GEOMETRY 195 COMPETENCY test of properties and facts of plane and solid shapes and their construction. problem-solving related to coordinates, Pythagoras' theorem and transformation with uses of their concepts. CHAPTERS 11. Lines and Angles 12. Plane Shapes 13. Similarity and Congruency 14. Solid Objects 15. Coordinates 16. Symmetry and Tessellation 17. Transformation 18. Bearing and Scale Drawing LEARNING OUTCOMES identify the angles formed when a transversal intersects two straight lines, test the relation of the angles formed when a transversal intersects two parallel lines, discover the properties of a triangle and test experimentally, discover the properties of the quadrilateral (rhombus, trapezium and kite) and test experimentally, construct rectangle, square and parallelogram using given conditions, identify the polygon and measure the interior and exterior angle of the polygon and then establish their relation, test the conditions of congruent triangles, solve the problems related to congruent triangles, recognize and discover similar shapes. recognize the relation between triangular prism and pyramid, construct the nets of cube, cuboid, tetrahedron, cone and cylinder, find the distance between two points, discover regular and irregular tessellation and prepare them, reflect the triangle in the graph, translate the point, line segment and triangle up and upward on the basis of coordinates, rotate the point, line segment and triangle around the origin through 90°, use the bearing and scale drawing in the map. GEOMETRY UNIT V Estimation Teaching Hours - 42 (Th. + Pr.)
196 The Leading Maths - 8 CHAPTER 11 LINES AND ANGLES Lesson Topics Pages 11.1 Review on Point, Line and Angle 197 11.2 Pair of Angles on Parallel Lines and Transversal 207 What are line and line segment ? Can you measure the line and line segment ? How are the edges of the room, board, book and ruler ? How are the corners of the board, book and classroom ? Can you find the measure of the angles ? Can you draw the given angles by using protractor ? How can you identify the greater and smaller angles ? Why do we use set-squares ? Why do we use compass ? WARM-UP
GEOMETRY 197 At the end of this topic, the student will be able to: ¾ solve the problems on different kinds of angles. Learning Objectives 11.1 Review on Point, Line and Angle I. Some Basic Terminology A point is dimensionless i.e. lengthless, breadthless and heightless. Such points form a line in the same or different direction. Picture Definition Notation P A point is a position in space that has no length, breadth and height. Point P A B A line is a straight path of the points that are in continuous infinitely. Two points determine a line. AB ↔ or BA ↔ P Q A ray is a part of a line that has one end point. PQ → A B A line segment is a part of a line that has two end pairs. AB or AB P Q R A plane is a flat surface that extends continuous infinitely. a A O B An angle is formed by two rays or lines with a common end point. The rays are sides of the angle, and the end point is the vertex. ∠ABC P O Q R S Intersecting lines meet or intersect at a point. m t Parallel lines are the lines in the same plane that never intersect. They are the same or constant distance at all corresponding points. t // m m n Perpendicular lines are the lines that intersect with a right angle. m ⊥ n
