Question

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?

Answer 1

`53`

Explanation:

The base has a surface area of `3 xx 9` (it is as if one corner of the `3 xx 9` rectangle has been cut off and attached to the other side).

There are `4` triangles that must be added to this. The basic formula for area is

`"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 `2` (the height of the peak) and the distance from the centre of the pyramid to one of the edges - which is half of `9` or half of `3` (depending on which direction you go).

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)` to the area of the triangles (keep in mind that there are two of both of them, therefore we must add `11.079...` and `1.875` twice).

`27 + (11.079 xx 2) +(1.875 xx 2) = 52.908... ~~ 53`