Question

Consider a regular pyramid A with a square base and a right circular cone B.

Answer 1

Answer:

Pyramid

Step-by-step explanation:

$\text{Volume of a Square Pyramid }=\frac{1}{3} \times l^2 \times Height\\\\ \text{Volume of a Cone }=\frac{1}{3} \pi r^2 \times Height$

Given that the two solids have the same volume

$\frac{1}{3} \times l^2 \times Height=\frac{1}{3} \pi r^2 \times Height$

If the length of a side of the square base of pyramid A is the same as the base radius of cone B. i.e l=r

$\frac{1}{3} \times l^2 \times Height of Pyramid= \frac{1}{3} \pi l^2 \times Height of cone \\\\ Cancel out \frac{1}{3} \times l^2 on both sides\\\\Height of Pyramid= \pi \times Height of cone$

If the height of the cone is 1

$H eight of Pyramid= \pi \times 1 \approx 3.14 units$

Therefore, the pyramid has a greater height.