198 The Leading Maths - 8 II. Measuring Angles and their Types An angle is measured in degree (°). Angles are classified according to their measures as follows: Name Measure of Angles x Figure Null Angle x = 0° x Right Angle x = 90° x Acute Angle 0° < x < 90° x Obtuse Angle 90° < x < 180° x Straight Angle x = 180° x Reflex Angle 180° < x < 360° x Complete Angle x = 360° x Two rays are said to be opposite if the angle between them is 180°. In the figure, OA → and OB → are opposite rays. Two angles are said to be equal, if they have the same measure. The measures of ∠EOD and ∠AOB are 30°, so they are equal angles. Read as ∠EOD equals ∠AOB. Write as ∠EOD = ∠AOB. B O A 0 180 20 160 30 150 40 140 50 130 60 120 70 110 80 100 90 100 80 110 70 120 60 130 50 140 40 150 30 160 20 170 10 10 170 180 0 E O A C B D
GEOMETRY 199 III. Types of Pair of Angles Some angles which are in pair with special name and character. Look at the table below: Special Name Definition Example Complementary Angles Two angles whose sum is equal to 90°. Each of the two angles are called complement. x + y = 90° y x A O B C Supplementary Angles Two angles whose sum is equal to 180°. Each of the two angles are called supplement. a + q = 180° O B a q A C Adjacent Angles Angles in the same plane that have (i) common vertex, (ii) a common side and (iii) the other areas are on opposite side of the common side are called the adjacent angles. C B A O B O A C Vertical angles or Vertically Opposite Angles Two non-adjacent angles formed by two intersecting lines (or line segments) are called vertically opposite angles. ∠AOC and ∠BOD ∠AOD and ∠BOC Linear Pair of Angles Adjacent angles whose noncommon sides are opposite rays are called linear pair of angles. The sum of their measure is 180°. ∠AOD and ∠DOB ∠DOB and ∠BOC ∠BOC and ∠AOC ∠COA and ∠AOD A C D B O A B D O C
200 The Leading Maths - 8 IV. Some Important Properties S.N. Properties Figure To prove 1. The complementary angles of the same angles are equal. A B C D O x y z ∠x = ∠z 2. The supplementary angles of the same angles are equal. a D C B A c b ∠a = ∠b. 3. When two straight lines intersect, the measure of the sum of adjacent angles is 180°. b D B A C a c O d (i) ∠a + ∠b = 180° (ii) ∠a + ∠d = 180° (iii) ∠c + ∠d = 180° (iv) ∠b + ∠c = 180° 4. When two straight lines intersect at a point, the pair of vertically opposite angles is equal. d b D C B A a c O (i) ∠a = ∠c (ii) ∠b = ∠d CLASSWORK EXAMPLES Example : 1 In the figure, the angles AOB and BOC are complementary angles. If ∠BOC = 22°. (a) Write the relation between ∠AOB and ∠BOC. (b) Find the size of ∠AOB marked x. Solution: (a) ∠AOB + ∠BOC = 90° A O C B x 22°
GEOMETRY 201 (b) Since angles AOB and BOC are complementary angles, so, ∠AOB + ∠BOC = 90° or, x + 22° = 90° [ Substitution] or, x = 90° – 22° [ Transposition] or, x = 68° Hence, ∠AOB = 68°. Example : 2 Angles AOB and COD are complementary to the same angle BOC. (a) Define supplementary angles. (b) Find the value of x. (c) Find the value of ∠AOB. (d) What is the measure of ∠BOC? Solution: (a) The sum of angles is 180°, they are called supplementary angles. (b) Given, ∠AOB = ∠COD [ Complementary angles of ∠BOC] or, 2x – 13° = x + 17° or, 2x – x = 17° + 13° or, x = 30° (c) ∠AOB = 2a – 13° = 2 × 30° – 13° = 60° – 13° = 47° (d) ∠BOC = 90° – ∠AOB = 90° – 47° = 43° Example : 3 The lines AB and CD intersect at O and ∠AOC = 28°. (a) Why are the angles x and z equal? (b) What is the value of x? (c) Find the measure of the angles y and z. Solution: (a) The angles x and z are equal in th given figure because they are vertically opposite angles. (b) Given, ∠AOC + ∠AOD = 180° [ Linear or Adjacent angles] or, 28° + x = 180° [ Substitution] or, x = 180°– 28° [ Transposition property] ∴ x = 152° A B D O x–17° C 2x–13° z 28° y A x D C B O
