Question

PLEASE HELP ASAP!!!!

A right triangle with vertices J (0,2), K (0,0), and L (9,0) is rotated around the -axis.

A. Use complete sentences to describe the resulting solid. Include the specific measurements for the solid’s dimensions such as height, width, length, radius, etc.
B. To the nearest tenth of a cubic unit, what is the volume of the resulting three-dimensional figure? Include all of your calculations.

Answer 1

A. The resulting solid is a right circular cone. The base of the cone is a circle with a radius of 9 units (which is the distance between points K and L), and the height of the cone is 2 units (which is the distance between points J and K). The axis of rotation is the -axis.

B. To find the volume of the cone, we can use the formula:

V = (1/3)πr^2h

where r is the radius of the base and h is the height of the cone.

In this case, r = 9 units and h = 2 units, so we can substitute these values into the formula:

V = (1/3)π(9^2)(2)

V = (1/3)π(81)(2)

V = (1/3)π(162)

V ≈ 170.8 cubic units

Therefore, the volume of the resulting three-dimensional figure is approximately 170.8 cubic units.

Answer 2

A) The solid resulting from rotating the right triangle around the x-axis is a cone. The height of the cone is equal to the horizontal distance between vertex K and vertex L, which is 9 units. The radius of the circular base of the cone is equal to the vertical distance between vertex K and vertex J, which is 2 units.

B) The volume of the resulting solid is 37.7 cubic units (to the nearest tenth).

Explanation:

Given a right triangle with vertices J (0, 2), K (0, 0), and L (9, 0) is rotated around the x-axis.

Part A

The solid resulting from rotating the right triangle around the x-axis is a cone.

The height of the cone is equal to the horizontal distance between vertex K and vertex L, which is 9 units.

`\begin{aligned} \implies \text{Height}&=x_L-x_K\\&=9-0\\&=9\;\sf units\end{aligned}`

The radius of the circular base of the cone is equal to the vertical distance between vertex K and vertex J, which is 2 units.

`\begin{aligned} \implies \text{Radius\;of\;circular\;base}&=y_J-y_K\\&=2-0\\&=2\;\sf units\end{aligned}`

Part B

To calculate the volume of the cone, we can use the formula:

`V = (1)/(3)\pi r^2 h`

where r is the radius of the circular base and h is the height of the cone.

From part A, the values are:

• r = 2 units
• h = 9 units

Substitute the values into the formula and solve for V:

`\implies V = (1)/(3)\pi (2)^2 (9)`

`\implies V = (1)/(3)\pi (4)(9)`

`\implies V = (1)/(3)\pi (36)`

`\implies V = 12\pi`

`\implies V = 37.7\; \sf cubic\;units\;(nearest\;tenth)`

Therefore, the volume of the resulting solid is 37.7 cubic units (to the nearest tenth).