10. Areas of Composite Figures
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Chapter 9
10.
Areas of Composite Figures
| Student Learning Objectives: |
|---|
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Why
This lesson explains how to calculate the area and perimeter of composite figures by combining basic geometric shapes. It provides step-by-step insights for understanding how these figures are formed and how to approach real-world problems involving irregular shapes.
Lesson Settings & Tools
| | 8 Theory slides |
| | 14 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Areas of Composite Figures
Slide of 8
Two or more geometric shapes can make up a plane figure. There are no standard procedures for finding the area or perimeter of the figure in these cases. However, breaking down the figure into known geometric shapes helps reveal that information. This lesson will use real-life problems to discuss calculating the area and perimeter of figures with different shapes.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
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Challenge
Finding the Area of a Pool
Tearrik began training at a newly opened gym in his neighborhood called Little Muscles. In the swimming area there are three pools. One is a circular pool with an area of 28.2 square meters. Another is a rectangular pool with an area of 54 square meters.
The shape of the third pool surprised Tearrik a little. It looks like the combination of half the circular pool with the rectangular pool.
What is the area of the third pool?
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Discussion
Composite Figures
A composite figure is a plane figure made of two or more two-dimensional figures combined in a way that creates a new figure. For example, the following figure is made of a rectangle, a triangle, and a semicircle.
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Example
Area and Perimeter of the Gym's Logo
The wrestling room of the gym where Tearrik works out has a circular carpet with the gym's logo inside.
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Hint
a Decompose the logo into familiar geometric shapes such as triangles, rectangles, trapezoids, and semicircles. Find the area of the involved figures and add them up.
b Use the Pythagorean Theorem to find the lengths of the slant sides. Notice that all these sides have the same length. The circumference of a semicircle is π times the radius.
Solution
a The gym's logo is a dumbbell formed by different geometric figures — a green rectangle, two purple trapezoids, two gray trapezoids, two gray rectangles, and two red semicircles. Thus, it is a composite figure.
The area of the logo is the sum of the areas of the involved figures. Then, start by recalling the formulas for finding the area of the corresponding figures.
| Figure | Area's Formula |
|---|---|
| Rectangle | A=l* w |
| Trapezoid | A=(b_1+b_2)h/2 |
| Semicircle | A=π r^2/2 |
Next, find the area of each of the pieces that make up the logo.
| Figure | Dimensions (cm) | Area (cm^2) |
|---|---|---|
| Green rectangle | l = 40 w=20 |
A_1=40* 20 = 800 |
| Purple trapezoid | b_1 = 100 b_2=20 h=30 |
A_2=(100+20)30/2 = 1800 |
| Gray trapezoid | b_1 = 180 b_2=100 h=30 |
A_3=(180+100)30/2 = 4200 |
| Gray rectangle | l = 100 w=10 |
A_4=100* 10 = 1000 |
| Red semicircle | r=20 | A_5=π (20)^2/2 ≈ 628.32 |
Since all the figures but the green rectangle appear twice in the logo, the corresponding areas must be multiplied by 2.
A = A_1+2* A_2+2* A_3+2* A_4+2* A_5
SubstituteValues
Substitute values
A = 800 + 2* 1800 + 2* 4200 + 2* 1000 + 2* 628.32
Multiply
Multiply
A = 800 + 3600 + 8400 + 2000 + 1256.64
AddTerms
Add terms
A = 16 056.64
Finally, choose the appropriate conversion factor to convert the area from square centimeters to square meters. Recall that 1 square meter is the same as 10 000 square centimeters.
