Asked by Kristine
Find the surface area of a right regular square pyramid with a side and a slant height of 9in. Help I know its tricky??
Answers
Answered by
Steve
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.
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.
Answered by
jeff
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