Question

find the volume of the given irregular pentagonal pyramid.

Explanation:

Step1: Recall the formula for the volume of a pyramid

The volume \( V \) of a pyramid is given by the formula \( V=\frac{1}{3}Bh \), where \( B \) is the area of the base and \( h \) is the height of the pyramid.

Step2: Determine the area of the pentagonal base

The base is an irregular pentagon, but from the diagram, it seems to be composed of a rectangle and a triangle (or maybe a trapezoid? Wait, looking at the dimensions: 12 cm, 8 cm, and 4 cm. Wait, maybe the base is a trapezoid? Wait, no, the base of the pyramid: let's re - examine. Wait, the base is a pentagon, but maybe it's a combination. Wait, another approach: maybe the base is a trapezoid with bases \( a = 12\) cm, \( b = 8\) cm and height \( h_{base}=4\) cm? Wait, no, the formula for the area of a trapezoid is \( A=\frac{(a + b)}{2}\times h\). Wait, but the height of the pyramid is given as 6 cm? Wait, maybe the base is a trapezoid with parallel sides 12 cm and 8 cm, and the distance between them (the height of the trapezoid) is 4 cm. Then the area of the trapezoidal base \( B=\frac{(12 + 8)}{2}\times4\)
\[

$$\begin{align*} B&=\frac{20}{2}\times4\\ &=10\times4\\ & = 40\space cm^{2} \end{align*}$$

\]

Step3: Calculate the volume of the pyramid

Now, the height of the pyramid \( h = 6\) cm. Using the volume formula \( V=\frac{1}{3}Bh\), substitute \( B = 40\) and \( h=6\)
\[

$$\begin{align*} V&=\frac{1}{3}\times40\times6\\ &=\frac{240}{3}\\ &=80\space cm^{3} \end{align*}$$

\]

Answer:

The volume of the irregular pentagonal pyramid is \( 80\space cm^{3}\)