# The base of a triangular pyramid is a triangle with corners at (6 ,2 ), (3 ,5 ), and (4 ,2 ). If the pyramid has a height of 9 , what is the pyramid's volume?

Jul 2, 2017

The volume is $9$ cubic units.

#### Explanation:

This answer is basically in two parts:

Part 1 -- area of the triangular base is half of any side times the height from that side to the opposite vertex.

Part 2 -- vume of the pyramid is one-third the area of the base times the height.

Now to the math.

Part 1 -- it helps to draw out the triangle on graph paper. Note that the first and third vertices are both on the horizontal line $y = 2$ and the third vertex is at $y = 5$. So we have a side that's $2$ units long (from $\left(4 , 2\right)$ to $\left(6 , 2\right)$) and the height to the opposite vertex is three units (from $y = 2$ along the entire side to $y = 5$ on the third vertex). So the area of the triangle is (1/2)×2×3=3 square units.

Part 2 -- The area of the base is $3$ square units and the height is $9$ units. Pugthose numbers into the formula given above for the volume of a pyramis: (1/3)×3×9=9 cubic units.

Jul 2, 2017

Volume of pyramid is $9$ cubic.unit

#### Explanation:

Vertices of triangular base are $\left(6 , 2\right) , \left(3 , 5\right) , \left(4 , 2\right)$

The area of the triangular base is ${A}_{b} = \frac{1}{2} \left({x}_{1} \left({y}_{2} - {y}_{3}\right) + {x}_{2} \left({y}_{3} - {y}_{1}\right) + {x}_{3} \left({y}_{1} - {y}_{2}\right)\right)$ or

${A}_{t} = \frac{1}{2} \left(6 \left(5 - 2\right) + 3 \left(2 - 2\right) + 4 \left(2 - 5\right)\right) = \frac{1}{2} \left(18 + 0 - 12\right) = 12 \cdot 6 = 3$sq.unit

Volume of pyramid is V= 1/3* A_b*h ; h= 9 , A_b=3  or

$V = \frac{1}{\cancel{3}} \cdot \cancel{3} \cdot 9 = 9$ cubic.unit [Ans]