202 The Leading Maths - 8 (c) ∠DOB = ∠AOC [ VOA] or, y = 28° and ∠BOC = ∠AOD [ VOA] or, z = 152° ∴ x = 152°, y = 28° and z = 152° as required. EXERCISE 11.1 Your mastery depends on practice. Practice like you play. Read, Understand, Think and Do Keeping Skill Sharp 1. (a) Define parallel lines. (b) What is the perpendicular distance between a pair of parallel lines? (c) What are perpendicular lines? Define. (d) Write the difference between acute angle and obtuse angle. (e) Write the relation between right angle and straight angle. (f) Why is supplementary angle different from complementary angle? (g) What is the relation between a pair of vertical opposite angles? (h) If A and B are two complementary angles and A = 30°, what is the value of B? 2. Circle ( ) the correct answers. (a) A line segment has fixed ................ i. direction ii. length iii. angle iv. both (i) and (ii) (b) A pair of parallel lines intersect at ................ i. only one point ii. two points iii. three points iv. no point (c) An obtuse angle is ................. acute angle. i. less than ii. greater than iii. equal to iv. not equal with (d) If x and y are supplementary angles and x = 48°, which is the value of y? i. 132° ii. 42° iii. 138° iv. 228° (e) If a and b are vertically opposite angles and b = 50°, what is the correct value of a? i. 40° ii. 130° iii. 50° iv. 100°
GEOMETRY 203 3. Observe the following complementary angles and answer the given questions. (i) A O x 62° C B (ii) D O x 60° C B (iii) G F E 35° O x (iv) H O I J x 72° (a) Write the definition of the complementary angle. (b) Calculate the measurement of unknown angles in each figure. 4. In each of the figures below x and y are the complementary angles to the angle BOC. (i) D O B A C 64°x y (ii) B 45° O D x x y C (iii) O 30° Q U B x y A (a) Find the sizes of the angles marked x and y. (b) Is x = y? Why ? 5. Observe the following supplementary angles. (i) A O C B 30° x (ii) C AO B 40° x (iii) A O C 90° x B (iv) B A O C 135° x (a) What is the supplementary angles ? Define them.. (b) Find the measure of ∠AOB.
204 The Leading Maths - 8 6. In the figure, angle AOC is supplementary to the angles marked x and y. (i) (ii) (iii) (a) Define vertically opposite angles. (b) Find the values of x and y. (c) Is x = y ? Why ? (d) What is your decision about the vertically opposite angles AOD and COB? 7. ∠COB = 140° in the figure. Can you find the size of angles? (a) ∠AOC (b) ∠AOD (c) ∠BOD? (d) What is the sum of angles at the point O around it? 8. If ∠AOB and ∠COD are complementary to the same angle BOC. (i) 2x – 17° x + 14° C B A O D (ii) 3x – 14° 2x – 1° A B C D O (iii) A O B C D 5x–26° 2x+16° (a) Find the value of x. (b) What is the measure of ∠COD? (c) Find the value of ∠BOC. x D C B A 35° y O A D C y 80° x B O A 135° C x y D B O A D C 140° B O
GEOMETRY 205 9. Study the following figures. (i) A B O C D 2x + 14° 130° (ii) A C O B D 150° 3x+18° (iii) A B C D 2x – 10° x + 10° O (iv) A C O B D 3x–9° 2x+7° (a) Find the value of x. (b) Find the measure of ∠AOC in each figure. 10. In each of the following figures, find the size of angles marked a, b and c. (a) A B C D 3a c b a (b) A B C D 3a c 2a b (c) A B C D 4a 2a 3b 2c (d) A B C D 5a c a b 11. (a) In the figure, ∠PQR = ∠PRQ. Prove with statements and reasons that ∠PQS = ∠PRT. (b) In the figure, ∠MNP = ∠MOQ. Prove that statements and reasons that ∠MNO = ∠MON. S Q P R T P N M O Q
