Question

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.

Answer 1

Answer:

The correct option is;

Yes, his calculations are correct and the volumes for figures are equal

Step-by-step explanation:

The volume, V, of a square  (of side s) pyramid of height h = $\frac{1}{3} \times s^2 \times h$

Where:

s = 9.7 inches

h = 9 inches

We have;

$V = \frac{1}{3} \times 9.7^2 \times 9 = \frac{1}{3} \times 94.09 \times 9 = \frac{1}{3} \times 846.81 = 282.27 \approx 282 \ in^3$

The volume, $V_c$, of a cylinder of radius r = Base area × Height, h = (π×r²)×h

Where:

Base area = π×r²

r = Radius of the cylinder = 5.47 inches

h = Height of the cylinder = 3 inches

$V_c$ = π × 5.47² × 3 = π × 29.9209 × 3 = π × 89.7627 = 281.998 ≈ 282 m³

Therefore, Jude is correct.