7.3: Solve Geometry Applications — Circles & Irregular Figures

In this section, we’ll continue working with geometry applications. We will add several new formulas to our collection of formulas. To help you as you do the examples and exercises in this section, we will show the problem-solving strategy for geometry applications here.

Problem-solving strategy for geometry applications:

  1. Read the problem and make sure you understand all the words and ideas. Draw the figure and label it with the given information.
  2. Identify what you are looking for.
  3. Name what you are looking for. Choose a variable to represent that quantity.
  4. Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
  5. Solve the equation using good algebra techniques.
  6. Check the answer to the problem and make sure it makes sense.
  7. Answer the question with a complete sentence.

Use the Properties of Circles

We will refer to the properties of circles as we use them to solve applications.

An image of a circle is shown. There is a line drawn through the widest part at the centre of the circle with a red dot indicating the centre of the circle. The line is labeled d. The two segments from the centre of the circle to the outside of the circle are each labeled r.
Figure 7.3.1

Properties of circles:

  • r is the length of the radius.
  • d is the length of the diametre.
  • [latex]d=2r[/latex]
  • Circumference is the perimeter of a circle. The formula for circumference is [latex]C=2\pi r[/latex]
  • The formula for the area of a circle is [latex]A=\pi {r}^{2}[/latex]

(see Figure 7.3.1)

Remember: We approximate [latex]\pi[/latex] with 3.14 or [latex]\tfrac{22}{7}[/latex] depending on whether the radius of the circle is given as a decimal or a fraction.

If you use the [latex]\pi[/latex] key on your calculator to do the calculations in this section, your answers will be slightly different from the answers shown. That is because the [latex]\pi[/latex]  key uses more than two decimal places.

Example A

A circular sandbox has a radius of 2.5 feet. Find the a) circumference and b) area of the sandbox.

Step 1: Read the problem. Draw the figure and label it with the given information (see Figure 7.3.2).

 

A circle with a radius of 2.5 ft.
Figure 7.3.2

a. Circumference

Step 2: Identify what you are looking for.

The circumference of the circle.

Step 3: Name. Choose a variable to represent it.

Let c = circumference of the circle

Step 4: Translate. Write the appropriate formula. Substitute. (Use 3.14 for [latex]\pi[/latex])

[latex]\begin{equation}\begin{split}C&=2\pi r \\ C&=2(3.14)(2.5) \end{split}\end{equation}[/latex]

Step 5: Solve the equation.

[latex]C\approx 15[/latex]

Step 6: Check. Does this answer make sense?

Yes. If we draw a square around the circle, its sides would be 5 ft (twice the radius), so its perimeter would be 20 ft (see Figure 7.3.3). This is slightly more than the circle’s circumference, 15.7 ft.

 

A circle with a radius of 2.5 ft surrounded by a square with 5 ft sides.
Figure 7.3.3

Step 7: Answer the question.

The circumference of the sandbox is 15.7 feet.

 

b. Area

Step 2: Identify what you are looking for.

The area of the circle.

Step 3: Name. Choose a variable to represent it.

Let A = the area of the circle

Step 4: Translate. Write the appropriate formula. Substitute. (Use 3.14 for [latex]\pi[/latex])

[latex]\begin{equation}\begin{split}A&=\pi r^2 \\ A&\approx (3.14)2.5^2 \end{split}\end{equation}[/latex]

Step 5: Solve the equation.

[latex]A\approx 19.625[/latex]

Step 6: Check. Does this answer make sense?

Yes. If we draw a square around the circle, its sides would be 5 ft, as shown in part a). So the area of the square would be 25 sq. ft. This is slightly more than the circle’s area, 19.625 sq. ft.

Step 7: Answer the question.

The area of the circle is 19.625 square feet.

Exercise 1

A circular mirror has a radius of 5 inches. Find the a) circumference and b) area of the mirror.

Exercise 1 Answers

  1. 31.4 in.
  2. 78.5 sq. in.

Exercise 2

A circular spa has a radius of 4.5 feet. Find the a) circumference and b) area of the spa.

