Question 1198258: 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? Round off the answer to the nearest integer.
Answer by onyulee(41)   (Show Source):
You can put this solution on YOUR website! Certainly, let's find the lateral surface area of the square pyramid.
**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.**
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