# Math Help - Area of a triangular pyramid

1. ## Area of a triangular pyramid

TA= .5Pl+B

The base of the triangle is an equilateral triangle with length of 6

The height of the pyramid is 9

How do I find the slant height of the pyramid?
I have attached a picture to show the pyramid.

I know that I have to do the Pyth. Thm. but am having trouble finding the length on the base of the triangle. I think that it is a 30-60-90[IMG]file:///Users/schinb64/Desktop/Picture%202.png[/IMG]

2. Originally Posted by IDontunderstand
TA= .5Pl+B

The base of the triangle is an equilateral triangle with length of 6

The height of the pyramid is 9

How do I find the slant height of the pyramid?
I have attached a picture to show the pyramid.

I know that I have to do the Pyth. Thm. but am having trouble finding the length on the base of the triangle. I think that it is a 30-60-90[IMG]file:///Users/schinb64/Desktop/Picture%202.png[/IMG]
Yes, take the distance from any corner of the base triangle to the
base of the perpendicular height line as

$3cos30^o$

then $s=\sqrt{9^2+3^2cos^230^o}$ is the length of an edge.

To get the length of the line from the midpoint of a side of the base to the top, use

$x=3tan30^o$

then apply Pythagoras' theorem.

You can also calculate the volume, if needed, by equating the base area to a circle area,
then calculating the volume of a cone of same height and base areas.

$(0.5)6^2sin60^o={\pi}r^2$

$V=\frac{{\pi}r^2h}{3}=\frac{18sin60^o(9)}{3}=54sin 60^o$

3. Hello, IDontunderstand!

If you don't know Trig, we can still work it out.

The base of the pyramid is an equilateral triangle with length of 6

The height of the pyramid is 9

How do I find the slant height of the pyramid?

The base is an equilateral triangle of side 6.

Code:
              A
*
/|\
/ | \
/  |  \
6 /   |h  \ 6
/    |    \
/     |     \
/      |      \
B * - - - + - - - * C
3   D   3

In right triangle $ADC\!:\;\;h^2 +3^2 \:=\:6^2 \quad\Rightarrow\quad h \:=\:3\sqrt{3}$

We want the "center" of the triangle (the centroid $G$).
It divides an altitude in the ratio 2:1

Code:
              A
*
/|\
/ | \
/  |  \
/   |   \
/    |  _ \
/     |2√3  \
/      |      \
/      G*       \
/        | _      \
/         |√3       \
/          |          \
B * - - - - - - - - - - - * C

This is the point directly below the top vertex of the pyramid $(V)$.

Now consider the side view of the pyramid.

Code:
      V
*
|\
| \
|  \
9 |   \ s
|    \
|     \
|      \
G * - -_- * A
2√3

We have: . $s^2 \:=\:9^2 + (2\sqrt{3})^2 \:=\:81 + 12 \:=\:93$

Therefore, the slant height is: . $s \:=\:\sqrt{93}$