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Sagot :
Let's analyze the problem step-by-step to understand the volumes of cone [tex]\( W \)[/tex] and square pyramid [tex]\( X \)[/tex].
1. Given Data:
- Radius of the cone [tex]\( W \)[/tex], [tex]\( r = 10 \)[/tex] cm.
- Height of the cone [tex]\( W \)[/tex], [tex]\( h = 5 \)[/tex] cm.
- The square pyramid [tex]\( X \)[/tex] has the same base area and height as the cone [tex]\( W \)[/tex].
2. Calculate the base area of cone [tex]\( W \)[/tex]:
- The base of cone [tex]\( W \)[/tex] is a circle with radius [tex]\( r = 10 \)[/tex] cm.
- The area of the base (circle) is given by the formula [tex]\( \text{Base Area}_{\text{cone}} = \pi r^2 \)[/tex].
[tex]\[ \text{Base Area}_{\text{cone}} = \pi (10)^2 = 100\pi \, \text{cm}^2 \][/tex]
3. Determine the volume of cone [tex]\( W \)[/tex]:
- The volume of a cone is given by [tex]\( \text{Volume}_{\text{cone}} = \frac{1}{3} \pi r^2 h \)[/tex].
[tex]\[ \text{Volume}_{\text{cone}} = \frac{1}{3} \pi (10)^2 (5) = \frac{1}{3} \pi (100) (5) = \frac{500}{3} \pi \, \text{cm}^3 \][/tex]
4. Base of the square pyramid [tex]\( X \)[/tex]:
- The square pyramid [tex]\( X \)[/tex] has the same base area as the cone [tex]\( W \)[/tex]. Hence, the base area of pyramid [tex]\( X \)[/tex] is also [tex]\( 100\pi \, \text{cm}^2 \)[/tex].
5. Height of the square pyramid [tex]\( X \)[/tex]:
- The height of the pyramid [tex]\( X \)[/tex] is given to be the same as the cone [tex]\( W \)[/tex], which is [tex]\( h = 5 \)[/tex] cm.
6. Calculate the volume of square pyramid [tex]\( X \)[/tex]:
- The volume of a square pyramid is given by [tex]\( \text{Volume}_{\text{pyramid}} = \frac{1}{3} \times \text{Base Area} \times \text{Height} \)[/tex].
[tex]\[ \text{Volume}_{\text{pyramid}} = \frac{1}{3} \times 100\pi \, \text{cm}^2 \times 5 \, \text{cm} = \frac{500}{3} \pi \, \text{cm}^3 \][/tex]
7. Compare the calculated volumes:
- Volume of cone [tex]\( W \)[/tex] = [tex]\( \frac{500}{3} \pi \, \text{cm}^3 \)[/tex].
- Volume of square pyramid [tex]\( X \)[/tex] = [tex]\( \frac{500}{3} \pi \, \text{cm}^3 \)[/tex].
Both volumes turn out to be exactly the same because they are derived using the same base area and height through the same volume formula.
Conclusion:
- Since the volumes are equal and neither Paul nor Manuel used an incorrect formula, the correct explanation is:
"Both Paul’s and Manuel’s arguments are correct."
1. Given Data:
- Radius of the cone [tex]\( W \)[/tex], [tex]\( r = 10 \)[/tex] cm.
- Height of the cone [tex]\( W \)[/tex], [tex]\( h = 5 \)[/tex] cm.
- The square pyramid [tex]\( X \)[/tex] has the same base area and height as the cone [tex]\( W \)[/tex].
2. Calculate the base area of cone [tex]\( W \)[/tex]:
- The base of cone [tex]\( W \)[/tex] is a circle with radius [tex]\( r = 10 \)[/tex] cm.
- The area of the base (circle) is given by the formula [tex]\( \text{Base Area}_{\text{cone}} = \pi r^2 \)[/tex].
[tex]\[ \text{Base Area}_{\text{cone}} = \pi (10)^2 = 100\pi \, \text{cm}^2 \][/tex]
3. Determine the volume of cone [tex]\( W \)[/tex]:
- The volume of a cone is given by [tex]\( \text{Volume}_{\text{cone}} = \frac{1}{3} \pi r^2 h \)[/tex].
[tex]\[ \text{Volume}_{\text{cone}} = \frac{1}{3} \pi (10)^2 (5) = \frac{1}{3} \pi (100) (5) = \frac{500}{3} \pi \, \text{cm}^3 \][/tex]
4. Base of the square pyramid [tex]\( X \)[/tex]:
- The square pyramid [tex]\( X \)[/tex] has the same base area as the cone [tex]\( W \)[/tex]. Hence, the base area of pyramid [tex]\( X \)[/tex] is also [tex]\( 100\pi \, \text{cm}^2 \)[/tex].
5. Height of the square pyramid [tex]\( X \)[/tex]:
- The height of the pyramid [tex]\( X \)[/tex] is given to be the same as the cone [tex]\( W \)[/tex], which is [tex]\( h = 5 \)[/tex] cm.
6. Calculate the volume of square pyramid [tex]\( X \)[/tex]:
- The volume of a square pyramid is given by [tex]\( \text{Volume}_{\text{pyramid}} = \frac{1}{3} \times \text{Base Area} \times \text{Height} \)[/tex].
[tex]\[ \text{Volume}_{\text{pyramid}} = \frac{1}{3} \times 100\pi \, \text{cm}^2 \times 5 \, \text{cm} = \frac{500}{3} \pi \, \text{cm}^3 \][/tex]
7. Compare the calculated volumes:
- Volume of cone [tex]\( W \)[/tex] = [tex]\( \frac{500}{3} \pi \, \text{cm}^3 \)[/tex].
- Volume of square pyramid [tex]\( X \)[/tex] = [tex]\( \frac{500}{3} \pi \, \text{cm}^3 \)[/tex].
Both volumes turn out to be exactly the same because they are derived using the same base area and height through the same volume formula.
Conclusion:
- Since the volumes are equal and neither Paul nor Manuel used an incorrect formula, the correct explanation is:
"Both Paul’s and Manuel’s arguments are correct."
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