Chapter 12, Problem 8CR

a.

To determine

To calculate: The area of trapezoid UVWX .

Answer to Problem 8CR

The area of trapezoid UVWX is $168$ .

Explanation of Solution

Given information:

Side UV=XW = 10,

Side UX = 15,

h = 8.

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 trapezoid: $A=\frac{1}{2}\times (b_1+b_2)\times h$

  $b_1$ and $b_2$ are bases of trapezoid.

h = height of trapezoid.

Calculation:

  

Side VA can be calculated by applying Pythagoras Theorem.

In right angled triangle VAU , we get

  $\begin{array}{l}{\left(VU\right)}^2={\left(VA\right)}^2+{\left(UA\right)}^2\\ {\left(10\right)}^2={\left(VA\right)}^2+{\left(8\right)}^2\\ {\left(VA\right)}^2=100-64\\ {\left(VA\right)}^2=36\\ VA=\sqrt{36}=6\end{array}$

△XBW is similar to △VAU.

  $\begin{array}{l}BW=VA=6\\ b_1=UX=15\\ b_2=VA+AB+BW\\ AB=UX\\ b_2=VA+UX+BW\\ b_2=6+15+6\\ b_2=VW=27\end{array}$

Area of trapezoid:

  $\begin{array}{l}A=\frac{1}{2}\times (b_1+b_2)\times h\\ A=\frac{1}{2}\times (UX+VW)\times h\\ A=\frac{1}{2}\times (15+27)\times 8\\ A=\frac{1}{2}\times (42)\times 8\\ A=168\end{array}$

b.

To determine

To find: The area of equilateral triangle ABC with sides as 10.

Answer to Problem 8CR

The area of equilateral triangle ABC is $43.30$ .

Explanation of Solution

Given information:

In △ABC ,

Side AB=BC=CA = 10.

Formula used:

Area of equilateral triangle: $A=\frac{s^2}{4}\sqrt{3}$

s = side of equilateral triangle

Calculation:

  

Area of equilateral triangle ABC :

  $\begin{array}{l}s=AB=10\\ A=\frac{{\left(10\right)}^2}{4}\sqrt{3}\\ A=\frac{100}{4}\sqrt{3}\\ A=25\sqrt{3}\\ A=43.30\end{array}$

c.

To determine

To calculate: The area of circle.

Answer to Problem 8CR

The area of circle is $169\pi$ .

Explanation of Solution

Given information:

In a circle with center O ,

Side AC = 24,

Side OB = 5.

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 a circle: $A=\pi r^2$

r = radius of circle

Calculation:

  

Side OA can be calculated by applying Pythagoras Theorem.

In right angled triangle OBA , we get

  $\begin{array}{l}{\left(OA\right)}^2={\left(OB\right)}^2+{\left(BA\right)}^2\\ BA=\frac{AC}{2}=\frac{24}{2}=12\\ {\left(OA\right)}^2={\left(5\right)}^2+{\left(12\right)}^2\\ {\left(OA\right)}^2=25+144\\ {\left(OA\right)}^2=169\\ OA=\sqrt{169}=13\end{array}$

Area of a circle:

  $\begin{array}{l}r=OA=13\\ A=\pi r^2\\ A=\pi {\left(13\right)}^2\\ A=169\pi \end{array}$