A regular pyramid has a square base of side 14 centimetres and vertical height 22 centimetres. The volume of this pyramid is equal to the volume of a solid sphere. What is the radius, in centimetres, of the sphere?

Introduction / Context:
This volumetric problem links the volume of a regular square pyramid to that of a sphere. We are given the base side and height of the pyramid and told that its volume equals the volume of a solid sphere. Using the standard volume formulas for both solids and equating them allows us to compute the radius of the sphere. Such comparisons are very common in aptitude tests where one shape is melted, recast, or equated to another shape.

Given Data / Assumptions:

• The pyramid has a square base with side length 14 cm.
• The vertical height of the pyramid is 22 cm.
• The pyramid is regular, so its apex is directly above the centre of the square base.
• The volume of the pyramid equals the volume of a solid sphere.
• We are asked to find the radius of the sphere in centimetres.
• Use the usual volume formulas with pi taken as 22 / 7 if needed.

Concept / Approach:
The volume of a square based pyramid is given by:
V_pyramid = (1 / 3) * (base area) * height.
The base area here is side^2. The volume of a sphere of radius r is:
V_sphere = (4 / 3) * pi * r^3.
Because the volumes are equal, we set V_pyramid = V_sphere and solve for r. Simplifying with pi = 22 / 7 gives an exact integer value for the radius.

Step-by-Step Solution:
Step 1: Base side of the pyramid = 14 cm, so base area = 14 * 14 = 196 cm^2. Step 2: Height of the pyramid h = 22 cm. Step 3: Volume of the pyramid is V_pyramid = (1 / 3) * 196 * 22. Step 4: Compute 196 * 22 = 4312, so V_pyramid = 4312 / 3 cubic centimetres. Step 5: Let the radius of the sphere be r. Then V_sphere = (4 / 3) * pi * r^3. Step 6: Set V_pyramid = V_sphere, so (4 / 3) * pi * r^3 = 4312 / 3. Step 7: Multiply both sides by 3: 4 * pi * r^3 = 4312. Step 8: Take pi = 22 / 7, so 4 * pi = 4 * 22 / 7 = 88 / 7. Step 9: Then r^3 = 4312 / (88 / 7) = 4312 * 7 / 88. Step 10: Simplify 4312 / 88 = 49, so r^3 = 49 * 7 = 343. Step 11: Since 343 = 7^3, we get r = 7 cm.

Verification / Alternative check:
We can verify by substituting r = 7 back into the sphere volume formula. With r = 7, V_sphere = (4 / 3) * pi * 7^3 = (4 / 3) * pi * 343. Using pi = 22 / 7, this becomes (4 / 3) * (22 / 7) * 343 = (4 / 3) * 22 * 49 = (4 * 22 * 49) / 3 = 4312 / 3 cubic centimetres, exactly the same as the previously computed pyramid volume. This confirms that r = 7 cm is correct.

Why Other Options Are Wrong:
493 and 983 suggest r^3 values that do not match the required 343 and would give volumes far larger than the pyramid. The options 9 and 14 also do not work: 9^3 and 14^3 lead to r^3 values that, when used in the sphere volume formula, do not produce 4312 / 3. Only r = 7 gives an exact match under the specified value of pi.

Common Pitfalls:
A frequent mistake is to forget that the base is square and to use the perimeter instead of the area. Another issue is misusing the formulas for volume, for example confusing the pyramid volume factor (1 / 3) with that of a prism. Some students also plug in approximate values for pi too early, making it harder to spot the exact cube of 7. Keeping all fractions symbolic until the last step and then substituting pi simplifies the algebra and leads cleanly to r^3 = 343.

Final Answer:
The radius of the sphere is 7 cm.