# The base of a triangular pyramid is a triangle with corners at (6 ,7 ), (2 ,5 ), and (3 ,1 ). If the pyramid has a height of 4 , what is the pyramid's volume?

Jan 13, 2018

Volume of a pyramid is $12$ cubic.unit.

#### Explanation:

Volume of a pyramid is $\frac{1}{3} \cdot$base area $\cdot$height.

$\left({x}_{1} , {y}_{1}\right) = \left(6 , 7\right) , \left({x}_{2} , {y}_{2}\right) = \left(2 , 5\right) , \left({x}_{3} , {y}_{3}\right) = \left(3 , 1\right) , h = 4$

Area of Triangle is

A_b = |1/2(x1(y2−y3)+x2(y3−y1)+x3(y1−y2))|

A_b = |1/2(6(5−1)+2(1−7)+3(7−5))| or

${A}_{b} = | \frac{1}{2} \left(24 - 12 + 6\right) | = | 9 | = 9$ sq.unit.

Volume of a pyramid is $\frac{1}{3} \cdot {A}_{b} \cdot h = \frac{1}{3} \cdot 9 \cdot 4 = 12$ cubic.unit [Ans]

Jan 13, 2018

$\text{volume } = 12$

#### Explanation:

$\text{the volume (V) of a pyramid is calculated using the formula}$

•color(white)(x)V=1/3xx"area of base "xx" height"

$\text{the area (A) of the triangle can be found using}$

•color(white)(x)A=1/2|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|

$\text{let } \left({x}_{1} , {y}_{1}\right) = \left(6 , 7\right) , \left({x}_{2} , {y}_{2}\right) = \left(2 , 5\right) , \left({x}_{3} , {y}_{3}\right) = \left(3 , 1\right)$

$A = \frac{1}{2} | 6 \left(5 - 1\right) + 2 \left(1 - 7\right) + 3 \left(7 - 5\right) |$

$\textcolor{w h i t e}{A} = \frac{1}{2} | 24 - 12 + 6 | = \frac{1}{2} | 18 | = 9$

$\Rightarrow V = \frac{1}{3} \times 9 \times 4 = 12$