SOLUTION: A square pyramid whose lateral edge is 3 ¼ times its base edge can hold 250 cubic inches of sand when level full. How many square inches are there in its lateral surface?

Algebra.Com
Question 1198241: A square pyramid whose lateral edge is 3 ¼ times its base edge can hold 250 cubic inches of sand when level full. How many square inches are there in its lateral surface?
Found 2 solutions by onyulee, ikleyn:
Answer by onyulee(41)   (Show Source): You can put this solution on YOUR website!
**1. Define Variables**
* Let 's' be the side length of the square base.
* Let 'l' be the length of the lateral edge (l = 3.25s).
* Let 'h' be the height of the pyramid.
**2. Find the Volume of the Pyramid**
The volume of a square pyramid is given by:
* Volume = (1/3) * Base Area * Height
* 250 = (1/3) * s^2 * h
**3. Find the Height of the Pyramid**
* We know that the lateral edge, base edge, and height form a right triangle.
* Using the Pythagorean Theorem:
* h^2 + (s/2)^2 = l^2
* h^2 + (s^2)/4 = (3.25s)^2
* h^2 = (3.25s)^2 - (s^2)/4
* h^2 = 10.5625s^2 - 0.25s^2
* h^2 = 10.3125s^2
* h = √(10.3125s^2)
* h = s√10.3125
**4. Substitute 'h' in the Volume Equation**
* 250 = (1/3) * s^2 * (s√10.3125)
* 250 = (√10.3125/3) * s^3
* s^3 = 250 / (√10.3125/3)
* s^3 ≈ 22.81
* s ≈ 2.84 inches
**5. Calculate the Lateral Edge Length**
* l = 3.25s
* l = 3.25 * 2.84
* l ≈ 9.21 inches
**6. Calculate the Slant Height**
* Let 'L' be the slant height.
* L^2 = h^2 + (s/2)^2
* L^2 = (s√10.3125)^2 + (s/2)^2
* L^2 = 10.3125s^2 + 0.25s^2
* L^2 = 10.5625s^2
* L = s√10.5625
* L = 2.84 * √10.5625
* L ≈ 9.31 inches
**7. Calculate the Lateral Surface Area**
* Lateral Surface Area = 4 * (1/2) * Base Edge * Slant Height
* Lateral Surface Area = 4 * (1/2) * s * L
* Lateral Surface Area = 4 * (1/2) * 2.84 * 9.31
* Lateral Surface Area ≈ 52.89 square inches
**Therefore, the lateral surface area of the square pyramid is approximately 53 square inches.**

Answer by ikleyn(52776)   (Show Source): You can put this solution on YOUR website!
.

The solution to this problem in the post by @onynlee is incorrect.

First error is in n.3, where they say

        "We know that the lateral edge, base edge, and height form a right triangle".

This statement is incorrect, and all associated with it calculations in n.3
are incorrect.   These three elements do not form any triangle.

They pull along with them incorrect calculations in the rest of the post.



RELATED QUESTIONS

A square pyramid whose lateral edge is 3 ¼ times its base edge can hold 250 cubic inches (answered by onyulee)
A square pyramid whose lateral edge is 3 ¼ times its base edge can hold 250 cubic inches (answered by greenestamps)
How do you find the slant height and base edge of a regular square pyramid when the... (answered by Theo)
Find the volume of a square pyramid if each base edge and each lateral edge equals 36... (answered by MathLover1,ikleyn)
1. The altitude of a regular square pyramid is three times its base edge. The volume of... (answered by ikleyn)
Lateral area and surface area of a regular square pyramid with base edge length 12 and... (answered by ewatrrr)
The volume of a square pyramid is 600^3 inches. The height of the pyramid is 8 in. What... (answered by MathLover1)
he volume of a pyramid varies jointly with the base area of the pyramid and its height.... (answered by josgarithmetic)
If the lateral surface area of a square pyramid is 84 ft2 and the base edge is 6, what is (answered by ikleyn)