16 056.64 cm^2 * 1m^2/10 000cm^2
▼
Evaluate
MoveLeftFacToNum
a*b/c= a* b/c
16 056.64 cm^2 * 1m^2/10 000cm^2
Cross out common units
16 056.64 cm^2 * 1m^2/10 000 cm^2
Cancel out common units
16 056.64* 1m^2/10 000
MultByOne
a * 1=a
16 056.64 m^2/10 000
CalcQuot
Calculate quotient
1.605664 m^2
RoundDec
Round to 1 decimal place(s)
1.6m^2
The gym's logo has an area of about 1.6 square meters.
b The perimeter of the gym's logo, which is a composite figure, is the distance around the logo. From the given diagram, almost all side lengths can be clearly identified.
Notice that all the slant segments have the same length because each of them is the hypotenuse of a right triangle with legs measuring 40 and 30 centimeters. Therefore, their lengths can be found by using the Pythagorean Theorem.
a^2+b^2=c^2
SubstituteValues
Substitute values
40^2 + 30^2 = c^2
▼
Solve for c
CalcPow
Calculate power
1600 + 900 = c^2
AddTerms
Add terms
2500 = c^2
SqrtEqn
sqrt(LHS)=sqrt(RHS)
sqrt(2500) = sqrt(c^2)
SqrtPowToNumber
sqrt(a^2)=a
sqrt(2500) = c
CalcRoot
Calculate root
50 = c
RearrangeEqn
Rearrange equation
c = 50
The length of each slant segment is 50 centimeters. The length of each semicircle is π times the radius. Since the radius is 20 centimeters, the length of each semicircle is 20π centimeters.
Finally, all the side lengths are added. There are six segments of 40 centimeters, eight segments of 50 centimeters, four segments of 10 centimeters, four segments of 30 centimeters, and two semicircles of 20π centimeters.
The perimeter of the logo, rounded to the nearest integer, is 926 centimeters.
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Example
Carpeting the Training Rooms
On the first floor of the gym, the weight room and the spinning room are connected by the locker room. The gym owner received many complaints about the current floor quality. The entire first floor, instead, will be covered using recycled yoga mats.
How many square meters of rubber matting does the gym owner have to buy?
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Hint
Add the areas of the two big rectangles and subtract the area of the locker room.
Solution
Notice that the three rooms together form a composite figure that is shaped as two overlapping rectangles.
The area of the entire floor is then the sum of the areas of the big rectangles minus the area of the locker room. The area of the locker room must be subtracted because it is common for both rectangles.
| Room | Figure | Dimensions (ft) | Area (ft^2) |
|---|---|---|---|
| Spinning | Upper Rectangle | l = 50 w=45 |
A_1 = 50* 45 = 2250 |
| Weight | Lower Rectangle | l = 55 w=40 |
A_2 = 55* 40 = 2200 |
| Locker | Small Rectangle | l = 55-35 = 20 w = 40-25=15 |
A_3 = 20* 15= 300 |
Finally, find the area of the entire floor.
A = A_1+ A_2- A_3
SubstituteValues
Substitute values
A = 2250+ 2200- 300
AddSubTerms
Add and subtract terms
A = 4150
The gym owner has to buy 4150 square feet of rubber matting to cover the entire floor.
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Example
Kettlebells
It is leg day and Tearrik wants to try a new exercise he discovered on social media — the single-kettlebell swing. Instead of the usual spherical shape, the kettleballs at this gym are polyhedrons.
What is the area of the front view of the kettlebell? Round to the nearest integer.
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Hint
The area of a regular polygon is half the perimeter times the apothem. The apothem is the distance from the center of the polygon to any of its sides. The handle is what is left when a small semicircle is removed from a large semicircle.
Solution
The front view of the kettlebell can be separated into two parts — the body and the handle. The body is a regular octagon and the handle looks like half a ring. The kettlebell's front view is then a composite figure whose area is the sum of these two areas.
A = A_(octagon) + A_(handle)
Area of the Octagon
The area of a regular polygon is half the perimeter times the apothem.
The perimeter of a regular polygon is equal to the side length times the number of sides. In this case, the number of sides is 8 and all of them have a length of 13.5 centimeters. p = 8* 13.5 ⇒ p = 108cm The apothem of a regular polygon is the distance from the center of the polygon to any of its sides. From the given diagram, the apothem of the kettlebell's front view is 16.3 centimeters.
