CMST Review Unit 2 Geometry Unit Summary This booklet is designed to help prepare CMST students for both the Core Math Test and the Core Math Tutorials. This unit is designed in particular to prepare students for the Geometry section of the CMST. In the Geometry Unit of the CMST, students will be tested on the topics of right triangles, area, circumference, perimeter, volume, similarity, and angle measures. Section Summary Section 1 (Area and Perimeter) Section 2 (Right Triangles and Similarity) Section 3 (Volume) Section 4 (Angles)

Section 1 Area and Perimeter Introduction This section introduces the concepts of area and perimeter. In this section, the students will learn some basic definitions and solve problems with practical applications of area and perimeter. At the end of each section, there will be some practice problems available for the students to practice. Key Terms and Definitions The area of an object is a physical quantity that expresses the size of a two dimensional object such as a square or rectangle. The perimeter of an object is the measured distance around the object. The circumference of a circle is a special case of perimeter that pertains to circles.

Key Formulas Object

Diagram

Rectangle

w = width

Area Formula

Perimeter

A = l⋅w

P = 2l + 2 w

l = length

Triangle

A= h = height

1 ⋅b⋅h 2

b = bas e

Square

A = s ⋅ s = s2

P = 4s

A = π r2

C = 2π r = π d

s = side

Circle r = radius

d = 2r

The Area of a Rectangle

The area of a rectangle is given by the formula A = l ⋅ w . To use this formula you simply substitute the values of the length and width into the formula. Now, let’s try an example of finding the area of a rectangle. Example 1 (Area of a Rectangle)

Find the area of rectangle with a length of 8 inches and a width of 5 inches.

w = 5 in l = 8 in

Solution: Substitute 8 inches in for the length and 5 inches in for the width into the area formula. A = lw = (8 in)(4 in) = 32 square inches

Example 2 (Area and Perimeter)

The area of a square is 49 square inches. What is the perimeter of the square? Solution: To find the perimeter of a square, you must first find the length of the side of the square. The length of the side of the square can be found by substituting 49 square inches in for the area of a square and solving the equation for s. A = s2

49 = s 2 To solve the equation for s you must take the square root of both sides of the equation. 49 = s 2 s = 7 inches Now, substitute 7 inches into the perimeter formula. P = 4(7 in) = 28 in

Here is an example that follows a similar process, but instead of finding the perimeter of the square you must find the area of the square. Example 3

The perimeter of a square is 40 square centimeters. What is the area of the square? Solution: To find the area of a square, you must first find the length of the side of the square. The length of the side of the square can be found by substituting 40 cm in for the perimeter of a square and solving the equation for s. P = 4s

40 = 4s 40 4s = 4 4 s = 10 cm Now, simply use the area of a square formula to find the area of the square. A = s 2 = (10 cm) 2 = 100 cm 2

Triangles Example 4

Find the area of the triangle

4

2

-5

5

-2

-4

-6

Solution: To find the area of the triangle, you can count the blocks to get the base and height of the triangle.

4

2

height = 3 units

-5

5

base = 6 units -2

-4

Now, use the area formula for the triangle.

A=

1 1 1 bh = (3)(6) = (18) = 9 square units 2 2 2

Area and Perimeter of a Circle

The circumference of a circle measures the distance around the circle. Both the area of a circle and the circumference of a circle are found by using a value called Pi. Pi is a nonterminating and non-repeating decimal that is approximately equal to 3.14. In most examples in this booklet, we use the Greek letter π to represent Pi. Now, let’s do a problem that uses Pi to find the area of a circle Example 5

Find the area of circle with a radius of 5 inches. Solution: To find the area of the circle, just substitute 5 units in for the radius of the circle and solve for the area A.

A = π r2 A = π (5) 2 A = 25π

Example 6

Find the area of the circle.

4

2

-5

5

10

-2

-4

Solution: First, find the radius by counting the blocks in the diagram.

4

2

-5

5

10

r = 3 units

-2

-4

Now, substitute 3 units for radius into the area formula A = π r 2 = π (3) 2 = 9π

Example 7

The following circle is inscribed in a square. Find the area of the circle if the area of the square is 16 square feet.

