Chapter 12.5, Problem 5PPS

To determine

To find: the lateral and surface area of each prism.

Answer to Problem 5PPS

The lateral area of the triangular prism is $3.84{\text{cm}}^2$ .

The surface area of the triangular prism is $4.8{\text{cm}}^2$ .

Explanation of Solution

Given:

  

Formula used:

The lateral area L of a triangular prism is, $L=P\cdot h$ .

The surface area S of a triangular prism is, S = L + 2B.

Calculation:

The lateral area L of a triangular prism is, $L=P\cdot h$ .

Here, P is the perimeter of the base of the triangle and h is the height the triangular prism.

The surface area S of a triangular prism is, S = L + 2B.

Here, L is the lateral area and B is the area of the triangular base.

The base of the prism is a triangle whose sides are 1cm, 1.2 cm and 1 cm.

Find the lateral area of the triangular prism.

The perimeter P of the base of the prism is $P=\left(l_1+l_2+l_3\right)$ .

  $\begin{array}{c}P=l_1+l_2+l_3\\ =1+1.2+1\\ =3.2\end{array}$

Substitute P = 3.2 and h = l .2 in the lateral area formula to find the lateral area of the prism.

  $\begin{array}{c}L=P\cdot h\\ =\left(3.2\right)\cdot \left(1.2\right)\\ =3.84\end{array}$

Therefore, the lateral area of the triangular prism is $3.84{\text{cm}}^2$ .

Find the surface area of the triangular prism as follows:

The area B of the base of the prism is, $B=\frac{1}{2}\cdot b\cdot a$

Here, b is the base and a is the height of the triangle.

  $\begin{array}{c}B=\frac{1}{2}\cdot b\cdot a\\ =\frac{1}{\cancel{2}}\cdot \left(1.2\right)\cdot \cancel{0}.{\cancel{8}}^{0.4}\\ =\left(1.2\right)\left(0.4\right)\\ =0.48\end{array}$

Substitute B = 0.48 and L = 3.84 in the surface area formula to find the surface area of the prism.

  $\begin{array}{c}S=L+2B\\ =3.84+2\left(0.48\right)\\ =3.84+0.96\\ =4.8\end{array}$

Therefore, the surface area of the triangular prism is $4.8{\text{cm}}^2$ .

Conclusion:

Therefore, the surface area of the triangular prism is $4.8{\text{cm}}^2$ and the lateral area of the triangular prism is $3.84{\text{cm}}^2$ .