Jude says that the volume of a square pyramid with base edges of 9.7 in and a height of 9 in is equal to the volume of a cylinder with a radius of 5.47 in and a height of 3 in. Jude rounded his answers to the nearest whole number. Examine Jude's calculations. Is he correct? Volume of Square Pyramid Volume of Cylinder V = one third B(h) V = πr2h V = one third(94.09)(9) V = π(5.472)(3) V = one third(846.81) V = π(29.9209)(3) V = 282 in3 V = π(89.7627) V ≈ 282 in3 Yes, his calculations are correct and the volumes for figures are equal.
No, he made a mistake in solving for the volume of the cylinder.
Yes, but he made a mistake in solving for the volume of the square pyramid.
No, he made a mistake in solving for the volume of both figures.

By approximation, Jude's calculation is correct and the correct option is; Yes, his calculations are correct and the volumes are equal

The given parameters are;

The length of the edges of the square pyramid, s = 9.7 inches

The height of the square pyramid, h = 9 inches

The radius of the cylinder, r = 5.47 inches

The height of the cylinder, h = 3 inches

The volume of a pyramid, = (1/3) × Base Area, B × Height, h

∴ =  (1/3) × B × h

The volume of a cylinder, = Base Area, B × Height, h

Base Area, B, of a cylinder = π·r²

Therefore;

The volume of a cylinder, =  π·r² × h = π·r²·h

Jude's calculations are presented as follows;

=  (1/3) × B × h =  (1/3) × s² × h

Therefore;

=  (1/3) × 9.7² × 9 = 282.27

≈ 282 in.³

=  π·r²·h

Therefore;

=  π × 5.47² × 3 ≈ 281.9978 ≈ 282

≈ 282 in.³

Therefore, Jude's calculation is correct by approximation, and the volumes of the cylinder and the pyramid are approximately equal

Learn more about volumes of cylinder and pyramids here;

the image of (6, 9) under a dilation is (4, 6). the scale factor is