Find the surface area of a right regular square pyramid with a side and a slant height of 9in. Help I know its tricky??

Answer 1

not tricky at all, if you take the time to draw a diagram.

Label the 4 base corners ABCD and the vertex P.

Drop an altitude to the center of the pryamid's base, point Q.

Draw a perpendicular line from Q to the center of one of the base sides. It meets at point R.

Then, triangle PQR is a right triangle, with legs 9 and 4.5, so its hypotenuse is s = 4.5√5, the slant height of the pyramid.

s is also the altitude of one of the pyramid's triangular faces. Now you have four triangles with base 9 and height s.

Calculate their areas, and add on the base of 9^2.

Answer 2

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