CMST Review Unit 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 partiular to prepare students for the Geometry setion of the CMST. In the Geometry Unit of the CMST, students will be tested on the topis of right triangles, area, irumferene, perimeter, volume, similarity, and angle measures. Setion Summary Setion 1 (Area and Perimeter) Setion (Right Triangles and Similarity) Setion 3 (Volume) Setion 4 (Angles) Setion 1 Area and Perimeter Introdution This setion introdues the onepts of area and perimeter. In this setion, the students will learn some basi definitions and solve problems with pratial appliations of area and perimeter. At the end of eah setion, there will be some pratie problems available for the students to pratie. Key Terms and Definitions The area of an objet is a physial quantity that expresses the size of a two dimensional objet suh as a square or retangle. The perimeter of an objet is the measured distane around the objet. The irumferene of a irle is a speial ase of perimeter that pertains to irles.
Key Formulas Objet Diagram Area Formula Perimeter Retangle w = width A = l w P = l + w l = length Triangle h = height 1 A = b h b = bas e Square = s s s P = 4s A = s = side Cirle r = radius A = π r C = π r = π d d = r The Area of a Retangle The area of a retangle 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 retangle. Example 1 (Area of a Retangle) Find the area of retangle with a length of 8 inhes and a width of 5 inhes. w = 5 in l = 8 in Solution: Substitute 8 inhes in for the length and 5 inhes in for the width into the area formula. A = lw = ( 8in)(4 in) = 3 square inhes
Example (Area and Perimeter) The area of a square is 49 square inhes. 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 an be found by substituting 49 square inhes in for the area of a square and solving the equation for s. A = s 49 = s To solve the equation for s you must take the square root of both sides of the equation. 49 = s s = 7 inhes Now, substitute 7 inhes into the perimeter formula. P = 4 (7in) = 8 in Here is an example that follows a similar proess, 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 entimeters. 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 an be found by substituting 40 m in for the perimeter of a square and solving the equation for s. P = 4s 40 = 4s 40 4s = 4 4 s = 10 m Now, simply use the area of a square formula to find the area of the square. A = s = ( 10m) = 100m
Triangles Example 4 Find the area of the triangle 4-5 5 - -4-6 Solution: To find the area of the triangle, you an ount the bloks to get the base and height of the triangle. 4 height = 3 units -5 5 base = 6 units - -4 Now, use the area formula for the triangle. 1 1 1 A = bh = (3)(6) = (18) = 9 square units
Area and Perimeter of a Cirle The irumferene of a irle measures the distane around the irle. Both the area of a irle and the irumferene of a irle are found by using a value alled Pi. Pi is a nonterminating and non-repeating deimal 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 irle Example 5 Find the area of irle with a radius of 5 inhes. Solution: To find the area of the irle, just substitute 5 units in for the radius of the irle and solve for the area A. A = π r A = π (5) A = 5π Example 6 Find the area of the irle. 4-5 5 10 - -4 Solution: First, find the radius by ounting the bloks in the diagram.
4-5 5 10 r = 3 units - -4 Now, substitute 3 units for radius into the area formula A = π r = π (3) = 9π Example 7 The following irle is insribed in a square. Find the area of the irle if the area of the square is 16 square feet. Solution: In order to find the area of the irle, you must first find the radius of the irle. The radius an be found by finding the diameter of the irle whih is equal to the length of the side of the square. The length of the side of the square an 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 = s 16 = s 16 = s s = 4 units
4 The radius of the irle is equal to half the diameter, so the radius would be = units. (See Diagram) 4 units units Now that you know the radius is units, use the area of irle to find the area of the irle. A = π r = π () = 4π Pratie Problems (Geometry Setion 1) 1) The area of a square is 64 square inhes. What is the perimeter of the square? ) The following irle is insribed in a square. Find the area of the irle if the area of the square is 36 square feet. 3) The perimeter of a square is 50 entimeters. What is the area of the square? 4) Find the area of the given triangle. Solutions: 1) 3 inhes ) 9 π 3) 100 square m 4) 15 square units
Setion Right Triangles and Similarity In this setion, you will learn some of the basi properties of right triangles. In partiular, you will learn to use the Pythagorean Theorem. You will also learn some of basi 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 speial names. The side opposite the right angle is alled the hypotenuse and the two other sides that form the right angle are alled the legs. = hypotenuse b = leg a = leg In a right triangle, there is a speial relationship between the sides of the right triangle alled 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. = a + b Now, let s try a few examples. Example 1 Find the missing side of the triangle in the figure below: =? b = 6 a = 8 Solution: Substitute the value of legs into the Pythagorean Theorem and solve for the hypotenuse.