Exercise 2 Answers

  1. 28.26 ft
  2. 63.585 sq. ft

We usually see the formula for circumference in terms of the radius r of the circle:

[latex]C=2\pi r[/latex]

However, since the diametre of a circle is two times the radius, we could write the formula for the circumference in terms [latex]\text{of}\phantom{\rule{0.2em}{0ex}}d[/latex].

[latex]\begin{array}{cccc}& & & C=2\pi r\hfill \\ \text{Using the commutative property, we get}\hfill & & & C=\pi \cdot2r\hfill \\ \text{Then substituting}\phantom{\rule{0.2em}{0ex}}d=2r\hfill & & & C=\pi \cdot\text{d}\hfill \\ \text{So}\hfill & & & C=\pi d\hfill \end{array}[/latex]

We will use this form of the circumference when we’re given the length of the diametre instead of the radius.

Example B

A circular table has a diametre of four feet. What is the circumference of the table?

Step 1: Read the problem. Draw the figure and label it with the given information (see Figure 7.3.4).

 

A circular wooden table with a diametre of 4 ft.
Figure 7.3.4

Step 2: Identify what you are looking for.

The circumference of the table.

Step 3: Name. Choose a variable to represent it.

Let c = the circumference of the table

Step 4: Translate. Write the appropriate formula. Substitute. (Use 3.14 for [latex]\pi[/latex])

[latex]\begin{equation}\begin{split}C&=\pi d \\ C&\approx (3.14)(4) \end{split}\end{equation}[/latex]

Step 5: Solve the equation.

[latex]C\approx 12.56[/latex]

Step 6: Check.

If we put a square around the circle, its side would be 4 (see Figure 7.3.5).

The perimeter would be 16. It makes sense that the circumference of the circle, 12.56, is a little less than 16.

 

A circle with a diametre of 4 ft surrounded by a square with 4 ft sides.
Figure 7.3.5

Step 7: Answer the question.

The diametre of the table is 12.56 square feet.

Exercise 3

Find the circumference of a circular fire pit whose diametre is 5.5 feet.

Exercise 3 Answer

17.27 ft

Exercise 4

If the diametre of a circular trampoline is 12 feet, what is its circumference?

Exercise 4 Answer

37.68 ft

Example C

Find the diametre of a circle with a circumference of 47.1 centimetres.

Step 1: Read the problem. Draw the figure and label it with the given information (see Figure 7.3.6).

 

A circle with a diametre of d and a circumference of 47.1 cm.
Figure 7.3.6

Step 2: Identify what you are looking for.

The diametre of the circle.

Step 3: Name. Choose a variable to represent it.

Let d = the diametre of the circle

Step 4: Translate. Write the appropriate formula. Substitute. (Use 3.14 for [latex]\pi[/latex])

[latex]\begin{equation}\begin{split}C&=\pi d \\ 47.1&\approx (3.14)d \end{split}\end{equation}[/latex]

Step 5: Solve.

[latex]\begin{equation}\begin{split}\dfrac{47.1}{3.14} &\approx \dfrac{3.14d}{3.14} \\ 15&\approx d \end{split}\end{equation}[/latex]

Step 6: Check.

[latex]\begin{equation}\begin{split}C&\approx \pi d \\ 47.1&=(3.14)(15) ? \\ 47.1&=47.1 \text{ ✔} \end{split}\end{equation}[/latex]

Step 7: Answer the question.

The diametre of the circle is approximately 15 centimetres.

Exercise 5

Find the diametre of a circle with a circumference of 94.2 centimetres.

Exercise 5 Answer

30 cm

Exercise 6

Find the diametre of a circle with a circumference of 345.4 feet.

Exercise 6 Answer

110 ft

Find the Area of Irregular Figures

So far, we have found the areas for rectangles, triangles, trapezoids, and circles. An irregular figure is a figure that is not a standard geometric shape. Its area cannot be calculated using any of the standard area formulas.

However, some irregular figures are made up of two or more standard geometric shapes. To find the area of one of these irregular figures, we can split it into figures whose formulas we know and then add the areas of the figures.

Example D

Find the area of the shaded region in Figure 7.3.7.