Solution: In order to find the area of the circle, you must first find the radius of the circle. The radius can be found by finding the diameter of the circle which is equal to the length of the side of the square. The length of the side of the square can be found by substituting the value of the area of square into the area of a square formula and solving the resulting equation for s.

A = s2 16 = s 2 ⇒ 16 = s 2 ⇒ s = 4 units

The radius of the circle is equal to half the diameter, so the radius would be

4 = 2 units. 2

(See Diagram) 4 units

2 units

Now that you know the radius is 2 units, use the area of circle to find the area of the circle. A = π r 2 = π (2) 2 = 4π Practice Problems (Geometry Section 1)

1) The area of a square is 64 square inches. What is the perimeter of the square? 2) The following circle is inscribed in a square. Find the area of the circle if the area of the square is 36 square feet.

3) The perimeter of a square is 50 centimeters. What is the area of the square? 4) Find the area of the given triangle

. Solutions: 1) 32 inches 2) 9π 3) 100 square cm 4) 15 square units

Section 2 Right Triangles and Similarity In this section, you will learn some of the basic properties of right triangles. In particular, you will learn to use the Pythagorean Theorem. You will also learn some of basic properties of similar triangles. The Pythagorean Theorem

A right triangle is a triangle that has one right angle. The sides of a right triangle have special names. The side opposite the right angle is called the hypotenuse and the two other sides that form the right angle are called the legs.

c = hypotenuse b = leg

a = leg

In a right triangle, there is a special relationship between the sides of the right triangle called the Pythagorean Theorem. The Pythagorean Theorem states that in a right triangle the sum of square of legs is equal to the square of the hypotenuse.

c2 = a2 + b2 Now, let’s try a few examples. Example 1

Find the missing side of the triangle in the figure below:

c =?

b=6

a=8

Solution: Substitute the value of legs into the Pythagorean Theorem and solve for c the hypotenuse.

c2 = a2 + b2 c 2 = 82 + 6 2 c 2 = 64 + 36 c 2 = 100 c 2 = 100 c = 10

Example 2

Given the lengths of the sides of a right triangle are 5 and 7, find the length of the hypotenuse.

c =? b=5

a=7

Solution: Since the hypotenuse is the longest side of the triangle, the other sides will be the legs of the right triangle. Therefore, plug in the values of 5 and 7 for a and b into the Pythagorean Theorem and solve for c.

c2 = a2 + b2 c 2 = 7 2 + 52 c 2 = 49 + 25 c 2 = 74 c 2 = 74 c = 74

Example 3

Find the length of the hypotenuse.

Solution: By using the diagram above and counting the blocks on the grid, you can find the lengths of the legs which are 12 units and 5 units. Next, use the Pythagorean Theorem to find the hypotenuse.

c =? 5 units

12 units

c2 = a2 + b2 c 2 = 12 2 + 5 2 c 2 = 144 + 25 c 2 = 169 c 2 = 169 c = 13

Example 4

Given that the length of hypotenuse of a right triangle is 17 inches and the length of one leg is 15 units, find the length of the missing side. Substitute the values of the leg and hypotenuse into the Pythagorean Theorem and solve for the missing leg (a) c2 = a2 + b2 17 2 = a 2 + 15 2 289 = a 2 + 225 289 − 225 = a 2 + 225 − 225 64 = a 2 a 2 = 64 a=8

Similar Triangles

Similar triangles have sides that are proportional. If the sides of a triangle are proportional, then there is a special ratio between the sides. (See diagram below) C

B

A

D

E

F

AB BC AC = = EF ED DF

In each the following examples, you will have to set up a ratio between the sides and solve for the missing value.

Example 4

Find the missing value. 3

5

?

9

15

18

Solution: Begin by using x as the missing value of the first triangle. Then, set up a ratio using the sides of the triangle and solve for the missing value.

