composite figures class assignment answers

Areas of Composite Figures

Catch-up and review.

Here are a few recommended readings before getting started with this lesson.

• Areas of Parallelograms, Triangles, and Trapezoids

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?

Composite Figures

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.

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.

Next, find the area of each of the pieces that make up the logo.

Substitute values

a*b/c= a* b/c

Cross out common units

Cancel out common units

Calculate quotient

Round to 1 decimal place(s)

Calculate power

sqrt(LHS)=sqrt(RHS)

sqrt(a^2)=a

Calculate root

Rearrange equation

Use a calculator

Round to nearest integer

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.

Add the areas of the two big rectangles and subtract the area of the locker room.

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.

Add and subtract terms

Kettlebells

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.

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.

p= 108, a= 16.3

a* 1/b= a/b

Area of the Handle

A_1= 71.57, A_2= 56.55

Subtract term

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.

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.

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.

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.

a-(- b)=a+b

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.

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.

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.

Consider the following figure.

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.

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.

Maya wants to carpet the floor of her bedroom and her dressing room.

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.

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.

Jordan is going to trim the vegetation in her backyard.

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.

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.

Consider the following composite figure.

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.

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.

Diego wants to paint the back wall of the attic.

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.

Recommended exercises

Composite Area Worksheets

Related Topics: More Math Worksheets More Printable Math Worksheets More Grade 7 Math Lessons

Printable “Area” worksheets: Area of Rectangles Word Problems Area of Rectangles Word Problems (interactive) Area of Composite Rectangles Composite Area Worksheets (include circles)

Examples, solutions, videos, and worksheets to help Grade 7 students review how to solve composite area problems.

How to solve composite area problems?

Composite area problems involve finding the total area of a shape made up of multiple simple geometric figures (e.g., rectangles, triangles, circles). The key is to break down the composite shape into simpler parts, calculate the area of each part, and then combine these areas appropriately.

Here are the steps to solve composite area problems:

• Identify the Simple Shapes: Break down the composite shape into recognizable simple shapes such as rectangles, triangles, circles, semi-circles, etc.
• Calculate Individual Areas: Use the appropriate formulas to find the area of each simple shape.
• Combine the Areas: Sum the areas of the simple shapes if they are additive (part of the whole), or subtract the areas if necessary (such as holes or gaps).

Click on the following worksheet to get a printable pdf document. Scroll down the page for more Composite Area Worksheets .

More Composite Area Worksheets

Printable (Answers on the second page.) Composite Area Worksheet #1 Composite Area Worksheet #2 Composite Area Worksheet #3 Composite Area Worksheet #4

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Composite Area Problems

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