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Pairs of Angles LESSON 7-1 Course 3 Problem of the Day Students at West Ridge Middle School are going on a trip to the museum. Nine have never gone before. Twelve have gone once, half as many as that have gone twice, and none have gone more than that. How many students are going on the trip? 27 Lesson Main Lesson 7-1 Feature Pairs of Angles LESSON 7-1 Course 3 Check Skills You’ll Need (For help, go to Lesson 1-6.) 1. Vocabulary Review What is the inverse operation of addition? Solve each equation. 2. a + 14 = 32 3. b – 5 = 26 4. 10 + c = –31 5. –48 = d – 19 Check Skills You’ll Need Lesson Main Lesson 7-1 Feature Pairs of Angles LESSON 7-1 Course 3 Check Skills You’ll Need Solutions 1. subtraction 2. 18 3. 31 4. –41 5. –29 Lesson Main Lesson 7-1 Feature Pairs of Angles LESSON 7-1 Course 3 Additional Examples Find the measure of the supplement of x° + m IGJ = 180° IGJ. The sum of the measures of supplementary angles is 180º. x° + 145° = 180° Substitute 145º for m DEF. x° + 145° – 145° = 180° – 145° Subtract 145º from each side. x° = 35° Simplify. The measure of the supplement of m IGJ is 35º. Lesson Main Lesson 7-1 Quick Check Feature Pairs of Angles LESSON 7-1 Course 3 Additional Examples Name a pair of adjacent angles and a pair of vertical angles in the figure. Find m HGK. The adjacent angles are JGI and IGH; HGK and IGH and The vertical angles are KGJ; KGJ and JGI; HGK. JGI and HGK; Since vertical angles are congruent, m HGI and HGK = m KGJ. JGI = 145°. Quick Check Lesson Main Lesson 7-1 Feature Pairs of Angles LESSON 7-1 Course 3 Additional Examples In this figure, if m GKJ and JKF. of m DKE + 90° = 180° m DKE = 90° Lesson Main DKH = 73°, find the measures DKE and FKE are supplementary. Subtract 90º from each side. Lesson 7-1 Feature Pairs of Angles LESSON 7-1 Course 3 Additional Examples (continued) m KHE and KHE + 73° = 90° DKH are complementary. m KHE = 17° m GKJ = m KHE = 17° GKJ and KHE are vertical angles. m JKF = m DKH = 73° JKF and DKH are vertical angles. So, the measure of Subtract 73º from each side. GKJ is 17° is and the measure of JKF is 73°. Quick Check Lesson Main Lesson 7-1 Feature Pairs of Angles LESSON 7-1 Course 3 Lesson Quiz Use the diagram to answer Questions 1 – 3. 1. List all pairs of vertical angles. AXD and BXC; AXB and 2. List any angles adjacent to AXD and BXC DXC CXD. 3. If m AXB = 110°, find m DXC. 110° 4. An angle measures 57°. What is the measure of its supplement? 123° Lesson Main Lesson 7-1 Feature Angles and Parallel Lines LESSON 7-2 Course 3 Problem of the Day Write the following sentence in mathematical symbols. Then indicate whether the sentence is true or false: Four and one sixth minus five eighths equals three and thirteen twenty fourths. 1 5 13 4 6 – 8 = 3 24 ; true Lesson Main Lesson 7-2 Feature Angles and Parallel Lines LESSON 7-2 Course 3 Check Skills You’ll Need (For help, go to Lesson 7-1.) 1. Vocabulary Review Which of the following pairs of angles are supplementary? 50° and 40°, 100° and 90°, 120° and 60°, 75° and 125° Find the measure of the supplement of each angle. 2. 48° 3. 119° 4. 67° 5. 131° Check Skills You’ll Need Lesson Main Lesson 7-2 Feature Angles and Parallel Lines LESSON 7-2 Course 3 Check Skills You’ll Need Solutions 1. 120° and 60° 2. 132° 3. 61° 4. 113° 5. 49° Lesson Main Lesson 7-2 Feature Angles and Parallel Lines LESSON 7-2 Course 3 Additional Examples Identify each pair of corresponding angles and each pair of alternate interior angles. 1 and angles. 