Question
To find the volume of the cone in the picture where S = 13 cm and R = 5 cm:
A light pink cone is depicted. A dashed line represents the circular base of the cone. A line labeled "R" extends from the base to a point on the side of the cone, indicating the radius. A vertical line labeled "H" extends from the base to the apex of the cone, indicating the height. A slanting line labeled "S" extends from the apex to a point on the base, indicating the slant height. The lines R, H, and S form a right angle at the base of the cone.
(3 points)
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A light pink cone is depicted. A dashed line represents the circular base of the cone. A line labeled "R" extends from the base to a point on the side of the cone, indicating the radius. A vertical line labeled "H" extends from the base to the apex of the cone, indicating the height. A slanting line labeled "S" extends from the apex to a point on the base, indicating the slant height. The lines R, H, and S form a right angle at the base of the cone.
(3 points)
Put responses in the correct input to answer the question. Select a response, navigate to the desired input and insert the response. Responses can be selected and inserted using the space bar, enter key, left mouse button or touchpad. Responses can also be moved by dragging with a mouse.
Step 1
Step 2
Step 3
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Answers
GPT-4o mini
To find the volume of the cone, we use the formula:
\[
V = \frac{1}{3} \pi R^2 H
\]
where:
- \( V \) is the volume,
- \( R \) is the radius of the base (5 cm),
- \( H \) is the height of the cone.
We are given:
- \( R = 5 \) cm
- The slant height \( S = 13 \) cm
First, we need to find the height \( H \) of the cone. We can use the Pythagorean theorem because the values of \( R \), \( H \), and \( S \) form a right triangle:
\[
S^2 = R^2 + H^2
\]
Substituting the known values:
\[
13^2 = 5^2 + H^2
\]
\[
169 = 25 + H^2
\]
\[
H^2 = 169 - 25
\]
\[
H^2 = 144
\]
\[
H = \sqrt{144} = 12 \, \text{cm}
\]
Now, substituting \( R \) and \( H \) back into the volume formula:
\[
V = \frac{1}{3} \pi (5)^2 (12)
\]
\[
V = \frac{1}{3} \pi (25)(12)
\]
\[
V = \frac{1}{3} \pi (300)
\]
\[
V = 100\pi
\]
Thus, the volume of the cone is:
\[
V \approx 314.16 \, \text{cm}^3 \quad (\text{using } \pi \approx 3.1416)
\]
In summary, the volume of the cone is \( 100\pi \, \text{cm}^3 \) or approximately \( 314.16 \, \text{cm}^3 \).
\[
V = \frac{1}{3} \pi R^2 H
\]
where:
- \( V \) is the volume,
- \( R \) is the radius of the base (5 cm),
- \( H \) is the height of the cone.
We are given:
- \( R = 5 \) cm
- The slant height \( S = 13 \) cm
First, we need to find the height \( H \) of the cone. We can use the Pythagorean theorem because the values of \( R \), \( H \), and \( S \) form a right triangle:
\[
S^2 = R^2 + H^2
\]
Substituting the known values:
\[
13^2 = 5^2 + H^2
\]
\[
169 = 25 + H^2
\]
\[
H^2 = 169 - 25
\]
\[
H^2 = 144
\]
\[
H = \sqrt{144} = 12 \, \text{cm}
\]
Now, substituting \( R \) and \( H \) back into the volume formula:
\[
V = \frac{1}{3} \pi (5)^2 (12)
\]
\[
V = \frac{1}{3} \pi (25)(12)
\]
\[
V = \frac{1}{3} \pi (300)
\]
\[
V = 100\pi
\]
Thus, the volume of the cone is:
\[
V \approx 314.16 \, \text{cm}^3 \quad (\text{using } \pi \approx 3.1416)
\]
In summary, the volume of the cone is \( 100\pi \, \text{cm}^3 \) or approximately \( 314.16 \, \text{cm}^3 \).