3.4: Solve Geometry Applications- Circles and Irregular Figures

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Learning Objectives

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.

How To

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

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

$r$ is the length of the radius
$d$ is the length of the diameter
$d = 2 r$
Circumference is the perimeter of a circle. The formula for circumference is
$C = 2 π r$
The formula for area of a circle is
$A = π r^{2}$

Remember, that we will use the $π$ 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 $2.5$ feet. Find the ⓐ circumference and ⓑ area of the sandbox. Round both answers to the nearest tenth.

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 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	$C = 2 π r$ $C = 2 π (2.5)$
Step 5. Solve the equation.	$C \approx 15.7 ft$
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	$A = π r^{2}$ $A = π {(2.5)}^{2}$
Step 5. Solve the equation.	$A \approx 19.6 sq. ft.$
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 $5$ inches. Find the ⓐ circumference and ⓑ area of the mirror.

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

$C = 2 π r$

But since the diameter of a circle is two times the radius, we could write the formula for the circumference in terms of $d$ .

	$C = 2 π r$
Using the commutative property, we can rewrite the formula.	$C = π 2 r$
Substituting $d$ for $2 r$	$C = π d$

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

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.	$C = π d$ $C = π (4)$
Step 5. Solve the equation.	$C \approx 12.57 feet$
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 $5.5$ feet.

Example 3.4.3

Find the diameter of a circle with a circumference of $47.1$ 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.	$C = π d$ $47.1 = π d$
Step 5. Solve by dividing both sides by pi.	$\frac{47.1}{π} = \frac{πd}{π}$ $d \approx 15$
Step 6. Check:	$C = π d$ $C = π 15$ $C \approx 47.1$
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 $94.2$ 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.

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

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 $12$ and a length of $4 .$ The red rectangle has a width of $2,$ 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 $4$ units long, the length of the red rectangle must be $6$ units.

$An image of a blue horizontal rectangle attached to a red vertical rectangle is shown. The top is labeled 12, the side of the blue rectangle is labeled 4. The whole side is labeled 10, the blue portion is labeled 4 and the red portion is labeled 6. The width of the red rectangle is labeled 2.$ $The first line says A sub figure equals A sub rectangle plus A sub red rectangle. Below this is A sub figure equals bh plus red bh. Below this is A sub figure equals 12 times 4 plus red 2 times 6. Below this is A sub figure equals 48 plus red 12. Below this is A sub figure equals 60.$

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.

Your Turn 3.4.4

Find the area of each shaded region:

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

Example 3.4.5

Find the area of the shaded region.

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

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 $8$ 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 $7 - 4$ .

The length of the rectangle is $8,$ so the base of the triangle will be $3$ , which is $8 - 4$ .

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

Now we can add the areas to find the area of the irregular figure.

$The top line reads A sub figure equals A sub rectangle plus A sub red triangle. The second line reads A sub figure equals lw plus one-half red bh. The next line says A sub figure equals 8 times 4 plus one-half times red 3 times red 3. The next line reads A sub figure equals 32 plus red 4.5. The last line says A sub figure equals 36.5 sq. units.$

The area of the figure is $36.5$ square units.

Your Turn 3.4.5

Find the area of each shaded region.

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

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 $105$ meters and width $68$ meters. Find the area enclosed by the track. Round your answer to the nearest hundredth.

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

Answer

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.

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

The rectangle has a length of $105$ m and a width of $68$ m. The semi-circles have a diameter of $68$ m, so each has a radius of $34$ m.

Area of figure = Area of rectangle + Area of two semicircles

Area of figure = $b h + 2 (\frac{1}{2} π r^{2})$

Area of figure = $106 • 68 + 2 (\frac{1}{2} π 34^{2})$

Area of figure $\approx 7140 + 3631.68 = 10771.68 square meters$

Your Turn 3.4.6

Find the area:

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

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Properties of Circles

Example 3.4.1

Your Turn 3.4.1

Example 3.4.2

Your Turn 3.4.2

Example 3.4.3

Your Turn 3.4.3

Example 3.4.4

Your Turn 3.4.4

Example 3.4.5

Your Turn 3.4.5

Example 3.4.6

Your Turn 3.4.6

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