Question 1198304
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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.<br>
So I will describe one method for solving the problem and tell you the final answer I got and let you do the calculations.<br>
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.<br>
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<br>
{{{250 = (1/3)(x^2)(h)}}} [1]<br>
We need to determine the height.<br>
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 {{{(1/2)(x*sqrt(2))}}}<br>
Use the Pythagorean Theorem to find the height of the pyramid in terms of x.<br>
Then use [1] above to find x, the side length of the square.<br>
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.<br>
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.<br>
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.<br>
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.<br>
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.<br>
My answer for the lateral surface area to the nearest whole integer number of square inches was 246.<br>
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.<br>