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Chem Tech makes paper cones used to filter liquids in lab experiments. To make the paper cones, the company starts with the net...Question
Chem Tech makes paper cones used to filter liquids in lab experiments. To make the paper cones, the company starts with the net of the shape, where the slant height is needed. The company wants the height of the cones to be 6 cm. If they make the diameter 13 cm, what must the slant height be? Use the Pythagorean Theorem and round the answer to the nearest tenth. (1 point) Responses 2.5 cm 2.5 cm 11.5 cm 11.5 cm 8.8 cm 8.8 cm 14.3 cm 14.3 cm
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Answered by
GPT-4o mini
To find the slant height of the cone, we can use the Pythagorean theorem. In a right triangle formed by the height, radius, and slant height of the cone, the following relationship holds:
\[
c^2 = a^2 + b^2
\]
where:
- \(c\) is the slant height,
- \(a\) is the height of the cone (6 cm),
- \(b\) is the radius of the base.
First, we need to find the radius. The diameter of the cone is given as 13 cm, so the radius \(b\) is:
\[
b = \frac{diameter}{2} = \frac{13 \, \text{cm}}{2} = 6.5 \, \text{cm}
\]
Now, applying the Pythagorean theorem:
\[
c^2 = a^2 + b^2
\]
Substituting the known values:
\[
c^2 = (6 \, \text{cm})^2 + (6.5 \, \text{cm})^2
\]
Calculating \(6^2\) and \(6.5^2\):
\[
c^2 = 36 + 42.25 = 78.25
\]
Now, solving for \(c\):
\[
c = \sqrt{78.25} \approx 8.83 \, \text{cm}
\]
Rounding to the nearest tenth, the slant height \(c\) is approximately:
\[
\boxed{8.8 \, \text{cm}}
\]
\[
c^2 = a^2 + b^2
\]
where:
- \(c\) is the slant height,
- \(a\) is the height of the cone (6 cm),
- \(b\) is the radius of the base.
First, we need to find the radius. The diameter of the cone is given as 13 cm, so the radius \(b\) is:
\[
b = \frac{diameter}{2} = \frac{13 \, \text{cm}}{2} = 6.5 \, \text{cm}
\]
Now, applying the Pythagorean theorem:
\[
c^2 = a^2 + b^2
\]
Substituting the known values:
\[
c^2 = (6 \, \text{cm})^2 + (6.5 \, \text{cm})^2
\]
Calculating \(6^2\) and \(6.5^2\):
\[
c^2 = 36 + 42.25 = 78.25
\]
Now, solving for \(c\):
\[
c = \sqrt{78.25} \approx 8.83 \, \text{cm}
\]
Rounding to the nearest tenth, the slant height \(c\) is approximately:
\[
\boxed{8.8 \, \text{cm}}
\]
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