3, 2 and 4, 5 and 7, 6 and 8 are pairs of corresponding 2 and 7, 3 and 6, are pairs of alternate interior angles. Quick Check Lesson Main Lesson 7-2 Feature Angles and Parallel Lines LESSON 7-2 Course 3 Additional Examples If p is parallel to q, and m 3 = 56º, find m 6. m 6=m m 6 = 56° 3 = 56° Alternate interior angles are congruent. Quick Check Lesson Main Lesson 7-2 Feature Angles and Parallel Lines LESSON 7-2 Course 3 Additional Examples In the diagram below, m 5 = m 6 = and m 7 = 80º. Explain why p and q are parallel and why s and t are parallel. p || q because 5 and 7 are congruent alternate interior angles. s || t because 6 and 7 are congruent corresponding angles. Quick Check Lesson Main Lesson 7-2 Feature Angles and Parallel Lines LESSON 7-2 Course 3 Lesson Quiz Use the diagram to answer the questions. 1. Classify 4 and 7 as alternate interior angles, corresponding angles, or neither. neither 2. Classify 2 and 8 as alternate interior angles, corresponding angles, or neither. alternate interior angles 3. If a || b and m 8 = 80°, find m 4. 4. Suppose m 5 = 100° and m 3 = 100°. What can you conclude about line a and line b? a || b 80° Lesson Main Lesson 7-2 Feature Congruent Polygons LESSON 7-3 Course 3 Problem of the Day Solve the proportion: 4 n = . 5 27 3 n = 21 5 Lesson Main Lesson 7-3 Feature Congruent Polygons LESSON 7-3 Course 3 Check Skills You’ll Need (For help, go to Lesson 4-4.) 1. Vocabulary Review Congruent angles have ? measures. Are the polygons similar? Explain. 2. Check Skills You’ll Need Lesson Main Lesson 7-3 Feature Congruent Polygons LESSON 7-3 Course 3 Check Skills You’ll Need Solutions 1. equal 2. LN ZY LM ; 5 =/ 4 ; not similar; corresponding sides are not in XZ 15 10 proportion. Lesson Main Lesson 7-3 Feature Congruent Polygons LESSON 7-3 Course 3 Additional Examples In the diagram below, list the congruent parts of the two figures. Then write a congruence statement. Congruent Angles A M B N C O D P Congruent Sides AB MN BC NO CD OP DA PM Since A corresponds to M, B corresponds to N, C corresponds to O, and D corresponds to P, a congruence statement is ABCD MNOP. Quick Check Lesson Main Lesson 7-3 Feature Congruent Polygons LESSON 7-3 Course 3 Additional Examples Show that each pair of triangles is congruent. a. b. Q QP Q QPR E EY Y Angle Side Angle EYT by ASA. SQ VT Q T QR TU SQR Side Angle Side VTU by SAS. Quick Check Lesson Main Lesson 7-3 Feature Congruent Polygons LESSON 7-3 Course 3 Additional Examples A surveyor drew the diagram below to find the distance from J to I across the canyon. Show that GHI KJI. Then find JK. J JI KIJ H Both are right angles. HI Both measure 48 ft. GIH They are vertical angles. So ∆GHI ∆ JI by ASA Corresponding parts of congruent triangles are congruent. JK corresponds to HG, so JK is 36 ft. Lesson Main Lesson 7-3 Quick Check Feature Congruent Polygons LESSON 7-3 Course 3 Lesson Quiz Use ABC and XYZ to answer the questions. 1.Suppose AC= XZ, AB = XY, and BC = YZ. Write a congruence statement for the figures. ∆ABC ∆XYZ 2. Suppose ABC and XYZ are congruent. If AB = 5 cm, BC = 8 cm, and AC = 10 cm, find XZ. 10 cm 3. Suppose ABC B Y, XYZ? A X, and AB XY. Why is ASA Lesson Main Lesson 7-3 Feature Congruent Polygons LESSON 7-3 Course 3 Lesson Quiz 4. Let AB = XY = 9 inches; BC = YZ = 24 inches; and m B = 85°, m Z = 35°, and m Y = 85°. Prove that the triangles are congruent and find m C. ∆ABC ∆XYZ by SAS; m C = 35° Lesson Main Lesson 7-3 Feature Classifying Triangles and Quadrilaterals LESSON 7-4 Course 3 Problem of the Day Ruth was born on February 29, 1984. If she insists on celebrating her birthday only on February 29, when will she celebrate her 12th birthday? 2032 Lesson Main Lesson 7-4 Feature Classifying Triangles and Quadrilaterals LESSON 7-4 Course 3 Check Skills You’ll Need (For help, go to the Skills Handbook page 640.) 