Chapter 9.6, Problem 15Q

Solution

To find: The volume of a right triangular prism.

The volume of the prism is $\text{200}\text{c}{\text{m}}^{\text{3}}$ .

Given data:

The base of a right triangular prism has sides of length 8 cm, 10 cm, and 13 cm. The height of the prism is 5 cm.

Method/Formula used:

The volume of the prism is product of the area of the base and height.

The angle A of a triangle ABC of sides a , b , and c shown as the base of prism in Fig.1, is given by

  $\cos A=\frac{b^2+c^2-a^2}{2bc}$ … (1)

  $\sin A=\sqrt{1-{\cos }^{2}A}$ … (2)

The area $\Delta$ of a triangle ABC is $\Delta =\frac{1}{2}bc\sin A$ …. (3)

Calculations:

  

The right triangular prism has sides of length 8 cm, 10 cm, 13 cm, and height 5 cm shown in Fig.1,

Denote sides of triangular base of the prism as: c equals 8 cm, side a equals 10 cm and side b as 13 cm.

Substitute 13 for b , 8 for c and 10 for a in the result (1), the angle A is,

  $\begin{array}{c}\cos A=\frac{{\left(13\right)}^2+{\left(8\right)}^2-{\left(10\right)}^2}{2\left(13\right)\left(8\right)}\\ =\frac{133}{208}\end{array}$

Substitute $\frac{133}{208}$ for $\cos A$ ,

  $\begin{array}{c}\sin A=\sqrt{1-{\cos }^{2}A}\\ =\sqrt{1-{\left(\frac{133}{208}\right)}^2}\\ =0.7689\end{array}$ ,

Therefore, the area of the triangular base ABC of the prism is

  ${\Delta }_b=\frac{1}{2}bc\sin A$

Substitute 13 for b , 8 for c and 0.7689 for $\sin A$

  $\begin{array}{c}{\Delta }_b=\frac{1}{2}\left(13\right)\left(8\right)\left(0.7689\right)\\ =40\end{array}$

Thus, area of the base of prism is $40\text{c}{\text{m}}^{\text{2}}$ .

Therefore, the volume V of the prism is,

  $\text{Volume=Area}\text{of}\text{base}\text{×height}$

Substitute 40 cm2 as the area of the base and 5 cm the height of the prism,

  $\begin{array}{c}\text{Volume=40×5}\\ \text{=200}\end{array}$

The volume of the prism is $\text{200}\text{c}{\text{m}}^{\text{3}}$ .