# The base of a triangular pyramid is a triangle with corners at (1 ,2 ), (5 ,2 ), and (7 ,5 ). If the pyramid has a height of 7 , what is the pyramid's volume?

Volume $= 14 \text{ }$cubic units

#### Explanation:

We compute for the area of the base which is a triangle first
Let ${P}_{1} \left({x}_{1} , {y}_{1}\right) = \left(1 , 2\right)$
Let ${P}_{2} \left({x}_{2} , {y}_{2}\right) = \left(5 , 2\right)$
Let ${P}_{3} \left({x}_{3} , {y}_{3}\right) = \left(7 , 5\right)$

Use the formula to compute for the area of the triangular base

Area $= \frac{1}{2} \left[\begin{matrix}{x}_{1} & {x}_{2} & {x}_{3} & {x}_{1} \\ {y}_{1} & {y}_{2} & {y}_{3} & {y}_{1}\end{matrix}\right] = \frac{1}{2} \left[\begin{matrix}1 & 5 & 7 & 1 \\ 2 & 2 & 5 & 2\end{matrix}\right]$

Area $= \frac{1}{2} \cdot \left({x}_{1} \cdot {y}_{2} + {x}_{2} \cdot {y}_{3} + {x}_{3} \cdot {y}_{1} - {x}_{2} \cdot {y}_{1} - {x}_{3} \cdot {y}_{2} - {x}_{1} \cdot {y}_{3}\right)$

Area $= \frac{1}{2} \cdot \left(2 + 25 + 14 - 10 - 14 - 5\right)$

Area $= 6$

Volume $= \frac{1}{3} \cdot A r e a \cdot h e i g h t$

Volume $= \frac{1}{3} \cdot 6 \cdot 7$

Volume $= 14 \text{ }$cubic units

Apr 6, 2016

$14 {\text{ units}}^{3}$

#### Explanation:

Area of the base is $\frac{b}{2} \times a {\text{ "->" " 2xx3 = 6 " units}}^{2}$

The volume of a triangular pyramid is:

$\text{base area" xx 1/3 xx "height}$

The height is given as 7

color(blue)("base area" xx 1/3 xx "height " ->" " 6xx1/3xx7 =14 " units"^3