1. Vocabulary Review How many degrees does a right angle have? Classify each angle as acute, right, obtuse, or straight. 2. 3. Check Skills You’ll Need Lesson Main Lesson 7-4 Feature Classifying Triangles and Quadrilaterals LESSON 7-4 Course 3 Check Skills You’ll Need Solutions 1. 90° 2. acute 3. obtuse Lesson Main Lesson 7-4 Feature Classifying Triangles and Quadrilaterals LESSON 7-4 Course 3 Additional Examples Classify LMN by its sides and angles. The triangle has two sides that are congruent and three acute angles. It is an isosceles acute triangle. Quick Check Lesson Main Lesson 7-4 Feature Classifying Triangles and Quadrilaterals LESSON 7-4 Course 3 Additional Examples What is the best name for figure WXYZ? Explain your choice. WXYZ has both pairs of opposite sides parallel, but adjacent sides are not equal, so it is a parallelogram. Quick Check Lesson Main Lesson 7-4 Feature Classifying Triangles and Quadrilaterals LESSON 7-4 Course 3 Lesson Quiz 1. Classify the triangle according to its angles and sides. obtuse isosceles triangles 2. A triangle’s sides are all congruent and its angles all measure 60°. Classify the triangle. equilateral, acute 3. Determine the best name for the quadrilateral. rhombus Lesson Main Lesson 7-4 Feature Classifying Triangles and Quadrilaterals LESSON 7-4 Course 3 Lesson Quiz 4. What is the best name for a figure that has four sides congruent, corresponding sides parallel, and all four angles congruent? square Lesson Main Lesson 7-4 Feature Angles and Polygons LESSON 7-5 Course 3 Problem of the Day A pound of turkey has 144 g of protein and will serve 4 people. If 4 people are served equal amounts, how many grams of protein will each receive? 36 g Lesson Main Lesson 7-5 Feature Angles and Polygons LESSON 7-5 Course 3 Check Skills You’ll Need (For help, go to the Lesson 1-1.) 1. Vocabulary Review How do you evaluate an algebraic expression? Evaluate each expression for a = 8. 2. 3(a + 1) 3. 5a + 8 a 4. (a – 2)6 Check Skills You’ll Need Lesson Main Lesson 7-5 Feature Angles and Polygons LESSON 7-5 Course 3 Check Skills You’ll Need Solutions 1. You replace each variable in the expression with a number and then simplify. 2. 27 3. 6 4. 36 Lesson Main Lesson 7-5 Feature Angles and Polygons LESSON 7-5 Course 3 Additional Examples Find the sum of the measures of the interior angles of an octagon. An octagon has 8 sides. (n – 2) 180º = (8 – 2) 180º Substitute 8 for n. = (6) 180º Subtract. = 1,080º Simplify. The sum of the interior angles of an octagon is 1,080. Quick Check Lesson Main Lesson 7-5 Feature Angles and Polygons LESSON 7-5 Course 3 Additional Examples Find the missing angle measure in the hexagon. Step 1 Find the sum of the measures of the interior angles of a hexagon. (n – 2) 180° = (6 – 2) 180° Substitute 6 for n. = 720° Lesson Main Simplify. Lesson 7-5 Feature Angles and Polygons LESSON 7-5 Course 3 Additional Examples (continued) Step 2 Write an equation. Let x = the missing angle measure. 720° = 120° + 115° + 136° + 80° + 147° + x° Write an equation. 720° = 598° + x° Add. 122° = x° Subtract 598º from each side. The missing angle measure is 122º. Quick Check Lesson Main Lesson 7-5 Feature Angles and Polygons LESSON 7-5 Course 3 Additional Examples A design on a tile is in the shape of a regular nonagon. Find the measure of each angle. (n – 2) 180° = (9 – 2) 180° Substitute 9 for n since a nonagon has 9 sides. Simplify. = 1,260° Divide the sum by the number of angles in a nonagon. 1,260° ÷ 9 = 140° Each angle of a regular nonagon has a measure of 140°. Quick Check Lesson Main Lesson 7-5 Feature Angles and Polygons LESSON 7-5 Course 3 Lesson Quiz 1. Find the sum of the measures of the interior angles of a polygon having 17 sides. 