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

Question 1198304: A square pyramid whose lateral edge is 3 ¼ times its base edge can hold 250 cubic inches of sand when leveling full. How many square inches are there in its lateral surface? Round off the answer to the nearest integer.
Answer by greenestamps(13209) (Show Source): You can put this solution on YOUR website!

There are a whole bunch of ugly calculations involved in solving this problem. You won't learn anything from the problem if we do them all for you.

So I will describe one method for solving the problem and tell you the final answer I got and let you do the calculations.

Let x be the side of the square base; then the length of each lateral edge of the pyramid is (13/4)x or 3.25x.

The volume of the pyramid is 250 cubic inches; the formula for the volume is one-third base times height. The base is a square with side x, so the area of the base is x^2. So

[1]

We need to determine the height.

To do that, form a right triangle with the height of the pyramid as one leg and a lateral edge of the pyramid as the hypotenuse. The other leg will be one-half the length of a diagonal of the square base, which is

Use the Pythagorean Theorem to find the height of the pyramid in terms of x.

Then use [1] above to find x, the side length of the square.

You will almost certainly be using a calculator by this time for the calculations. Be sure to keep at least 3 or 4 decimal places in all your calculations to ensure that your final answer will be sufficiently accurate.

At this point, it would be a good idea to use the numbers you got for the side length and height of the pyramid in [1] to verify that the volume you calculate is close to the actual volume of 250.

The number we are looking for in the end is the lateral surface area, which consists of 4 isosceles triangles with the side of the square base of the pyramid as the base and lateral edges of the pyramid as the congruent sides.

To find the area of each of those isosceles triangles, we need to find the slant height. That can be found as the hypotenuse of a right triangle in which one leg is the height of the pyramid and the other leg is half the width of the square base -- i.e., (1/2)x.

Then use the standard formula for the area of a triangle (one-half base times height) to find the area of each of the triangular faces of the pyramid and then finally the lateral surface area of the pyramid.

My answer for the lateral surface area to the nearest whole integer number of square inches was 246.

If you try the steps described and need help completing the problem, post a response to this response of mine showing the work you have done and I will be happy to help you along.

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