206 The Leading Maths - 8 (c) In the figure, ∠FBC = ∠GCB. Prove with statement and reasons that ∠EBC = ∠ECB. 12. (a) In the figure, if ∠PQS = 90°, prove with statements and reasons that ∠PQR + ∠TQS = 90°. (b) In the figure, QT and QU are bisectors of the angles PQR and RQS. Prove that ∠TQU is a right angle. ANSWERS 3. (i) 28° (ii) 30° (iii) 55° (iv) 18° 4. (i) x = 64°, y = 64° ∴ x = y (ii) x = 45°, y = 45°, ∴ x = y (iii) x = 60°, y = 60°∴ x = y 5. (i) 150° (ii) 140° (iii) 90° (iv) 45° 6. (i) 145°, 145° (ii) 100°, 100° (iii) 45°, 45° (c) Yes, x = y (d) Vertically opposite angles AOD & COB are equal. 8. (i) (a) 31° (b) 45° (c) 45° (ii) (a) 13° (b) 25° (c) 25° (iii) (a) 16° (b) 48° (c) 42° 9. (i) (a) 18° (b) 50° (ii) (a) 4° (b) 30° (iii) (a) 20° (b) 30° (iv) (a) 16° (b) 39° 10. (a) a = 45°, b = 45°, c = 135°, (c) a = 30°, b = 40°, c = 30° (b) a = 36°, b = 72°, c = 108° (d) a = 30°, b = 30°, c = 150° A B F G D C E R S P Q T P Q S U R T
GEOMETRY 207 At the end of this topic, the student will be able to: ¾ solve the problems related to the pair of angles made by transversal with lines. Learning Objectives 11.2 Pair of Angles Made by Transversal with Lines I. Transversal A line which intersects two or more lines (not necessarily parallel) in a plane at different points is called a transversal. t s p Line t is a transversal. Line s is a transversal. Line p is not a transversal. In the figure on the right, the two lines l and s are intersected by the transversal t. Eight angles are formed. They are ∠a, ∠b, ∠c, ∠d, ∠e, ∠f, ∠g and ∠h. II. Types of Pair of Angles Made by Transversal Special names Definition Examples Exterior angles Angles formed (out-side) the two lines (l and s) ∠a, ∠b, ∠g and ∠h Interior angles Angles formed between the two lines (l and s) ∠c, ∠d, ∠e and ∠f Co-interior (consecutive co- formed angles) interior angles(C-Formed) Interior angles on the same side of the transversal ∠c and ∠f; ∠d and ∠e Co-exterior angles Exterior angles on the same side of line transversal ∠g and ∠b; ∠h and ∠a Alternate exterior angles Non-adjacent exterior angles on two sides of the transversal ∠b and ∠h, ∠land ∠g l s t h g e a b c d f
208 The Leading Maths - 8 Special names Definition Examples Alternative interior angles (z-formed angles) Non-adjacent interior angles on two sides of two transversals ∠c and ∠e; ∠d and ∠f Corresponding angles (F-formed angles) Non-adjacent pair of angles one interior and another exterior angle on the same side of the transversal. ∠a and ∠e, ∠b and ∠f, ∠d and ∠h, ∠c and ∠g III. Pair of Angles on Parallel Lines and Transversal Activity 1 In a paper draw a pair of parallel lines and the transversal. Measure the size of angles, a, b, c, d, e, f, g and h. Angles ∠a ∠b ∠c ∠d ∠e ∠f ∠g ∠h Measure (a) Are the corresponding angles equal? (b) Are the alternate angles equal? (c) Are the co-interior angles adding up to 180°? Write your conclusion about (a) Corresponding angles (b) Alternative interior angles (c) Sum of co-interior angles and (d) Alternate exterior angles When the two parallel lines are intersected by a transversal, the corresponding angles postulate (CAP) as follows: The height or perpendicular between two parallel lines is always constant and the angle between them is zero. When these two parallel lines coincide, the angles ∠b and ∠f also coincide. Hence, ∠b = ∠f. THEOREM - 1 The corresponding angles are equal, if pair of parallel lines are intersected by a transversal. Corresponding Angles Theorem (CAT) In the figure, p//q and t is the transversal. Use the figure and give reason why the following statements are true. a e d h b p t q f c g a e d h b p t f q c g