 

An image of an attached horizontal rectangle and a vertical rectangle is shown. The top is labeled 12, the side of the horizontal rectangle is labeled 4. The side is labeled 10, the width of the vertical rectangle is labeled 2.
Figure 7.3.7

The given figure is irregular, but we can break it into two rectangles. The area of the shaded region will be the sum of the areas of both rectangles (see Figure 7.3.8).

 

An image of an attached horizontal rectangle and a vertical rectangle is shown. The top is labeled 12, the side of the horizontal rectangle is labeled 4. The side is labeled 10, the width of the vertical rectangle is labeled 2.
Figure 7.3.8

The blue rectangle has a width of 12 and a length of 4. The red rectangle has a width of 2, but its length is not labelled. The right side of the figure is the length of the red rectangle plus the length of the blue rectangle. Since the right side of the blue rectangle is 4 units long, the length of the red rectangle must be 6 units (see Figure 7.3.9).

 

An image of a blue horizontal rectangle attached to a red vertical rectangle is shown. The top is labeled 12, the side of the blue rectangle is labeled 4. The whole side is labeled 10, the blue portion is labeled 4 and the red portion is labeled 6. The width of the red rectangle is labeled 2.
Figure 7.3.9

[latex]\begin{equation}\begin{split}{A}_{\text{figure}}&={A}_{\textcolor{blue}{\text{rectangle}}}+{A}_{\textcolor{red}{\text{rectangle}}} \\ {A}_{\text{figure}}&=\textcolor{blue}{bh}+\textcolor{red}{bh} \\ {A}_{\text{figure}}&=\textcolor{blue}{12 \cdot 4}+\textcolor{red}{2 \cdot 6} \\ {A}_{\text{figure}}&=\textcolor{blue}{48}+\textcolor{red}{12} \\ {A}_{\text{figure}}&= 60 \end{split}\end{equation}[/latex]

The area of the figure is 60 square units.

Is there another way to split this figure into two rectangles? Try it, and make sure you get the same area.

Exercise 7

Find the area of each shaded region in Figure 7.3.10.

 

A blue geometric shape is shown. It looks like a horizontal rectangle attached to a vertical rectangle. The top is labeled as 8, the width of the horizontal rectangle is labeled as 2. The side is labeled as 6, the width of the vertical rectangle is labeled as 3.
Figure 7.3.10

Exercise 7 Answer

28 sq. units

Exercise 8

Find the area of each shaded region in Figure 7.3.11.

 

A blue geometric shape is shown. It looks like a horizontal rectangle attached to a vertical rectangle. The top is labeled as 14, the width of the horizontal rectangle is labeled as 5. The side is labeled as 10, the width of the missing space is labeled as 6.
Figure 7.3.11

Exercise 8 Answer

110 sq. units

Example E

Find the area of the shaded region in Figure 7.3.12.

 

A blue geometric shape is shown. It looks like a rectangle with a triangle attached to the top on the right side. The left side is labeled 4, the top 5, the bottom 8, the right side 7.
Figure 7.3.12

Step 1: We can break this irregular figure into a triangle and a rectangle. The area of the figure will be the sum of the areas of the triangle and rectangle.

The rectangle has a length of 8 units and a width of 4 units.

We need to find the base and height of the triangle.

Since both sides of the rectangle are 4, the vertical side of the triangle is 3, which is [latex]7-4[/latex].

The length of the rectangle is 8, so the base of the triangle will be 3, which is [latex]8-4[/latex] (see Figure 7.3.13).

 

A geometric shape is shown. It is a blue rectangle with a red triangle attached to the top on the right side. The left side is labeled 4, the top 5, the bottom 8, the right side 7. The right side of the rectangle is labeled 4. The right side and bottom of the triangle are labeled 3.
Figure 7.3.13

Step 2: Now, we can add the areas to find the area of the irregular figure.