= a = 8 = 64 + 36 = 100 = 10 + b + 6 = 100 Example Given the lengths of the sides of a right triangle are 5 and 7, find the length of the hypotenuse. =? b = 5 a = 7 Solution: Sine 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. = a = 7 + b + 5 = 49 + 5 = 74 = 74 = 74
Example 3 Find the length of the hypotenuse. Solution: By using the diagram above and ounting the bloks on the grid, you an find the lengths of the legs whih are 1 units and 5 units. Next, use the Pythagorean Theorem to find the hypotenuse. 5 units =? 1 units = 1 = 144 + 5 = 169 = a = 13 + b + 5 = 169
Example 4 Given that the length of hypotenuse of a right triangle is 17 inhes 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) 17 = a 89 = a 89 5 = a 64 = a = a + b + 15 + 5 + 5 5 a a = 8 = 64 Similar Triangles Similar triangles have sides that are proportional. If the sides of a triangle are proportional, then there is a speial ratio between the sides. (See diagram below) C B A D E F AB EF = BC ED = AC DF In eah 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. 5 3? 15 9 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 ross produt whih will result in: 15 x = 5(18) Now, solve for x. 15x = 90 15x 90 = 15 15 x = 6
Example 5 Triangle 1 and are similar triangles. If triangle 1 has sides 6,6,10 and 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 ross produt whih will result in: 6 x = 9(10) Now, solve for x. 6x = 90 6x 90 = 6 6 x = 15 Pratie Problems (Geometry Unit ) 1) Given the lengths of the sides of a right triangle are 1 and 16, find the length of the hypotenuse. ) 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 1,1,x, then find x? 1) 0 ) 85 3) 18
Setion 3 Volume The main topi of this setion is volume. You will speifially look at how to find the volume of various three dimension geometri objets suh as retangular solids and ylinders. The volume of an objet is the amount spae oupied by the objet. The proedure for finding volume of an objet is similar to finding the area of an objet in that it is simply a matter of substituting orret values into the proper formula. Here is a list of some of the key formulas used to find volume. Objet Shape Formula to Find Volume Retangular Solid V = l w h Cylinder V = π r h Cube 3 V = s Now, let s find the volume of some three dimensional objets.
Example 1 Find the volume of a ylinder with a radius of 3 inhes and height of 4 inhes. 3 in 4 in Solution: Substitute the values of the radius and the height into the volume formula. V = π r h V = π (3m) V = π (9 m V = 36π m )(4 m) 3 (4 m) Example Find the volume of retangular solid that is 3 feet by feet by 4 feet. 4 feet feet 3 feet Solution: Substitute the values of the length, width, and height into the volume of retangular solid formula. V = l w h = = 3 ( 3 ft)( ft)(4 ft) = (6 ft )(4 ft) 4 ft
Example 3 Find the volume of a retangular solid with a square base if its height is 6 inhes and the length of its base is 4 inhes. 6 feet 4 feet 4 feet Solution: Sine the base is square, the value of the length and width are equal. Therefore, the base of the retangular solid is 4 feet by 4 feet. So, to get the volume of the retangular solid just substitute the valuesl = 4 ft, w = 4 ft, and h = 6 ft into the formula V = l w h V V V V = l w h = (4 ft)(4 ft)(6 ft) = (16 ft = 96 ft 3 )(6 ft)
Pratie Problems (Geometry Unit ) 1) Find the volume of the following ylinder. 4 in 6 in ) Find the volume of a retangular solid that is 6 inhes by 5 inhes by 3 inhes. 3) Find the volume of a retangular solid with a square base if its height is 3 inhes and the length of its base is 6 inhes. Solutions: 1) 96 π ) 90 square inhes 3) 54 ubi inhes
Setion 4 Angle Measure The main objetive in this setion of the Geometry Unit is to find angle measures and estimate angle measures. By the end of the setion, you should be able to estimate an angle from a diagram given to you. In partiular, you should be able to estimate an angle from irular shaped diagram that is divided up into equal setions. In this situation, you just divide 360 by number of setions in the irle diagram. The reason you use 360 to make this alulation is that there is 360 degrees in a irle. Now try this example. Example 1 Find the measure of angle A A Solution: Sine a irle has 360 degrees and there are 4 setions in this irle, you simply divide 360 by 4 360 Angle measure = = 15 4 0
Example Find the measure of angle A A Solution: Sine a irle has 360 degrees and there are 6 setions in this irle, you simply divide 360 by 6. 360 Angle measure = = 0 60 6 Pratie Problems (Geometry Unit 4) 1) Find the measure of A. A ) Find the measure of A. A Solutions: 1) 0 30 ) 0 45