Question

determine (a) the volume and (b) the surface area of the three-dimensional figure. use the \\(\pi\\) key on your calculator.

a) the volume is
(simplify your answer. type an integer or a decimal rounded to the nearest hundredth as needed.)

b) the surface area is
(simplify your answer. type an integer or a decimal rounded to the nearest hundredth as needed.)

Explanation:

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<reasoning>

Identify the given dimensions

We are given a cone with:

• Diameter of the base \(d = 5\text{ ft}\), which means the radius is:

\[
r = \frac{d}{2} = 2.5\text{ ft}
\]

• Height of the cone \(h = 23\text{ ft}\).

Calculate the volume of the cone

Using the Volume of Cone knowledge point
\[

$$\begin{aligned} V &= \frac{1}{3}\pi r^2 h \\ V &= \frac{1}{3}\pi (2.5)^2 (23) \\ V &= \frac{1}{3}\pi (6.25) (23) \\ V &= \frac{143.75}{3}\pi \approx 150.5339\text{ ft}^3 \end{aligned}$$

\]

Find the slant height of the cone

Using the Pythagorean Theorem knowledge point
\[

$$\begin{aligned} l &= \sqrt{r^2 + h^2} \\ l &= \sqrt{2.5^2 + 23^2} \\ l &= \sqrt{6.25 + 529} \\ l &= \sqrt{535.25} \approx 23.13548\text{ ft} \end{aligned}$$

\]

Calculate the surface area of the cone

Using the Surface Area of Cone knowledge point
\[

$$\begin{aligned} A &= \pi r^2 + \pi r l \\ A &= \pi (2.5)^2 + \pi (2.5)(23.13548) \\ A &= 6.25\pi + 57.8387\pi \\ A &= 64.0887\pi \approx 201.34\text{ ft}^2 \end{aligned}$$

\]
</reasoning>

<answer>

Question 1

a) The volume is <blank>150.53</blank> \(\text{ft}^3\)

Question 2

b) The surface area is <blank>201.34</blank> \(\text{ft}^2\)
</answer>

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"question_type": "Multi-part",
"knowledge_point": [
"Mathematics",
"Geometry",
"Volume of Cone"
]
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</post_analysis>

Answer:

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<reasoning>

Identify the given dimensions

We are given a cone with:

• Diameter of the base \(d = 5\text{ ft}\), which means the radius is:

\[
r = \frac{d}{2} = 2.5\text{ ft}
\]

• Height of the cone \(h = 23\text{ ft}\).

Calculate the volume of the cone

Using the Volume of Cone knowledge point
\[

$$\begin{aligned} V &= \frac{1}{3}\pi r^2 h \\ V &= \frac{1}{3}\pi (2.5)^2 (23) \\ V &= \frac{1}{3}\pi (6.25) (23) \\ V &= \frac{143.75}{3}\pi \approx 150.5339\text{ ft}^3 \end{aligned}$$

\]

Find the slant height of the cone

Using the Pythagorean Theorem knowledge point
\[

$$\begin{aligned} l &= \sqrt{r^2 + h^2} \\ l &= \sqrt{2.5^2 + 23^2} \\ l &= \sqrt{6.25 + 529} \\ l &= \sqrt{535.25} \approx 23.13548\text{ ft} \end{aligned}$$

\]

Calculate the surface area of the cone

Using the Surface Area of Cone knowledge point
\[

$$\begin{aligned} A &= \pi r^2 + \pi r l \\ A &= \pi (2.5)^2 + \pi (2.5)(23.13548) \\ A &= 6.25\pi + 57.8387\pi \\ A &= 64.0887\pi \approx 201.34\text{ ft}^2 \end{aligned}$$

\]
</reasoning>

<answer>

Question 1

a) The volume is <blank>150.53</blank> \(\text{ft}^3\)

Question 2

b) The surface area is <blank>201.34</blank> \(\text{ft}^2\)
</answer>

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