[latex]\begin{equation}\begin{split}{A}_{\text{figure}}&={A}_{\textcolor{blue}{\text{rectangle}}}+{A}_{\textcolor{red}{\text{triangle}}} \\ {A}_{\text{figure}}&=\textcolor{blue}{lw}+\dfrac{1}{2}\textcolor{red}{bh} \\ {A}_{\text{figure}}&=\textcolor{blue}{8 \cdot 4}+\dfrac{1}{2}\cdot \textcolor{red}{3 \cdot 3} \\ {A}_{\text{figure}}&=\textcolor{blue}{32}+\textcolor{red}{4.5} \\ {A}_{\text{figure}}&=36.5\text{ sq. units}\end{split}\end{equation}[/latex]

The area of the figure is 36.5 square units.

Exercise 9

Find the area of each shaded region in Figure 7.3.14.

 

A blue geometric shape is shown. It looks like a rectangle with a triangle attached to the lower right side. The base of the rectangle is labeled 8, the height of the rectangle is labeled 4. The distance from the top of the rectangle to where the triangle begins is labeled 3, the top of the triangle is labeled 3.
Figure 7.3.14

Exercise 9 Answer

36.5 sq. units

Exercise 10

Find the area of each shaded region in Figure 7.3.15.

 

A blue geometric shape is shown. It looks like a rectangle with an equilateral triangle attached to the top. The base of the rectangle is labeled 12, each side is labeled 5. The base of the triangle is split into two pieces, each labeled 2.5.
Figure 7.3.15

Exercise 10 Answer

70 sq. units

Example F

A high school track is shaped like a rectangle with a semi-circle (half a circle) on each end (see Figure 7.3.16). The rectangle has a length of 105 metres and a width of 68 metres. Find the area enclosed by the track. Round your answer to the nearest hundredth.

 

A track is shown, shaped like a rectangle with a semi-circle attached to each side.
Figure 7.3.16

We will break the figure into a rectangle and two semi-circles. The area of the figure will be the sum of the areas of the rectangle and the semi-circles.

The rectangle has a length of 105 m and a width of 68 m. The semi-circles have a diametre of 68 m, so each has a radius of 34 m (see Figure 7.3.17).

 

A blue geometric shape is shown. It looks like a rectangle with a semi-circle attached to each side. The base of the rectangle is labeled 105 m. The height of the rectangle and diametre of the circle on the left is labeled 68 m.
Figure 7.3.17

[latex]\begin{equation}\begin{split}{A}_{\text{figure}}&={A}_{\textcolor{blue}{\text{rectangle}}}+{A}_{\textcolor{red}{\text{semi-circles}}} \\ {A}_{\text{figure}}&=\textcolor{blue}{bh}+\textcolor{red}{2(\dfrac{1}{2} \pi \cdot r^2)} \\ {A}_{\text{figure}} & \approx \textcolor{blue}{105 \cdot 68} + \textcolor{red}{2(\dfrac{1}{2} \cdot 3.14 \cdot 34^2)} \\ {A}_{\text{figure}} & \approx \textcolor{blue}{7140}+\textcolor{red}{3629.84} \\ {A}_{\text{figure}} & \approx 10\,769.84 \text{ square metres}\end{split}\end{equation}[/latex]

The area of the track is 10769.84 m2.

Exercise 11

Find the area of Figure 7.3.18.

 

A shape is shown. It is a blue rectangle with a portion of the rectangle missing. There is a red circle the same height as the rectangle attached to the missing side of the rectangle. The top of the rectangle is labeled 15, the height is labeled 9.
Figure 7.3.18

Exercise 11 Answer

103.2 sq. units

Exercise 12

Find the area of Figure 7.3.19.

 

A blue geometric shape is shown. It appears to be two trapezoids with a semicircle at the top. The base of the semicircle is labeled 5.2. The height of the trapezoids is labeled 6.5. The combined base of the trapezoids is labeled 3.3.
Figure 7.3.19

Exercise 12 Answer

38.24 sq. units

Additional online resources:

Key Concepts

Problem Solving Strategy for Geometry Applications

    1. Read the problem and make sure you understand all the words and ideas. Draw the figure and label it with the given information.
    2. Identify what you are looking for.
    3. Name what you are looking for. Choose a variable to represent that quantity.
    4. Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
    5. Solve the equation using good algebra techniques.
    6. Check the answer to the problem and make sure it makes sense.
    7. Answer the question with a complete sentence.

