To examine Jude's calculations, let's break down the volumes of both the square pyramid and the cylinder step-by-step and compare the results rounded to the nearest whole number.
### Volume of the Square Pyramid
The formula for the volume [tex]\( V \)[/tex] of a square pyramid is:
[tex]\[ V = \frac{1}{3} B h \][/tex]
where [tex]\( B \)[/tex] is the area of the base and [tex]\( h \)[/tex] is the height.
1. Calculate the area of the base ([tex]\( B \)[/tex]):
- The base is a square with edge length [tex]\( 9.7 \)[/tex] inches.
- Area ([tex]\( B \)[/tex]) = edge^2 = [tex]\( 9.7 \times 9.7 = 94.09 \)[/tex] square inches.
2. Calculate the volume ([tex]\( V \)[/tex]) of the pyramid:
- Height ([tex]\( h \)[/tex]) = 9 inches.
- Volume [tex]\( V = \frac{1}{3} \times 94.09 \times 9 = \frac{1}{3} \times 846.81 \approx 282.27 \)[/tex] cubic inches.
### Volume of the Cylinder
The formula for the volume [tex]\( V \)[/tex] of a cylinder is:
[tex]\[ V = \pi r^2 h \][/tex]
where [tex]\( r \)[/tex] is the radius and [tex]\( h \)[/tex] is the height.
1. Calculate the volume ([tex]\( V \)[/tex]) of the cylinder:
- Radius ([tex]\( r \)[/tex]) = 5.47 inches.
- Height ([tex]\( h \)[/tex]) = 3 inches.
- Volume [tex]\( V = \pi \times (5.47)^2 \times 3 = \pi \times 29.9209 \times 3 \approx 281.99783888638444 \)[/tex] cubic inches.
### Rounding to the Nearest Whole Number
- The volume of the square pyramid, when rounded to the nearest whole number, is [tex]\( 282 \)[/tex] cubic inches.
- The volume of the cylinder, when rounded to the nearest whole number, is [tex]\( 282 \)[/tex] cubic inches.
### Conclusion
Jude’s calculations are correct. Upon rounding to the nearest whole number, both the square pyramid and the cylinder volumes are equal to [tex]\( 282 \)[/tex] cubic inches.
Hence, the correct answer is:
Yes, his calculations are correct and the volumes for figures are equal.