2,700° 2. Five angles of a hexagon measure 128°, 190°, 112°, 154°, and 90°. Find the measure of the missing angle. 46° 3. Find the measure of each angle of a regular polygon having 24 sides. 165° 4. A regular figure has an interior angle measure of 135°. How many sides does it have? 8 Lesson Main Lesson 7-5 Feature Areas of Polygons LESSON 7-6 Course 3 Problem of the Day Lucy predicted that her final average for math class would be at least a 93. Her test grades were 88, 90, 92, 97. What is the lowest test score she can make on the last test to make this true? 98 Lesson Main Lesson 7-6 Feature Areas of Polygons LESSON 7-6 Course 3 Check Skills You’ll Need (For help, go to Lesson 2-6.) 1. Vocabulary Review What is a formula? Find the area of each figure. 2. 3. Check Skills You’ll Need Lesson Main Lesson 7-6 Feature Areas of Polygons LESSON 7-6 Course 3 Check Skills You’ll Need Solutions 1. A formula is a rule that shows the relationship between two or more quantities. 2. A = • w = 10 • 8 A = 80 cm2 Lesson Main 3. A = s2 = 72 A = 49 ft2 Lesson 7-6 Feature Areas of Polygons LESSON 7-6 Course 3 Additional Examples Find the area of the triangular part of the doghouse. 1 A = 2 bh Use the area of a triangle formula. 1 = 2 • 36 • 21 Substitute 36 for b and 21 for h. = 378 Multiply. The area of the triangular part of the doghouse is 378 in.2. Quick Check Lesson Main Lesson 7-6 Feature Areas of Polygons LESSON 7-6 Course 3 Additional Examples Find the area of the trapezoid. A = 1 h (b1 + b2) 2 Use the formula. = 1 (4.4) (6.7 + 9.3) Substitute 4.4 for h, 6.7 for b1, and 9.3 for b2. = 35.2 Simplify. 2 The area of the trapezoid is 35.2 in.2. Lesson Main Lesson 7-6 Quick Check Feature Areas of Polygons LESSON 7-6 Course 3 Lesson Quiz 1. Find the area. 14 cm2 2. An architect is designing a restaurant with a triangular entrance. The base of the triangle is 10 ft wide. The entrance is 14 ft tall. Find the area of the triangle. 70 ft2 Lesson Main Lesson 7-6 Feature Areas of Polygons LESSON 7-6 Course 3 Lesson Quiz 3. Find the area. 56 ft2 4. If both bases of the figure in Exercise 3 are doubled, what is the new area of the trapezoid? 112 ft2 Lesson Main Lesson 7-6 Feature Circumference and Area of a Circle LESSON 7-7 Course 3 Problem of the Day Opal, Charles, Jean, and Scott had an earthworm-catching contest. Jean caught one fourth as many worms as Opal and twice as many as Charles. Opal caught 3 times as many worms as Scott. How many worms did each of the other three contestants catch if Scott caught 8 worms? Opal, 24; Jean, 6; Charles, 3 Lesson Main Lesson 7-7 Feature Circumference and Area of a Circle LESSON 7-7 Course 3 Check Skills You’ll Need (For help, go to Lesson 7-6.) 1. Vocabulary Review Explain the difference between perimeter and area. Find the area of the figure below. 2. Check Skills You’ll Need Lesson Main Lesson 7-7 Feature Circumference and Area of a Circle LESSON 7-7 Course 3 Check Skills You’ll Need Solutions 1. Perimeter is the distance around a figure. Area is the number of square units a figure encloses. 2. A = 1 bh 2 = 1 • 13 • 7 2 A = 45.5 ft² Lesson Main Lesson 7-7 Feature Circumferences and Areas of Circles LESSON 7-7 Course 3 Additional Examples The diameter of a tractor tire is 125 cm. Find the circumference and area. Round to the nearest tenth. C= = d Use the formula for circumference. (125) Substitute 125 for d. 125 Use a calculator. 392.6990817 The circumference is about 392.7 cm. Lesson Main Lesson 7-7 Feature Circumferences and Areas of Circles LESSON 7-7 Course 3 Additional Examples (continued) A= = r2 Use the formula for the area of a circle. (62.5) 2 The radius is 125 ÷ 2, or 62.5. Substitute 62.5 for r. 62.5 Use a calculator. 