Properties of Circles

A circle's diametre is equivalent to 2 times its radius.
Figure 7.3.20
  • [latex]d=2r[/latex]
  • Circumference: [latex]C=2\pi[/latex] r or [latex]C=\pi d[/latex]
  • Area: A[latex]=\pi {r}^{2}[/latex]
  • (see Figure 7.3.20)

Glossary

  • Irregular figure — An irregular figure is a figure that is not a standard geometric shape. Its area cannot be calculated using any of the standard area formulas.

7.3: Practice Questions

1. Use the Properties of Circles

In the following exercises, solve using the properties of circles.

  1. The lid of a paint bucket is a circle with a radius of 7 inches. Find the a) circumference and b) area of the lid.
  2. An extra-large pizza is a circle with a radius of 8 inches. Find the a) circumference and b) area of the pizza.
  3. A farm sprinkler spreads water in a circle with a radius of 8.5 feet. Find the a) circumference and b) area of the watered circle.
  4. A circular rug has a radius of 3.5 feet. Find the a) circumference and b) area of the rug.
  5. A reflecting pool is in the shape of a circle with a diametre of 20 feet. What is the circumference of the pool?
  6. A turntable is a circle with a diametre of 10 inches. What is the circumference of the turntable?
  7. A circular saw has a diametre of 12 inches. What is the circumference of the saw?
  8. A round coin has a diametre of 3 centimetres. What is the circumference of the coin?
  9. A barbecue grill is a circle with a diametre of 2.2 feet. What is the circumference of the grill?
  10. The top of a pie tin is a circle with a diametre of 9.5 inches. What is the circumference of the top?
  11. A circle has a circumference of 163.28 inches. Find the diametre.
  12. A circle has a circumference of 59.66 feet. Find the diametre.
  13. A circle has a circumference of 17.27 metres. Find the diametre.
  14. A circle has a circumference of 80.07 centimetres. Find the diametre.

In the following exercises, find the radius of the circle with the given circumference.

  1. A circle has a circumference of 150.72 feet.
  2. A circle has a circumference of 251.2 centimetres.
  3. A circle has a circumference of 40.82 miles.
  4. A circle has a circumference of 78.5 inches.

2. Find the Area of Irregular Figures

In the following exercises, find the area of the irregular figure. Round your answers to the nearest hundredth.