A_(octagon) = 1/2pa
SubstituteII
p= 108, a= 16.3
A_(octagon) = 1/2* 108* 16.3
▼
Simplify
Multiply
Multiply
A_(octagon) = 1/2* 1760.4
MoveLeftFacToNumOne
a* 1/b= a/b
A_(octagon) = 1760.4/2
CalcQuot
Calculate quotient
A_(octagon) = 880.2
The area of the regular octagon is 880.2 square centimeters.
Area of the Handle
The handle has the shape of half a ring which is what is left when a small semicircle is removed from a large semicircle.
The inner semicircle has a diameter of 12 centimeters so its radius is r_1 = 6 centimeters. The outer semicircle has a diameter of 13.5 centimeters so its radius is r_2 = 6.75 centimeters. The area of a semicircle is half the product of π and the radius squared. A_(semicircle) = 1/2π r^2 Use this formula to find the area of the semicircles.
| Semicircle | Radius (cm) | A=1/2π r^2 | Area (cm^2) |
|---|---|---|---|
| Outer | 6.75 | A_1 = 1/2π(6.75)^2 | A_1 = 71.57 |
| Inner | 6 | A_2 = 1/2π(6)^2 | A_2 = 56.55 |
The area of the handle is the difference between the area of the outer semicircle and the area of the inner semicircle.
A_(ring) = A_1-A_2
SubstituteII
A_1= 71.57, A_2= 56.55
A_(ring) = 71.57 - 56.55
SubTerm
Subtract term
A_(ring) = 15.02
The area of the front view of the handle is 15.02 square centimeters.
Area of the Kettlebell's Front View
Finally, add the areas of the regular octagon and the area of the handle. A = 880.2 + 15.02 ⇓ A = 895.22 The area of the front view of the kettlebell is about 895 square centimeters.
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Example
New Training Zone
Tearrik noticed a poster on the wall saying that the gym would soon open its second floor with a boxing room and a multipurpose room. The blueprint for the second floor is shown next.
Find the perimeter and area of the of the entire floor. Round the answer to the nearest integer. Every unit in the blueprint corresponds to 4 meters.
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Hint
Use the coordinate plane to find the side lengths of the composite figure. The perimeter is the sum of the lengths of the exterior sides. Break down the figure into known figures and find the sum of their areas.
Solution
The first step involves finding the perimeter of the entire floor. After that, the area of the floor will be calculated.
Perimeter of The Floor
Start by identifying the coordinates of the vertices of the blueprint and the lengths of the horizontal and vertical sides. Let m be the length of the slant side of the entrance.
The value of m can be calculated by substituting ( -4, 1) and ( 0, 4) into the Distance Formula.
d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
SubstituteValues
Substitute values
m=sqrt(( 0-( -4))^2+( 4- 1)^2)
m = 5
The slant side is 5 units long. The length of the curved wall at the back of the stage is the length of a semicircle whose radius is 3 units. The length of a semicircle is π times the radius. C = π* 3 ⇒ C ≈ 9.42 The wall at the back of the stage is about 9.42 units long. By adding the lengths of the exterior sides of the blueprint, the perimeter of the entire floor can be calculated. P = 5 + 9 + 9.42 + 3 + 2 + 6 + 2 + 4 + 3 ⇓ P = 43.42 The perimeter of the entire floor in the blueprint is 43.42 units. Convert this to meters by using the fact that every unit in the blueprint is the same as 4 meters. 43.42units * 4m/1unit = 173.68m The entire floor has a perimeter of about 174 meters.
Area of The Floor
Notice that the entire floor can be broken down into some known figures.
