 + 178.92268) \text{ cm}^3$$ $$V_{total} \approx 3.14159 (668.92268) \text{ cm}^3$$ $$V_{total} \approx 2101.07 \text{ cm}^3$$

The volume of the figure is approximately:

$$\text{2101.07 cm}^3$$