12271.8463 The area is about 12,271.8 cm2. Quick Check Lesson Main Lesson 7-7 Feature Circumferences and Areas of Circles LESSON 7-7 Course 3 Additional Examples Find the area of the unshaded region of the square tile with a circle inside of it, as shown below. Round to the nearest tenth. You can separate the figure into a circle and a square. Lesson Main Lesson 7-7 Feature Circumferences and Areas of Circles LESSON 7-7 Course 3 Additional Examples (continued) Step 1 Find the area of the square. Area of square = s2 A = (12)2 Substitute 12 for s. = 144 Simplify. Step 2 Find the area of the circle. Area of circle = A= r2 (6)2 113.1 Lesson Main Substitute 6 for r. Multiply. Round to the nearest tenth. Lesson 7-7 Feature Circumferences and Areas of Circles LESSON 7-7 Course 3 Additional Examples (continued) Step 3 Subtract the area of the circle from the area of the square. The area of the shaded region is about 144 cm2 – 113.1 cm2 = 30.9 cm2. Quick Check Lesson Main Lesson 7-7 Feature Circumferences and Areas of Circles LESSON 7-7 Course 3 Lesson Quiz 1. Find the circumference and area of the circle. Round to the nearest tenth. 18.8 km; 28.3 km2 2. The diameter of the lid of a can of soup is 5 cm. Find its circumference and area. Round to the nearest tenth. 15.7 cm; 19.6 cm2 Lesson Main Lesson 7-7 Feature Circumferences and Areas of Circles LESSON 7-7 Course 3 Lesson Quiz 3. Find the area. Round to the nearest tenth. 273.0 in.2 Lesson Main Lesson 7-7 Feature Constructions LESSON 7-8 Course 3 Problem of the Day Estimate in years the age of someone who is a million minutes old. about 2 yr Lesson Main Lesson 7-8 Feature Constructions LESSON 7-8 Course 3 Check Skills You’ll Need (For help, go to Lesson 7-3.) 1. Vocabulary Review How are congruent polygons the same? List the congruent parts of each pair of congruent figures. 2. ∆PQR ∆TUV 3. ABCD LMNO Check Skills You’ll Need Lesson Main Lesson 7-8 Feature Constructions LESSON 7-8 Course 3 Check Skills You’ll Need Solutions 1. Congruent polygons have the same size and shape. 2. P PQ 3. A AB T; Q TU; QR L; B LM; BC Lesson Main U; R V; UV; RP VT M; C MN; CD N; D NO; DA Lesson 7-8 O; OL Feature Constructions LESSON 7-8 Course 3 Additional Examples Use a protractor to draw a 40º angle and label it Construct N congruent to B. B. Step 1 Draw a ray with endpoint N. Step 2 Put the compass tip at B and draw an arc that intersects the sides of B. Label the points of intersection A and C. Step 3 Keep the compass open to the same width. Put the compass tip at N. Draw an arc that intersects the ray at a point O. Lesson Main Lesson 7-8 Feature Constructions LESSON 7-8 Course 3 Additional Examples (continued) Step 4 Adjust the compass width So that the tip is at C and the pencil is at A. Step 5 Keep the compass open to the same width. Put the compass tip on O and draw an arc that intersects the first arc at point M. Step 6 Draw NM. N is congruent to Lesson Main Quick Check B. Lesson 7-8 Feature Constructions LESSON 7-8 Course 3 Additional Examples Draw a line f. Construct a line parallel to line f. Step 1 Draw line f. Step 2 Draw line k that intersects line f at D. Label the angle formed 1. Then label point E on line k. Step 3 Construct an angle at E that is congruent to 1. EC is parallel to line f. Quick Check Lesson Main Lesson 7-8 Feature Constructions LESSON 7-8 Course 3 Lesson Quiz For Exercises 1–2, use the diagram below. 1. Construct an angle that is congruent to R. 2. Construct a line parallel to RS. Lesson Main Lesson 7-8 Feature