Chapter 12.1, Problem 3PSA

a.

To determine

To calculate: The lateral area of a right triangular prism with dimensions as l is 11, a is 17, b is 17 and c is 16.

Answer to Problem 3PSA

The lateral area of a right triangular prism is $550$ .

Explanation of Solution

Given information:

l =11, a = 17, b = 17 and c =16.

Formula used:

Area of rectangle: $A=l\times w$

l = length of rectangle

w= width of rectangle

Calculation:

  

  $\begin{array}{l}A({▭}_1)=l\times a=11\times 17=187\\ A({▭}_2)=l\times b=11\times 17=187\\ A({▭}_3)=l\times c=11\times 16=176\end{array}$

Lateral Area:

  $\begin{array}{l}A=A({▭}_1)+A({▭}_2)+A({▭}_3)\\ A=187+187+176\\ A=550\end{array}$

b.

To determine

To find: The area of one base in a right triangular prism.

Answer to Problem 3PSA

The area of one base in a right triangular prism is $120$ .

Explanation of Solution

Given information:

l =11, a = 17, b = 17 and c =16.

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:

  

Draw an altitude AD.

  $DC=\frac{BC}{2}=\frac{16}{2}=8$

In right angled triangle ADE , we get

  $\begin{array}{l}{\left(AC\right)}^2={\left(AD\right)}^2+{\left(CD\right)}^2\\ {\left(17\right)}^2={\left(AD\right)}^2+{\left(8\right)}^2\\ {\left(AD\right)}^2=289-64\\ {\left(AD\right)}^2=225\\ AD=\sqrt{225}\\ AD=15\end{array}$

A right triangle formed is $8-15-17\Delta$

Area of one base:

  $\begin{array}{l}A=\frac{1}{2}\times BC\times AD\\ A=\frac{1}{2}\times 16\times 15\\ A=120\end{array}$

c.

To determine

To find: The total area of a right triangular prism.

Answer to Problem 3PSA

The total area of a right triangular prism is $790$ .

Explanation of Solution

Given information:

l =11, a = 17, b = 17 and c =16.

Formula used:

Area of rectangle: $A=l\times w$

l = length of rectangle

w= width of rectangle

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:

  

  $\begin{array}{l}A({▭}_1)=l\times a=11\times 17=187\\ A({▭}_2)=l\times b=11\times 17=187\\ A({▭}_3)=l\times c=11\times 16=176\end{array}$

Lateral Area:

  $\begin{array}{l}A=A({▭}_1)+A({▭}_2)+A({▭}_3)\\ A=187+187+176\\ A=550\end{array}$

Draw an altitude AD.

  $DC=\frac{BC}{2}=\frac{16}{2}=8$

In right angled triangle ADE , we get

  $\begin{array}{l}{\left(AC\right)}^2={\left(AD\right)}^2+{\left(CD\right)}^2\\ {\left(17\right)}^2={\left(AD\right)}^2+{\left(8\right)}^2\\ {\left(AD\right)}^2=289-64\\ {\left(AD\right)}^2=225\\ AD=\sqrt{225}\\ AD=15\end{array}$

A right triangle formed is $8-15-17\Delta$

Area of one base:

  $\begin{array}{l}A=\frac{1}{2}\times BC\times AD\\ A=\frac{1}{2}\times 16\times 15\\ A=120\end{array}$

Total Area = Lateral Area + 2 (Area of Base)

Total Area $=550+2\left(120\right)$

Total Area $=550+240$

Total Area $=790$