Chapter 12.2, Problem 3PSA

a.

To determine

To find: The pyramid is regular with rectangular dimensions as 40 and 30.

Answer to Problem 3PSA

The pyramid is not regular because all the lateral faces are not equal.

Explanation of Solution

Given information:

A pyramid has rectangular base with dimensions as 40 and 30.

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 altitude perpendicular to base.

The altitude AD can be calculated by applying Pythagoras Theorem.

In right angle triangle ABC , we get

  $\begin{array}{l}{\left(AD\right)}^2+{\left(DC\right)}^2={\left(AC\right)}^2\\ {\left(AD\right)}^2+{\left(7\right)}^2={\left(25\right)}^2\\ {\left(AD\right)}^2+49=625\\ {\left(AD\right)}^2=625-49\\ {\left(AD\right)}^2=576\\ AD=\sqrt{576}=24\end{array}$

The altitude AD drawn perpendicular divides the triangle face into two right triangles of $7-24-25$ family.

Lateral face is triangle.

Area of lateral face 1 $=\frac{1}{2}\times 14\times 24$

Area of lateral face 1 $=168$

In right angle triangle AFE , we get

  $\begin{array}{l}{\left(AF\right)}^2+{\left(FE\right)}^2={\left(AE\right)}^2\\ {\left(AF\right)}^2+{\left(15\right)}^2={\left(25\right)}^2\\ {\left(AF\right)}^2=625-225\\ {\left(AF\right)}^2=400\\ AF=\sqrt{400}=20\end{array}$

The altitude AF drawn perpendicular divides the triangle face into two right triangles of $20-15-25$ family.

Lateral face is triangle.

Area of lateral face 2 $=\frac{1}{2}\times 30\times 20$

Area of lateral face 2 $=300$

Area of lateral face 1 $\ne$ Area of lateral face 2

The pyramid is not regular because all the lateral faces are not equal.

b.

To determine

To calculate: The lateral area of pyramid with rectangular dimensions as 40 and 30.

Answer to Problem 3PSA

The lateral area of pyramid is $936$ .

Explanation of Solution

Given information:

A pyramid has rectangular base with dimensions as 40 and 30.

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 altitude perpendicular to base.

The altitude AD can be calculated by applying Pythagoras Theorem.

In right angle triangle ABC , we get

  $\begin{array}{l}{\left(AD\right)}^2+{\left(DC\right)}^2={\left(AC\right)}^2\\ {\left(AD\right)}^2+{\left(7\right)}^2={\left(25\right)}^2\\ {\left(AD\right)}^2+49=625\\ {\left(AD\right)}^2=625-49\\ {\left(AD\right)}^2=576\\ AD=\sqrt{576}=24\end{array}$

The altitude AD drawn perpendicular divides the triangle face into two right triangles of $7-24-25$ family.

Lateral face is triangle.

Area of lateral face 1 $=\frac{1}{2}\times 14\times 24$

Area of lateral face 1 $=168$

In right angle triangle AFE , we get

  $\begin{array}{l}{\left(AF\right)}^2+{\left(FE\right)}^2={\left(AE\right)}^2\\ {\left(AF\right)}^2+{\left(15\right)}^2={\left(25\right)}^2\\ {\left(AF\right)}^2=625-225\\ {\left(AF\right)}^2=400\\ AF=\sqrt{400}=20\end{array}$

The altitude AF drawn perpendicular divides the triangle face into two right triangles of $20-15-25$ family.

Lateral face is triangle.

Area of lateral face 2 $=\frac{1}{2}\times 30\times 20$

Area of lateral face 2 $=300$

Lateral area of pyramid = (2 $\times$ Area of lateral face 1) + (2 $\times$ Area of lateral face 2)

Lateral area of pyramid $=336+600$

Lateral area of pyramid $=936$

c.

To determine

To find: The total area of pyramid with rectangular dimensions as 40 and 30.

Answer to Problem 3PSA

The total area of pyramid is $1356$ .

Explanation of Solution

Given information:

A pyramid has rectangular base with dimensions as 40 and 30.

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

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

l = length of rectangle

w= width of rectangle Total Area = Lateral Area + Area of Rectangular Base

Calculation:

  

Draw altitude perpendicular to base.

The altitude can be calculated by applying Pythagoras Theorem.

In right angle triangle ABC , we get

  $\begin{array}{l}{\left(AD\right)}^2+{\left(DC\right)}^2={\left(AC\right)}^2\\ {\left(AD\right)}^2+{\left(7\right)}^2={\left(25\right)}^2\\ {\left(AD\right)}^2+49=625\\ {\left(AD\right)}^2=625-49\\ {\left(AD\right)}^2=576\\ AD=\sqrt{576}=24\end{array}$

The altitude AD drawn perpendicular divides the triangle face into two right triangles of $7-24-25$ family.

Lateral face is triangle.

Area of lateral face 1 $=\frac{1}{2}\times 14\times 24$

Area of lateral face 1 $=168$

In right angle triangle AFE , we get

  $\begin{array}{l}{\left(AF\right)}^2+{\left(FE\right)}^2={\left(AE\right)}^2\\ {\left(AF\right)}^2+{\left(15\right)}^2={\left(25\right)}^2\\ {\left(AF\right)}^2=625-225\\ {\left(AF\right)}^2=400\\ AF=\sqrt{400}=20\end{array}$

The altitude AF drawn perpendicular divides the triangle face into two right triangles of $20-15-25$ family.

Lateral face is triangle.

Area of lateral face 2 $=\frac{1}{2}\times 30\times 20$

Area of lateral face 2 $=300$

Lateral area of pyramid = ( 2 $\times$ Area of lateral face 1) + (2 $\times$ Area of lateral face 2)

Lateral area of pyramid $=336+600$

Lateral area of pyramid $=936$

Area of rectangular base $=14\times 30$

Area of rectangular base $=420$

Total Area = Lateral Area + Area of Rectangular Base

Total Area $=936+420$

Total Area $=1356$