GEOMETRY 209 Statements Reasons 1. ∠a = ∠c 1. Vertically opposite angles (VOA) 2. p//q 2. Given 3. ∠a = ∠e 3. Corresponding angles property (CAP) 4. ∠b + ∠c = 180° 4. Sum of linear pairs of angles 5. ∠h = ∠d 5. - - 6. ∠c = ∠g 6. - - 7. ∠e = ∠g 7. - - 8. ∠f + ∠g = 180° 8. - - We shall use corresponding angles property as axiom and prove the following theorems. To prove this theorem, we have to be sure about several things we are going to do. The following are the steps we have to follow one after another. Understanding the statement of theorem Draw figure according to the statement of the theorem. List the conditions that are given. List the items that need to be proved. Demonstrate the proof with statements/arguments and reasons. Let us prove theorem-1 step by step as suspected above. THEOREM - 2 If a pair of parallel lines are intersected by a transversal, each pair of alternate interior angles are equal. Alternate Interior Angle Theorem (AIAT) Given: Draw p//q and t is the transversal so that angles d and f, and c and e are the pairs of alternate interior angles. To prove: (a) ∠d = ∠f (b) ∠c = ∠e Proof: Statements Reasons 1. ∠d = ∠b 1. Vertically opposite angles (VOA) a e d h b p t q f c g
210 The Leading Maths - 8 Statements Reasons 2. ∠b = ∠f 2. Corresponding angles property 3. ∠d = ∠f 3. Transitive property of equal angles ( stat. 2) 4. Similarly, ∠c = ∠e 4. As above This completes the proof. THEOREM - 3 If a pair of parallel lines are intersected by a transversal, each pair of cointerior angles are supplementary. Co-interior Angles Theorem (CIAT) Given: Draw p//q and t is the transversal so that ∠c and ∠f, and ∠d and ∠e are the pairs of co-interior angles. To prove: (i) ∠c + ∠f = 180° (ii) ∠d + ∠e = 180° Proof: Statements Reasons 1. ∠b + ∠c = 180° 1. Linear pair of angles 2. ∠b = ∠f 2. Corresponding angle property 3. ∠c + ∠f = 180° 3. Replacing ∠f for ∠b 4. Similarly, ∠d + ∠e = 180° 4. As above This completes the proof. CLASSWORK EXAMPLES Example : 1 Observe the given figure. (a) Define corresponding angles. (b) Find the values of a, b and c. Solution: (a) Non-adjacent pair of interior and exterior angles of the same side of the transversal of two straight lines are called corresponding angles. a e d h b p t q f c g a 55° 5b 2c 3b
GEOMETRY 211 (b) From the given figure, (i) a + 55° = 180° [Why ?] or, a = 180° – 55° = 125° (ii) a = 5b [Why ?] or, 125° = 5b or, b = 125° 5 = 25° (iii) 3b = 3 × 25° = 75° (iv) 2c = 3b or, 2c = 75° or, c = 75° 2 or, c = 37.5° Example : 2 In the figure, p//q and t is the transversal and ∠b = 60°. (a) Write one pair of alternate angles. Are they equal? (b) Find the size of the angles marked a, b, c, d, e, f, g. Solution: (a) The angles d and f are the alternate angles. They are equal because they form in the pair of parallel lines p and q. (b) From the given figure, (i) ∠a + ∠b = 180° [ Linear pair of angles] or, ∠a + 60° = 180° [ Substitution] or, ∠a = 180° – 60° = 120° (ii) ∠c = ∠a = 120° [ VOA] (iii) ∠d = ∠b = 60° [ VOA] (iv) ∠e = ∠a = 120° [ CAP] (v) ∠h = ∠d = 60° [ CAP ] (vi) ∠f = ∠b = 60° [ CAP] (vii) ∠g = ∠c = 120° [ CAP] a e d h b p t q f c g
212 The Leading Maths - 8 EXERCISE 11.2 Your mastery depends on practice. Practice like you play. Read, Understand, Think and Do Keeping Skill Sharp 1. (a) Define co-interior angles between two lines intersected by a transversal. (b) What are alternate angles? Define. (c) What types of pair of angles between two parallel lines intersected by a transversal are called corresponding angles? (d) What is the relation between the alternate angles made by a transversal on a pair of parallel lines? (e) What is the relation between the co-interior angles made by a transversal on a pair of parallel lines? 2. Circle ( ) the correct answers. (a) Which pair of angles on a pair of parallel lines are made by a transversal? i. co-interior angles ii. complementary angles iii. adjacent angles iv. supplementary angles (b) What is the value of a in the given figure? i. 120° ii. 30° iii. 60° iv. 90° (c) Which is the co-interior angle of x? i. w ii. y iii. u iv. z (d) Which is the corresponding angle of ∠AGF? i. ∠ BGF ii. ∠ EHD iii. ∠ CHF iv. ∠ CHE (e) If a pair of alternate angles in two parallel lines are 3p and 60°, what is the value of p? i. 60° ii. 20° iii. 180° iv. 30° a 60° z u v y x w C A F B E D G H
