Chapter 11, Problem 25RP

a.

To determine

To calculate: The area of shaded region with sides of larger triangles as 8 and 10.

Answer to Problem 25RP

The area of shaded region is $33.6$ .

Explanation of Solution

Given information:

Sides of larger triangle are 8 and 10.

Sides of smaller triangle are x and 4.

Formula used:

The below property is used:

The corresponding sides of similar triangles are congruent.

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

b = base of triangle

h = height of triangle

Calculation:

  

The larger triangle and smaller triangle are similar triangles.

The corresponding sides of similar triangles are congruent.

  $\begin{array}{l}\frac{4}{x}=\frac{10}{8}\\ 10x=32\\ x=\frac{32}{10}=3.2\end{array}$

  $\begin{array}{l}{A}_{LongerTriangle}=\frac{1}{2}\times 8\times 10\\ A_{LongerTriangle}=40\\ A_{SmallerTriangle}=\frac{1}{2}\times 3.2\times 4\\ A_{SmallerTriangle}=6.4\end{array}$

The shaded area is difference between longer triangle and smaller triangle.

  $\begin{array}{l}{A}_{ShadedArea}=A_{LongerTriangle}-A_{SmallerTriangle}\\ A_{ShadedArea}=40-6.4\\ A_{ShadedArea}=33.6\end{array}$

b.

To determine

To find: The area of shaded region with radius of circle as 3.

Answer to Problem 25RP

The area of shaded region is $18.49$ .

Explanation of Solution

Given information:

Radius of circle r = 3.

Formula used:

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

s = side of equilateral triangle.

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

r = radius of circle

Calculation:

  

  ${30}^{\circ }{60}^{\circ }{90}^{\circ }\Delta$ is formed with radius of circle and triangle.

Side opposite to ${30}^{\circ }$ angle = 3,

  $\begin{array}{l}\tan {30}^{\circ }=\frac{oppositeside}{adjacentside}\\ \frac{1}{\sqrt{3}}=\frac{3}{adjacentside}\\ adjacentside=3\sqrt{3}\end{array}$

Side opposite to ${60}^{\circ }$ angle $=3\sqrt{3}$ ,

Side of equilateral triangle $=6\sqrt{3}$ .

  $\begin{array}{l}{A}_{\text{Equilateral }Triangle}=\frac{s^2}{4}\sqrt{3}\\ A_{\text{Equilateral }Triangle}=\frac{{\left(6\sqrt{3}\right)}^2}{4}\sqrt{3}\\ A_{\text{Equilateral }Triangle}=27\sqrt{3}\\ A_{Circle}=\pi r^2\\ A_{Circle}=\pi {\left(3\right)}^2\\ A_{Circle}=9\pi \end{array}$

The shaded area is difference between equilateral triangle and circle.

  $\begin{array}{l}{A}_{ShadedArea}=A_{\text{Equilateral }Triangle}-A_{Circle}\\ A_{ShadedArea}=27\sqrt{3}-9\pi \\ A_{ShadedArea}=18.49\end{array}$

c.

To determine

To find: The area of shaded region which contains triangle of hypotenuse 5.

Answer to Problem 25RP

The area of shaded region is $2.85$ .

Explanation of Solution

Given information:

A triangle of hypotenuse 5.

Formula used:

The below theorems are used:

Two tangent theorem states that if two tangent segments are drawn to one circle from the same external point, then they are congruent.

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

r = radius of circle

Calculation:

  

By two tangent theorem, we get

  $\begin{array}{l}AB=AF=2,\\ ED=EF=3,\\ BC=DC=x.\end{array}$

Side x can be calculated by applying Pythagoras Theorem.

In right angled triangle ACE , we get

  $\begin{array}{l}{\left(AE\right)}^2={\left(AC\right)}^2+{\left(CE\right)}^2\\ {\left(5\right)}^2={\left(2+x\right)}^2+{\left(x+3\right)}^2\\ x=1\end{array}$

The circumscribed triangle is $3-4-5\Delta$

  

The shaded area is difference between triangle and circle.

  $\begin{array}{l}{A}_{ShadedArea}=A_{Triangle}-A_{Circle}\\ A_{ShadedArea}=6-\pi \\ A_{ShadedArea}=2.85\end{array}$