The area of the floor is the sum of the areas of the involved figures.
| Place | Figure | Dimensions | Area |
|---|---|---|---|
| Entrance and locker room | Trapezoid | b_1 = 6 b_2=3 h=4 |
A_1 = (6+3)4/2= 18 |
| Seating area, hall, and boxing room | Rectangle | l = 8 w=6 |
A_2 = 8* 6= 48 |
| Multipurpose room without the stage | Rectangle | l = 6 w=3 |
A_3 = 6* 3= 18 |
| Stage | Semicircle | r = 3 | A_4 = 1/2π (3)^2 ≈ 14.14 |
Next, add these four areas. A = 18+48+18+14.14 ⇓ A = 98.14 The area of the entire floor in the blueprint is about 98.14 square units. Finally, convert this to square meters. 98.14units * (4m/1unit)^2 = 1570.24m^2 The entire floor has an area of about 1570 square meters.
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Example
Composite Figures in Real Life
Composite figures are everywhere, it is a matter of looking closely. For instance, a running track combines rectangles and semicircles.
Many traffic signs can also be decomposed into known figures. The traffic sign indicating the direction of a street is formed by a rectangle and a triangle. The traffic sign corresponding to no U-turn
combines two rectangles, a semicircle, and a triangle.
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Areas of Composite Figures
Exercises
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The perimeter of a composite figure is the distance around the figure. In other words, we need to add all the side lengths. Let's begin by writing them all.
We are ready to find the perimeter. P = 6 + 4 + 2 + 1 + 2 + 1 + 2 + 4 ⇓ P = 22 The figure has a perimeter of 22 inches.
The given figure is the combination of two rectangles.
The area of the entire figure is the sum of the areas of the two rectangles. A = A_(r_1) +A_(r_2) The big rectangle is 6 inches long and 4 inches wide. The small rectangle is 2 inches long and 1 inch wide. A_(r_1) &= 6* 4 = 24in^2 A_(r_2) &= 2* 1 = 2in^2 The composite figure has an area of 24+2=26 square inches.
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The perimeter of a composite figure is the distance around the whole figure. In this case, the given figure is formed by a rectangle and two semicircles. Let C_1 and C_2 be the distances around the semicircles.
Notice that the two semicircles have the same diameter. Therefore, together they form a full circle. Let's break down the figure and connect the two semicircles.
Therefore, the perimeter of the given figure is equal to the circumference of a circle plus twice the length of the rectangle. P = 2π r + 2l The rectangle is 92 yards long and the radius of the semicircles is half the width of the rectangle. Thus, the radius is 622=31 yards.
The perimeter of the given figure, rounded to the nearest integer, is 379 yards.
As we said in the previous part, the given figure can be broken down into a rectangle and a full circle.
The area of the composite figure is the sum of the area of the circle and the area of the rectangle. A = A_(circle) + A_(rectangle) The diameter of the circle is 62 yards. This means that the circle has a radius of 31 yards. Let's use this to find its area.
The rectangle is 92 yards long and 62 yards wide.
We are ready to find the area of the given figure. A = 3017.54 + 5704 ⇓ A = 8721.54 The area of the given figure, rounded to the nearest integer, is 8722 square yards.
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Maya wants to carpet the floor of her bedroom and her dressing room.
If the rug is sold in whole numbers of square meters, how much rug does Maya have to buy?
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We can see that the bedroom and the walk-in closet together look like a rectangle with a triangle at the left. Alternatively, the entire room can be split into a trapezoid and a rectangle.
We can use any of the two combinations to find the area of the room. For instance, let's use the first one. A = A_(rectangle) + A_(triangle) The rectangle is 5 meters long and 3 meters wide. The triangle has a base of 4-3=1 meter and it is 5-3.4=1.6 meters high.
| Area of Rectangle | Area of Triangle | |
|---|---|---|
| Formula | A = l* w | A = bh/2 |
| Substitute values | A = 5* 3 | A = 1* 1.6/2 |
| Calculate | A = 15 | A=0.8 |
Finally, we add the two areas. A = 15 + 0.8 ⇓ A = 15.8 The entire room has an area of 15.8 square meters. Since the rug is sold in whole numbers of square meters, Maya has to buy 16 square meters.
