The square pyramid has base edge length 14 in. The height of the pyramid is 24 in. Find the surface area of the square pyramid.

Question

I can't figure this out for a math problem for school

Answer 1

inside the square pyramid construct a right triangle

one leg is the height of 24 in and the other leg is the distance from the center of the square perpendicular to one side which is half the length of one side which is therefore 7 in

the hypotenuse is the slant height of the pyramid

find the slant height using the Pythagorean Theorem

s^2=24^2+7^2

s^2=576+49

s^2=625

s=√625

s=25 in

the base of one of the four triangles is 14 in and the height is 25 in

A=(1/2)bh

A=(1/2)(14)(25)

A=7*25

A=175 sq in

there are four triangles

4*175=700 sq in

find the area of the square base

A=s^2

A=14^2

A=196 sq in

add the areas together

700+196=896 sq in for the surface area

Answer 2

Square base = 14 * 14 = 196.

 

Each of four triangular faces = 1/2 * 14 * 25 = 175.

So total surface area = 896.

So, how did I get the height (25) of the triangular faces?

 

Think of it as finding the hypotenuse of a right triangle whose legs are 24 inches and 7 inches long.  24 is given as the height of the pyramid, and 7 is half the length of a side, or the distance from the center of the base to the nearest point on any of the edges.  24^2 + 7^2 = 576 + 49 = 625; square root of 625 = 25.