Chapter 12.2, Problem 25PPS

To determine

The volume value of chocolate bar by referring the given figure:

  

Answer to Problem 25PPS

The volume of the chocolate bar is $2.055i{n}^3$ .

Explanation of Solution

Given:

  

  $\begin{array}{l}b=1.5in\\ w_3=1in\\ l_1=l_2=l_3=3in\\ h_1=h_2=0.5in\end{array}$

Concept Used:

The volume of the rectangular prism is given by the following expression:

  $V=l\times w\times h$

Where,

Length of rectangular prism: $l$

Width of rectangular prism: $w$

Height of rectangular prism: $h$

The volume of the triangular prism is given by the following expression:

  $V=\frac{1}{2}\left(l\times b\times h\right)$

Where,

Length of triangular prism: $l$

Base of triangular prism: $b$

Height of triangular prism: $h$

Calculation:

In the given figure there are two triangular prism and one rectangular prism so number each volume as:

  

From diagram, understands that there is no base value, let’s calculate it first:

On the both sides of rectangular prism there is similar triangular prism so:

Total base of the triangular prism is equal to the subtraction of base of chocolate bar and base of rectangular prism:

  $b^{\text{'}}=b-b_3=1.5-1=0.5in$

So,

Base of the every triangular prism is calculated as:

  $b_1=b_2=\frac{b^{\text{'}}}{2}=\frac{0.5in}{2}=0.25in$

Now, the height of the rectangular prism is calculated by using the hypotenuse formula:

  $\begin{array}{l}{h}_3{}^2=h_2{}^2+b_2{}^2\\ h_3{}^2={\left(0.5\right)}^2+{\left(0.25\right)}^2\\ h_3{}^2=0.3125\\ h_3{}^2=0.5590\\ h_3\approx 0.56in\end{array}$

Volume of triangular prism 1:

  $\begin{array}{l}{V}_1=\frac{1}{2}\left(l_1\times b_1\times h_1\right)\\ V_1=\frac{1}{2}\left(3\times 0.25\times 0.5\right)\\ V_1=0.1875i{n}^3\end{array}$

Volume of triangular prism 2 is same as volume of triangular prism 1:

  $V_2=V_1=0.1875i{n}^3$

Volume of the rectangular prism 3:

  $\begin{array}{l}{V}_3=l\times w_3\times h_3\\ V_3=3\times 1\times 0.56\\ V_3=1.68i{n}^3\end{array}$

So, total volume of the chocolate bar is:

  $\begin{array}{l}V=V_1+V_2+V_3\\ V=0.1875i{n}^3+0.1875i{n}^3+1.68i{n}^3\\ V=2.055i{n}^3\end{array}$

Conclusion:

Hence, the volume of the chocolate bar is $2.055i{n}^3$