  1. A geometric shape is shown. It is a horizontal rectangle attached to a vertical rectangle. The top is labeled 6, the height of the horizontal rectangle is labeled 2, the distance from the edge of the horizontal rectangle to the start of the vertical rectangle is 4, the base of the vertical rectangle is 2, the right side of the shape is 4.
  2. A geometric shape is shown. It is an L-shape. The base is labeled 10, the right side 1, the top and left side are each labeled 4.
  3. A geometric shape is shown. It is a sideways U-shape. The top is labeled 6, the left side is labeled 6. An inside horizontal piece is labeled 3. Each of the vertical pieces on the right are labeled 2.
  4. A geometric shape is shown. It is a U-shape. The base is labeled 7. The right side is labeled 5. The two horizontal lines at the top and the vertical line on the inside are all labeled 3.
  5. A geometric shape is shown. It is a rectangle with a triangle attached to the bottom left side. The top is labeled 4. The right side is labeled 10. The base is labeled 9. The vertical line from the top of the triangle to the top of the rectangle is labeled 3.
  6. A trapezoid is shown. The bases are labeled 5 and 10, the height is 5.
  7. Two triangles are shown. They appear to be right triangles. The bases are labeled 3, the heights 4, and the longest sides 5.
  8. A geometric shape is shown. It appears to be composed of two triangles. The shared base of both triangles is 8, the heights are both labeled 6.
  9. A geometric shape is shown. It is a trapezoid attached to a triangle. The base of the triangle is labeled 6, the height is labeled 5. The height of the trapezoid is 6, one base is 3.
  10. A geometric shape is shown. It is a rectangle with a triangle and another rectangle attached. The left side is labeled 8, the bottom is 8, the right side is 13, and the width of the smaller rectangle is 2.
  11. A geometric shape is shown. It is a rectangle with a triangle and another rectangle attached. The left side is labeled 12, the right side 7, the base 6. The width of the smaller rectangle is labeled 1.
  12. A geometric shape is shown. It is a rectangle attached to a semi-circle. The base of the rectangle is labeled 5, the height is 7.
  13. A geometric shape is shown. It is a rectangle attached to a semi-circle. The base of the rectangle is labeled 10, the height is 6. The portion of the rectangle on the left of the semi-circle is labeled 5, the portion on the right is labeled 2.
  14. A geometric shape is shown. A triangle is attached to a semi-circle. The base of the triangle is labeled 4. The height of the triangle and the diametre of the circle are 8.
  15. A geometric shape is shown. A triangle is attached to a semi-circle. The height of the triangle is labeled 4. The base of the triangle, also the diametre of the semi-circle, is labeled 4.
  16. A geometric shape is shown. It is a rectangle attached to a semi-circle. The base of the rectangle is labeled 5, the height is 7.
  17. A geometric shape is shown. A trapezoid is shown with a semi-circle attached to the top. The diametre of the circle, which is also the top of the trapezoid, is labeled 8. The height of the trapezoid is 6. The bottom of the trapezoid is 13.
  18. A geometric shape is shown. It is a rectangle with a triangle attached to the top on the left side and a circle attached to the top right corner. The diametre of the circle is labeled 5. The height of the triangle is labeled 5, the base is labeled 4. The height of the rectangle is labeled 6, the base 11.
  19. A geometric shape is shown. It is a trapezoid with a triangle attached to the top, and a circle attached to the triangle. The diametre of the circle is 4. The height of the triangle is 5, the base of the triangle, which is also the top of the trapezoid, is 6. The bottom of the trapezoid is 9. The height of the trapezoid is 7.

In the following exercises, solve.

  1. A city park covers one block plus parts of four more blocks, as shown in Figure 7.3.21. The block is a square with sides 250 feet long, and the triangles are isosceles right triangles. Find the area of the park.
    A square is shown with four triangles coming off each side.
    Figure 7.3.21
  2. A gift box will be made from a rectangular piece of cardboard measuring 12 inches by 20 inches, with squares cut out of the corners of the sides, as shown in Figure 7.3.22. The sides of the squares are 3 inches. Find the area of the cardboard after the corners are cut out.
    A rectangle is shown. Each corner has a gray shaded square. There are dotted lines drawn across the side of each square attached to the next square.
    Figure 7.3.22
  3. Perry needs to put in a new lawn. His lot is a rectangle with a length of 120 feet and a width of 100 feet. The house is rectangular and measures 50 feet by 40 feet. His driveway is rectangular and measures 20 feet by 30 feet, as shown in Figure 7.3.23. Find the area of Perry’s lawn.
    A rectangular lot is shown. In it is a home shaped like a rectangle attached to a rectangular driveway.
    Figure 7.3.23
  4. Denise is planning to put a deck in her backyard. The deck will be a 20-ft by 12-ft rectangle with a semicircle of diametre 6 feet, as shown in Figure 7.3.24. Find the area of the deck.
    A picture of a deck is shown. It is shaped like a rectangle with a semi-circle attached to the top on the left side.
    Figure 7.3.24

3. Everyday Math

  1. Area of a Tabletop: Yuki bought a drop-leaf kitchen table. The rectangular part of the table is a 1-ft by 3-ft rectangle with a semicircle at each end, as shown in Figure 7.3.25.
    An image of a table is shown. There is a rectangular portion attached to a semi-circular portion. There is another semi-circular leaf folded down on the other side of the rectangle.
    Figure 7.3.25
    1. Find the area of the table with one leaf up.
    2. Find the area of the table with both leaves up.
  2. Painting: Leora wants to paint the nursery in her house. The nursery is an 8-ft by 10-ft rectangle, and the ceiling is 8 feet tall. There is a 3-ft by 6.5-ft door on one wall, a 3-ft by 6.5-ft closet door on another wall, and one 4-ft by 3.5-ft window on the third wall. The fourth wall has no doors or windows. If she will only paint the four walls and not the ceiling or doors, how many square feet will she need to paint?

