Question
Raj, Amit, Parul, and Lara were asked to find the volume of the following cones.
Raj
Amit
Parul
Lara
A cone with diameter of 5 centimeters and height of 3 centimeters. A cone with a radius of 3 centimeters and height of 4 centimeters. A cone with a height of 7 centimeters and Base = 4 pi centimeters squared. A cone with a height of 6 centimeters and Base = 3 pi centimeters squared.
Whose cone has a volume closest to 9 pi centimeters cubed?
Raj
Amit
Parul
Lara
Raj
Amit
Parul
Lara
A cone with diameter of 5 centimeters and height of 3 centimeters. A cone with a radius of 3 centimeters and height of 4 centimeters. A cone with a height of 7 centimeters and Base = 4 pi centimeters squared. A cone with a height of 6 centimeters and Base = 3 pi centimeters squared.
Whose cone has a volume closest to 9 pi centimeters cubed?
Raj
Amit
Parul
Lara
Answers
GPT-4o mini
To find whose cone has a volume closest to \(9\pi\) cm³, we will calculate the volume of each cone using the formula for the volume of a cone:
\[
V = \frac{1}{3} \pi r^2 h
\]
where \(V\) is the volume, \(r\) is the radius, and \(h\) is the height.
**1. Raj's cone:**
- Diameter = 5 cm, so the radius \(r = \frac{5}{2} = 2.5\) cm
- Height \(h = 3\) cm
Now, calculate the volume:
\[
V = \frac{1}{3} \pi (2.5)^2 (3) = \frac{1}{3} \pi (6.25) (3) = \frac{1}{3} \pi (18.75) = 6.25\pi \text{ cm}^3
\]
**2. Amit's cone:**
- Radius \(r = 3\) cm
- Height \(h = 4\) cm
Now, calculate the volume:
\[
V = \frac{1}{3} \pi (3^2)(4) = \frac{1}{3} \pi (9)(4) = \frac{1}{3} \pi (36) = 12\pi \text{ cm}^3
\]
**3. Parul's cone:**
- Height \(h = 7\) cm
- Base area = \(4\pi\) cm²
To find the radius from the base area, we use the area of the base:
\[
\text{Area} = \pi r^2 \implies 4\pi = \pi r^2 \implies r^2 = 4 \implies r = 2 \text{ cm}
\]
Now, calculate the volume:
\[
V = \frac{1}{3} \pi (2^2)(7) = \frac{1}{3} \pi (4)(7) = \frac{1}{3} \pi (28) = \frac{28}{3}\pi \text{ cm}^3 \approx 9.33\pi \text{ cm}^3
\]
**4. Lara's cone:**
- Height \(h = 6\) cm
- Base area = \(3\pi\) cm²
Again, using the area of the base to find the radius:
\[
\text{Area} = \pi r^2 \implies 3\pi = \pi r^2 \implies r^2 = 3 \implies r = \sqrt{3} \text{ cm}
\]
Now, calculate the volume:
\[
V = \frac{1}{3} \pi (\sqrt{3})^2(6) = \frac{1}{3} \pi (3)(6) = \frac{1}{3} \pi (18) = 6\pi \text{ cm}^3
\]
**Summary of volumes:**
- Raj: \(6.25\pi\) cm³
- Amit: \(12\pi\) cm³
- Parul: \(\frac{28}{3}\pi \approx 9.33\pi\) cm³
- Lara: \(6\pi\) cm³
**Comparing with \(9\pi\) cm³:**
- Raj's: \(6.25\pi\) (difference: \(9 - 6.25 = 2.75\))
- Amit's: \(12\pi\) (difference: \(12 - 9 = 3\))
- Parul's: \(9.33\pi\) (difference: \(9.33 - 9 = 0.33\))
- Lara's: \(6\pi\) (difference: \(9 - 6 = 3\))
The closest volume to \(9\pi\) cm³ is from **Parul**, with a volume of approximately \(9.33\pi\) cm³.
\[
V = \frac{1}{3} \pi r^2 h
\]
where \(V\) is the volume, \(r\) is the radius, and \(h\) is the height.
**1. Raj's cone:**
- Diameter = 5 cm, so the radius \(r = \frac{5}{2} = 2.5\) cm
- Height \(h = 3\) cm
Now, calculate the volume:
\[
V = \frac{1}{3} \pi (2.5)^2 (3) = \frac{1}{3} \pi (6.25) (3) = \frac{1}{3} \pi (18.75) = 6.25\pi \text{ cm}^3
\]
**2. Amit's cone:**
- Radius \(r = 3\) cm
- Height \(h = 4\) cm
Now, calculate the volume:
\[
V = \frac{1}{3} \pi (3^2)(4) = \frac{1}{3} \pi (9)(4) = \frac{1}{3} \pi (36) = 12\pi \text{ cm}^3
\]
**3. Parul's cone:**
- Height \(h = 7\) cm
- Base area = \(4\pi\) cm²
To find the radius from the base area, we use the area of the base:
\[
\text{Area} = \pi r^2 \implies 4\pi = \pi r^2 \implies r^2 = 4 \implies r = 2 \text{ cm}
\]
Now, calculate the volume:
\[
V = \frac{1}{3} \pi (2^2)(7) = \frac{1}{3} \pi (4)(7) = \frac{1}{3} \pi (28) = \frac{28}{3}\pi \text{ cm}^3 \approx 9.33\pi \text{ cm}^3
\]
**4. Lara's cone:**
- Height \(h = 6\) cm
- Base area = \(3\pi\) cm²
Again, using the area of the base to find the radius:
\[
\text{Area} = \pi r^2 \implies 3\pi = \pi r^2 \implies r^2 = 3 \implies r = \sqrt{3} \text{ cm}
\]
Now, calculate the volume:
\[
V = \frac{1}{3} \pi (\sqrt{3})^2(6) = \frac{1}{3} \pi (3)(6) = \frac{1}{3} \pi (18) = 6\pi \text{ cm}^3
\]
**Summary of volumes:**
- Raj: \(6.25\pi\) cm³
- Amit: \(12\pi\) cm³
- Parul: \(\frac{28}{3}\pi \approx 9.33\pi\) cm³
- Lara: \(6\pi\) cm³
**Comparing with \(9\pi\) cm³:**
- Raj's: \(6.25\pi\) (difference: \(9 - 6.25 = 2.75\))
- Amit's: \(12\pi\) (difference: \(12 - 9 = 3\))
- Parul's: \(9.33\pi\) (difference: \(9.33 - 9 = 0.33\))
- Lara's: \(6\pi\) (difference: \(9 - 6 = 3\))
The closest volume to \(9\pi\) cm³ is from **Parul**, with a volume of approximately \(9.33\pi\) cm³.