5 x = 15 18 After setting up a ratio, take the cross product which will result in: 15 x = 5(18) Now, solve for x. 15 x = 90 15 x 90 = 15 15 x=6

Example 5

Triangle 1 and 2 are similar triangles. If triangle 1 has sides 6,6,10 and 2 has sides 9,9,x, then find x? Solution: Set up a ratio using the sides of the triangle solve for the missing value.

6 10 = 9 x After setting up a ratio, take the cross product which will result in:

6 x = 9(10) Now, solve for x. 6 x = 90 6 x 90 = 6 6 x = 15

Practice Problems (Geometry Unit 2)

1) Given the lengths of the sides of a right triangle are 12 and 16, find the length of the hypotenuse. 2) Find the length of the hypotenuse.

3) Triangle A and B are similar triangles. If triangle A has sides 4,4,6 and B has sides 12,12,x, then find x? 1) 20 2)

85 3) 18

Section 3 Volume The main topic of this section is volume. You will specifically look at how to find the volume of various three dimension geometric objects such as rectangular solids and cylinders. The volume of an object is the amount space occupied by the object. The procedure for finding volume of an object is similar to finding the area of an object in that it is simply a matter of substituting correct values into the proper formula. Here is a list of some of the key formulas used to find volume.

Object

Shape

Formula to Find Volume

V = l ⋅ w⋅h

Rectangular Solid

Cylinder

V = π r 2h

Cube

V = s3

Now, let’s find the volume of some three dimensional objects.

Example 1

Find the volume of a cylinder with a radius of 3 inches and height of 4 inches.

3 in

4 in

Solution: Substitute the values of the radius and the height into the volume formula.

V = π r 2h V = π (3 cm ) 2 ( 4 cm ) V = π (9 cm 2 )( 4 cm ) V = 36π cm 3

Example 2

Find the volume of rectangular solid that is 3 feet by 2 feet by 4 feet. 4 feet

2 feet

3 feet

Solution: Substitute the values of the length, width, and height into the volume of rectangular solid formula.

V = l ⋅ w ⋅ h = (3 ft )(2 ft )(4 ft ) = (6 ft 2 )(4 ft ) = 24 ft 3

Example 3

Find the volume of a rectangular solid with a square base if its height is 6 inches and the length of its base is 4 inches.

6 feet

4 feet

4 feet

Solution: Since the base is square, the value of the length and width are equal. Therefore, the base of the rectangular solid is 4 feet by 4 feet. So, to get the volume of the rectangular solid just substitute the values l = 4 ft , w = 4 ft , and h = 6 ft into the formula V = l ⋅ w ⋅ h V = l ⋅w⋅h V = (4 ft )(4 ft )(6 ft ) V = (16 ft 2 )(6 ft ) V = 96 ft 3

Practice Problems (Geometry Unit 2)

1) Find the volume of the following cylinder.

4 in

6 in

2) Find the volume of a rectangular solid that is 6 inches by 5 inches by 3 inches. 3) Find the volume of a rectangular solid with a square base if its height is 3 inches and the length of its base is 6 inches. Solutions: 1) 96π 2) 90 square inches 3) 54 cubic inches

Section 4 Angle Measure The main objective in this section of the Geometry Unit is to find angle measures and estimate angle measures. By the end of the section, you should be able to estimate an angle from a diagram given to you. In particular, you should be able to estimate an angle from circular shaped diagram that is divided up into equal sections. In this situation, you just divide 360 by number of sections in the circle diagram. The reason you use 360 to make this calculation is that there is 360 degrees in a circle. Now try this example. Example 1

Find the measure of angle A

A

Solution: Since a circle has 360 degrees and there are 24 sections in this circle, you simply divide 360 by 24

Angle measure =

360 = 15 0 24

Example 2

Find the measure of angle A

A

Solution: Since a circle has 360 degrees and there are 6 sections in this circle, you simply divide 360 by 6.

Angle measure =

360 = 60 0 6

Practice Problems (Geometry Unit 4)

1) Find the measure of A.

A

2) Find the measure of A.

A

Solutions: 1) 30 0 2) 45 0