4. Writing Exercises

  1. Describe two different ways to find the area of Figure 7.3.26, and then show your work to make sure both ways give the same area.
    A geometric shape is shown. It is a vertical rectangle attached to a horizontal rectangle. The width of the vertical rectangle is 3, the left side is labeled 6, the bottom is labeled 9, and the width of the horizontal rectangle is labeled 3. The top of the horizontal rectangle is labeled 6, and the distance from the top of that rectangle to the top of the other rectangle is labeled 3.
    Figure 7.3.26
  2. A circle has a diametre of 14 feet. Find the area of the circle
    1. Using 3.14 for [latex]\pi[/latex].
    2. Using [latex]\tfrac{22}{7}[/latex] for [latex]\pi[/latex].
    3. Which calculation to do prefer? Why?

7.3: Practice Answers

  1. Use the properties of circles
    1. C = 43.96 in.
      A = 153.86 sq. in.
    2. C = 53.38 ft
      A = 226.865 sq. ft
    3. 62.8 ft
    4. 37.68 in.
    5. 6.908 ft
    6. 52 in.
    7. 5.5 m
    8. 24 ft
    9. 6.5 mi

     

  2. Find the area of irregular figures
    1. 16 sq. units
    2. 30 sq. units
    3. 57.5 sq. units
    4. 12 sq. units
    5. 67.5 sq. units
    6. 89 sq. units
    7. 44.81 sq. units
    8. 41.12 sq. units
    9. 35.13 sq. units
    10. 95.625 sq. units
    11. 187,500 sq. ft
    12. 9400 sq. ft

     

  3. Everyday math.
    1. Area of a Tabletop.
      1. 6.5325 sq. ft
      2. 10.065 sq. ft

     

  4. Writing exercises
    1. Answers will vary.

References

Mathispower4u. (2011a, June 17). Example: Determine the area of an L-shaped polygon using decimals [Video]. YouTube. https://youtu.be/roRp3BjHwbU?si=gFXn80n90TVE2tM6.

Mathispower4u. (2011b, June 21). Example: Determine an area involving a rectangle and circle [Video]. YouTube. https://youtu.be/085t_Fmje4o?si=p8ElBx_v3oqEPsrw.

Mathispower4u. (2011c, June 21). Example: Determine perimeter involving a rectangle and circle [Video]. YouTube. https://youtu.be/fW2Ly8WFHzU?si=2codTCp-I45piJ3d.

Mathispower4u. (2011d, June 21). Example: Determine the area of a circle [Video]. YouTube. https://youtu.be/SIKkWLqt2mQ?si=1ggAEqaNrQbdZ1DI.

Mathispower4u. (2011e, June 21). Examples: Determine the circumference of a circle [Video]. YouTube. https://youtu.be/sHtsnC2Mgnk?si=L9jRfnkFQm2U7ClR.

Mathispower4u. (2011f, August 5). Ex: Find the area of an L-shaped polygon involving whole numbers [Video]. YouTube. https://youtu.be/lx8vweYTPpg?si=PET4FPjfwI-p_xyZ.

Attributions

All figures in this chapter are from 3.4 Solve Geometry Applications: Circles and Irregular Figures in Introductory Algebra by Izabela Mazur, via BCcampus.

This chapter has been adapted from 3.4 Solve Geometry Applications: Circles and Irregular Figures in Introductory Algebra (BCcampus) by Izabela Mazur (2021), which is under a CC BY 4.0 license.

The original chapter was adapted from 9.5 Solve Geometry Applications: Circles and Irregular Figures in Prealgebra 2e (OpenStax) by Lynn Marecek, MaryAnne Anthony-Smith, and Andrea Honeycutt Mathis (2020), which is under a CC BY 4.0 license. Adapted by Izabela Mazur.

License

Icon for the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License

Intermediate PreAlgebra: Building Success Copyright © 2024 by Kim Tamblyn, TRU Open Press is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

Share This Book