3.4: Solve Geometry Applications- Circles and Irregular Figures
- Page ID
- 152038
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)After completing this section, you should be able to:
- Use the properties of circles
- Find the area of 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
- Read the problem and make sure you understand all the words and ideas. Draw the figure and label it with the given information.
- Identify what you are looking for.
- Name what you are looking for. Choose a variable to represent that quantity.
- Translate into an equation by writing the appropriate formula or model for the situation. Substitute in the given information.
- Solve the equation using good algebra techniques.
- Check the answer in the problem and make sure it makes sense.
- Answer the question with a complete sentence.
Use the Properties of Circles
Circles are defined as the set of points equidistant from a center point.
Properties of Circles
- is the length of the radius
- is the length of the diameter
- Circumference is the perimeter of a circle. The formula for circumference is
- The formula for area of a circle is
Remember, that we will use the button in our calculators to get an accurate answer to each problem. Make sure to read and follow rounding directions carefully.
Example 3.4.1
A circular sandbox has a radius of feet. Find the ⓐ circumference and ⓑ area of the sandbox. Round both answers to the nearest tenth.
- Answer
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ⓐ
Step 1. Read the problem. Draw the figure and label it with the given information.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
Step 5. Solve the equation. 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. This is slightly more than the circle's circumference, 15.7 ft.
Step 7. Answer the question. The circumference of the sandbox is 15.7 feet. ⓑ
Step 1. Read the problem. Draw the figure and label it with the given information.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
Step 5. Solve the equation. Step 6. Check.
Yes. If we draw a square around the circle, its sides would be 5 ft, as shown in part ⓐ. So the area of the square would be 25 sq. ft. This is slightly more than the circle's area, 19.6 sq. ft. Step 7. Answer the question. The area of the circle is 19.6 square feet.
Your Turn 3.4.1
A circular mirror has radius of inches. Find the ⓐ circumference and ⓑ area of the mirror.
We usually see the formula for circumference in terms of the radius of the circle:
But since the diameter of a circle is two times the radius, we could write the formula for the circumference in terms of .
Using the commutative property, we can rewrite the formula. | |
Substituting for |
We will use this form of the circumference when we’re given the length of the diameter instead of the radius.
Example 3.4.2
A circular table has a diameter of four feet. What is the circumference of the table? Round your answer to the nearest hundredth.
- Answer
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Step 1. Read the problem. Draw the figure and label it with the given information. 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 for the situation.
Substitute.
Step 5. Solve the equation. Step 6. Check: If we put a square around the circle, its side would be 4.
The perimeter would be 16. It makes sense that the circumference of the circle, 12.57, is a little less than 16.
Step 7. Answer the question. The diameter of the table is 12.57 square feet.
Your Turn 3.4.2
Find the circumference of a circular fire pit whose diameter is feet.
Example 3.4.3
Find the diameter of a circle with a circumference of centimeters. Round to the nearest whole number.
- Answer
-
Step 1. Read the problem. Draw the figure and label it with the given information. Step 2. Identify what you are looking for. the diameter of the circle Step 3. Name. Choose a variable to represent it. Let d = the diameter of the circle Step 4. Translate.
Write the formula.
Substitute.
Step 5. Solve by dividing both sides by pi.
Step 6. Check:
Step 7. Answer the question. The diameter of the circle is approximately 15 centimeters.
Your Turn 3.4.3
Find the diameter of a circle with circumference of centimeters.
Find the Area of Irregular Figures
So far, we have found area 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. But 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 3.4.4
Find the area of the shaded region.
- Answer
-
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.
The blue rectangle has a width of and a length of The red rectangle has a width of but its length is not labeled. 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 units long, the length of the red rectangle must be units.
The area of the figure is square units.
Is there another way to split this figure into two rectangles? Try it, and make sure you get the same area.
Your Turn 3.4.4
Find the area of each shaded region:
Example 3.4.5
Find the area of the shaded region.
- Answer
-
We can break this irregular figure into a triangle and rectangle. The area of the figure will be the sum of the areas of triangle and rectangle.
The rectangle has a length of units and a width of units.
We need to find the base and height of the triangle.
Since both sides of the rectangle are the vertical side of the triangle is , which is .
The length of the rectangle is so the base of the triangle will be , which is .
Now we can add the areas to find the area of the irregular figure.
The area of the figure is square units.
Your Turn 3.4.5
Find the area of each shaded region.
Example 3.4.6
A high school track is shaped like a rectangle with a semi-circle (half a circle) on each end. The rectangle has length meters and width meters. Find the area enclosed by the track. Round your answer to the nearest hundredth.
- Answer
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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 semicircles.
The rectangle has a length of m and a width of m. The semi-circles have a diameter of m, so each has a radius of m.
Area of figure = Area of rectangle + Area of two semicircles
Area of figure =
Area of figure =
Area of figure
Your Turn 3.4.6
Find the area: