Join IDNLearn.com and start getting the answers you've been searching for. Explore a wide array of topics and find reliable answers from our experienced community members.
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."
Thank you for participating in our discussion. We value every contribution. Keep sharing knowledge and helping others find the answers they need. Let's create a dynamic and informative learning environment together. Your search for solutions ends at IDNLearn.com. Thank you for visiting, and we look forward to helping you again.
