Chapter 12.1, Problem 9PSC

To determine

To calculate: The lateral area and total area of the prism.

Answer to Problem 9PSC

The lateral area is $616$ and total area is $712$ .

Explanation of Solution

Given information:

A prism with rectangle of dimensions 11 by 13 and triangle with dimensions 4, 13 and 15.

Formula used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  

In right angle triangle,

  $a^2+b^2\text{= }{c}^2$

Area of triangle: $A=\frac{1}{2}\times b\times h$

b = base of triangle

h = height of triangle

Calculation:

  

Lateral Area $=2\left(11\times 13\right)+2\left(11\times 15\right)$

Lateral Area $=286+330$

Lateral Area $=616$

We consider sides x and y for our calculation.

A right angle is formed by sides x , y and 13.

A right angle is formed by sides x , y + 4 and 15.

Applying Pythagoras Theorem, we get

  $\begin{array}{l}{x}^2+y^2={\left(13\right)}^2\\ x^2+{\left(y+4\right)}^2={\left(15\right)}^2\\ x^2+y^2=169----Equation1\\ x^2+y^2+8y+16=225----Equation2\end{array}$

Multiplying Equation 1 by $-1$ , we get

Adding Equation 3 and Equation 4, we get

  $\begin{array}{l}8y+16=56\\ 8y=40\\ y=\frac{40}{8}=5\end{array}$

Substituting value of y in Equation 1, we get

  $\begin{array}{l}{x}^2+y^2={\left(13\right)}^2\\ x^2+{\left(5\right)}^2={\left(13\right)}^2\\ x^2={\left(13\right)}^2-{\left(5\right)}^2\\ x^2=169-25\\ x^2=144\\ x=\sqrt{144}\\ x=12\end{array}$

Area of base $=\frac{1}{2}\times 24\times 9-\frac{1}{2}\times 24\times 5$

Area of base $=108-60$

Area of base $=48$

Total Area= Lateral Area + 2 $\times$ Area of base

Total Area $=616+\left(2\times 48\right)$

Total Area $=712$