Chapter 11.6, Problem 14PSB

a.

To determine

To calculate:Thearea of shaded part which contains circle and square.

Answer to Problem 14PSB

The area of shaded part is $21.46{\text{ units}}^2$ .

Explanation of Solution

Given information:

A square with side as 10 and circle with diameter as 10.

Formula used:

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

r = radius of a circle.

Area of a square: $A=s^2$

s = side

  $A_{shadedpart}=A_{square}-A_{circle}$

Calculation:

  

  $\begin{array}{l}d=10\\ r=\frac{d}{2}=\frac{10}{2}=5\end{array}$

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

  $A=\pi {(5)}^2=25\pi$

Area of a square: $A=s^2$

  $A={(10)}^2=100$

  $\begin{array}{l}{A}_{shadedpart}=A_{square}-A_{circle}\\ A_{shadedpart}=100-25\pi \\ A_{shadedpart}=21.46{\text{ units}}^2\end{array}$

b.

To determine

To calculate: The area of shaded part which contains circle and triangle.

Answer to Problem 14PSB

The area of shaded part is $25\sqrt{3}-\frac{25}{3}\pi$ .

Explanation of Solution

Given information:

An equilateral triangle with each side10.

Formula used:

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

r = radius of a circle.

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

s = side of equilateral triangle.

  $A_{shadedpart}=A_{triangle}-A_{circle}$

Calculation:

  

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

  $A=\frac{{(10)}^2}{4}\sqrt{3}=25\sqrt{3}$

ΔBOD is $30°{60}^{\circ }{90}^{\circ }\Delta$

  $BD=5,soOD=\frac{5}{\sqrt{3}}=\frac{5\sqrt{3}}{3}$

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

  $A=\pi {(\frac{5\sqrt{3}}{3})}^2=\frac{25}{3}\pi$

  $\begin{array}{l}{A}_{shadedpart}=A_{triangle}-A_{circle}\\ A_{shadedpart}=25\sqrt{3}-\frac{25}{3}\pi \end{array}$

c.

To determine

To find: The area of shaded part which contains sectors and square.

Answer to Problem 14PSB

The area of shaded part is $100-25\pi$ .

Explanation of Solution

Given information:

A square with side as 10 and radius as 5.

Formula used:

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

r = radius of a circle.

Area of Sector $=\left(\frac{measureofarc}{360^o}\right)\pi r^2$

r = radius of a circle.

  $A_{shadedpart}=A_{square}-A_{sectors}$

Calculation:

  

Area of square: $A=s^2$

  $A={(10)}^2=100$

  $r=\frac{1}{2}*side=\frac{10}{2}=5$

  $\begin{array}{l}{A}_{sector}=(\frac{measureofarc}{360^o})\pi r^2\\ $ =(\frac{90^o}{360^o})\pi {(5)}^2$=\frac{1}{4}*25\pi \\ =\frac{25\pi }{4}\end{array}$

There are 4 sectors.

  $\begin{array}{l}{A}_{sectors}=4*A_{sector}\\ =4*\frac{25\pi }{4}\\ =25\pi \end{array}$

  $\begin{array}{l}{A}_{shadedpart}=A_{square}-A_{sectors}\\ A_{shadedpart}=100-25\pi \end{array}$

d.

To determine

To find the area of shaded part which contains sectors and triangle.

Answer to Problem 14PSB

The area of shaded part is $25\sqrt{3}-\frac{25\pi }{2}$ .

Explanation of Solution

Given information:

A triangle with side as 10.

Formula used:

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

s = side of equilateral triangle.

Area of Sector $=\left(\frac{measureofarc}{360^o}\right)\pi r^2$

r = radius of a circle.

  $A_{Shadedpart}=A_{Square}-A_{Sectors}$

Calculation:

  

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

  $A=\frac{{(10)}^2}{4}\sqrt{3}=25\sqrt{3}$

All the sectors are congruent. The radius of each sector is 5.(Two tangent theorem)

  $\begin{array}{l}r=\frac{1}{2}*side=\frac{10}{2}=5\\ A_{sector}=\left(\frac{measureofarc}{360^o}\right)\pi r^2\\ $ =(\frac{60^o}{360^o})\pi {(5)}^2$=\frac{1}{6}*25\pi \\ =\frac{25\pi }{6}\end{array}$

There are 3 sectors.

  $\begin{array}{l}{A}_{sectors}=3*A_{sector}\\ =3*\frac{25\pi }{6}\\ =\frac{25\pi }{2}\end{array}$

  $\begin{array}{l}{A}_{shadedpart}=A_{triangle}-A_{sectors}\\ A_{shadedpart}=25\sqrt{3}-\frac{25\pi }{2}\end{array}$

e.

To determine

To find: The area of shaded part which contains circle and square.

Answer to Problem 14PSB

The area of shaded part is $50\pi -100$ .

Explanation of Solution

Given information:

A square with side as 10.

Formula used:

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

r = radius of a circle.

Area of a square: $A=s^2$

s = side

  $A_{shadedpart}=A_{square}-A_{circle}$

Calculation:

  

  $\begin{array}{l}Diagonal=side\sqrt{2}\\ Diagonal=10\sqrt{2}\\ Diameter=Diagonal=10\sqrt{2}\\ d=10\sqrt{2}\\ r=\frac{d}{2}=\frac{10\sqrt{2}}{2}=5\sqrt{2}\end{array}$

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

  $A=\pi {(5\sqrt{2})}^2=50\pi$

Area of a square: $A=s^2$

  $A={(10)}^2=100$

  $\begin{array}{l}{A}_{shadedpart}=A_{circle}-A_{square}\\ A_{shadedpart}=50\pi -100\end{array}$

f.

To determine

To calculate: The area of shaded part which contains circle and triangle.

Answer to Problem 14PSB

The area of shaded part is $150\sqrt{3}-75\pi$ .

Explanation of Solution

Given information:

An equilateral triangle with each side10.

Formula used:

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

r = radius of a circle.

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

s = side of equilateral triangle.

  $A_{Shadedpart}=A_{Hexagon}-A_{circle}$

Calculation:

  

A hexagon can be divided into six equilateral triangles.

  $\begin{array}{l}{A}_{hexagon}=6(\frac{s^2}{4}\sqrt{3})\\ A_{hexagon}=6(\frac{{(10)}^2}{4}\sqrt{3})=6(25\sqrt{3})=150\sqrt{3}\end{array}$

Radius of circle is equal to apothem hexagon which is equal to altitude of a triangle.

The altitude divides the equilateral triangle into two $30°{60}^{\circ }{90}^{\circ }\Delta \text{'}s$ .

  $\begin{array}{l}Thebase=\frac{1}{2}*10\\ Altitude=base\sqrt{3}=5\sqrt{3}\\ Altitude\cong radius=5\sqrt{3}\end{array}$

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

  $A=\pi {(5\sqrt{3})}^2=75\pi$

  $\begin{array}{l}{A}_{shadedpart}=A_{hexagon}-A_{circle}\\ A_{shadedpart}=150\sqrt{3}-75\pi \end{array}$