A pyramid has a parallelogram shaped base and a peak directly above its center. Its base's sides have lengths of #3 # and #9 # and the pyramid's height is #2 #. If one of the base's corners has an angle of #(3pi)/8#, what is the pyramid's surface area?
1 Answer
Explanation:
The base has a surface area of
There are
#"area" = 0.5 xx "base" xx "height"#
There are two triangle sizes and two copies of each.
We need to find the sloped length of the pyramid, which requires the use of Pythagoras' theorem (please look this up on google images as it is relatively simple however is difficult to explain in words).
The two smaller lengths are
This tells us the lengths of the slanted surface of the triangle which is
#sqrt(6.25) = 2.5 -># for#3#
#sqrt(24.25) -># for#9#
Now we substitute the numbers into the equation for a triangle;
#"area" = 0.5 xx 4.5 xx sqrt(24.25) = 11.079...#
#"area" = 0.5 xx 1.5 xx sqrt(6.25) = 1.875#
Now we add the area of the base
#27 + (11.079 xx 2) +(1.875 xx 2) = 52.908... ~~ 53#