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Jordan is going to trim the vegetation in her backyard.
She thinks she can trim 3 square yards of vegetation per minute. How long will it take to trim the whole backyard?
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We need to know the area of the backyard to determine the time it will take Jordan to trim. The backyard is a composite figure formed by two rectangles and a triangle.
The area of the backyard is then the sum of the areas of these three polygons. A = A_(r_1) + A_(r_2) + A_(△) The base of the triangle is 6 yards and the height is 4 yards. Both rectangles are 5 yards wide. One of them is 14 yards long and the other is 10 yards long.
| Area of Rectangle 1 | Area of Rectangle 2 | Area of Triangle | |
|---|---|---|---|
| Formula | A_(r_1) = l* w | A_(r_2) = l* w | A_(△) = bh/2 |
| Substitute values | A_(r_1) = 14* 5 | A_(r_2) = 10* 5 | A_(△) = 6* 4/2 |
| Calculate | A_(r_1) = 70 | A_(r_2) = 50 | A_(△) = 12 |
Let's add the three areas. A = 70 + 50 + 12 ⇓ A = 132 The area of the entire backyard is 132 square yards. Finally, let's divide the area of the backyard by the rate at which Jordan trims the vegetation, 3 square yards per minute.
It will take Jordan 44 minutes to trim the vegetation of the entire backyard.
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The perimeter of a composite figure is the distance around the figure. We know the lengths of the straight sides, 8 and 10 centimeters. Let C be the length of the semicircle at the right.
The length of a semicircle is π times the radius. C = π r The diameter of the semicircle is 6 centimeters. Thus, its radius is 3 centimeters.
The semicircle is about 9.42 centimeters long. We are ready to find the perimeter of the given figure. P = 8 + 10 + 9.42 ⇓ P = 27.42 The perimeter of the given figure, rounded to the nearest integer, is 27 centimeters.
The given figure is the combination of a right triangle and a semicircle.
The area is then the sum of the corresponding two areas. A = A_(triangle) + A_(semicircle) The area of the triangle is half the base times the height. The area of the semicircle is half the product of π and the radius squared. The base of the triangle is 8 and the height is 6. The radius of the semicircle is 3.
| Area of Triangle | Area of Semicircle | |
|---|---|---|
| Formula | A = 1/2 b h | A = 1/2π r^2 |
| Substitute values | A = 1/2* 8* 6 | A = 1/2π ( 3)^2 |
| Calculate | A = 24 | A ≈ 14.14 |
Finally, let's add the two areas. A = 24 + 14.14 ⇓ A = 38.14 The area of the composite figure, rounded to the nearest integer, is 38 square centimeters.
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Diego wants to paint the back wall of the attic.
He knows he can cover 2.5 square meters with 225 milliliters of paint. How much paint does Diego need?
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We can determine how much paint Diego needs by knowing how many square meters he has to paint. Thus, we need to find the area of the wall. Let's calculate it! The wall is a composite figure made of a right trapezoid with a triangle on top. The base of the triangle is the minor base of the trapezoid.
The area of the wall is the sum of the areas of these two polygons. A = A_(trapezoid) + A_(triangle) The major base of the trapezoid measures 5 meters, the minor base measures 2.5 meters, and the height is 2 meters. Let's use these values to find its area.
The triangle has a base of 2.5 meters. The height of the triangle is the height of the entire wall minus the height of the trapezoid. Thus, the height of the triangle is 3.6-2 = 1.6 meters.
We are ready to find the area of the wall. A = 7.5 + 2 ⇓ A = 9.5 The entire wall has an area of 9.5 square meters. Now, we can find the amount of paint Diego needs by dividing the wall's area by his painting rate, which is 2.5 square meters per 225 milliliters.
Diego needs 855 milliliters of paint.
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Areas of Composite Figures
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