GEOMETRY 213 3. Refer to figure on the right, answer the following according to the reform of ∠d. (a) Which are the non-adjacent angles of ∠d. (b) Is ∠d an interior angle? (c) Which angle is the alternate interior of ∠d? (d) Which angle is the co-interior angle of ∠d? (e) Which angle is its corresponding angle? (f) Which of the exterior angles that are not its corresponding angles? 4. Refer to figure on the right, answer the following questions: (a) Which are the corresponding angles of ∠e? (b) Which are the alternate angles of ∠c? (c) Give the special pair of angles if it has (i) ∠a and ∠b (ii) ∠d and ∠f (iii) ∠f and ∠c (iv) ∠e and ∠f (v) ∠a and ∠c 5. Observe the following figures. (a) Write the definition of alternate angles. (b) Find the values of a, b, c and d. (i) (ii) 6. In each of the following figure, the parallel lines and the transversal are given. (a) In which condition does a pair of corresponding angles equal? (b) Find the size of the missing angles x, y, z, p, q, r and so on. (i) w 120° q z r x p t q p y s (ii) w b q y 120° c r x a p z d s l1 l2 t h g c a b e f d a b c f d f l 2a 2d 60° b+20° t b m c 130° 5a b l m c d t 2a
214 The Leading Maths - 8 (iii) x r s z q y p w 135° (iv) i k h j d e g a f c l 40° b (v) x a w 80° d y b z c (vi) a d b c p w s z q x r y 110° (vii) a w d z b x c y 55° (viii) w z x y 110° a p c r b q d s 7. Observe the following figures: (a) In which condition does a pair of co-interior angles equal to two right angles? (b) Find the size of missing angles numbered a, b, c, ........ etc. (i) 30° 40° a (ii) a 110° b c d e (iii) a 20° 30° (iv) a 60° b d c (v) a 130° 120° (vi) a d m e i l h b g n k c 115° f o j
GEOMETRY 215 (vii) a b e f c d g 60° 40° (viii) 2a p x q a y w r 8. In the figure p//q and t is the transversal. Measure the size of angles a, b, c, d and tabulate the result. (a) Angles ∠a ∠b ∠c ∠d Conclusion Measure ∠a =∠........, ∠b = ∠ ....... Repeat similar activities for the pair of alternate angles for the figures given below: (b) (c) Write your conclusion for the pair of alternate angles if when a pair of parallel lines are intersected by the transversal. 9. p is parallel to q and t is the transversal. (a) Measure the size of co-interior angles, find their sum and tabulate the result. Angles ∠b ∠c ∠b + ∠c ∠a ∠d ∠a + ∠d Conclusion Repeat similar activities for the pair of co-interior angles and tabulate the reset for the figures given below. (b) (c) p a d b e t q p w z x y t q p p r q s t q p a d b c t q p p s q r t q p w z x y t q
216 The Leading Maths - 8 Write your conclusion in the form of theorem statement for the pair of co-interior angles. 10. Repeat the steps like in question 3 or 4 and verify experimentally that : (a) the alternate exterior angles are equal (b) the co-exterior angles are supplementary. 11. Complete the following proof. Theorem statement: When a pair of parallel lines are intersected by a transversal, the pair of alternate exterior angles are equal. (AEA theorem). Given: p//q and t is the transversal. To prove: ∠b = ∠h ∠a = ∠g Proof Statement Reason 1. ∠b = ∠d 1. 2. ∠d = ∠h 2. 3. ∠b = ∠h 3. 4. Similarly, ∠a = ∠g 4. This completes the proof. 12. (a) Exactly the same way as in question 6, prove with statements and reasons that the same of co-exterior angles is supplementary. (b) Prove with statements and reasons that a line perpendicular to one of the parallel lines is perpendicular to the other parallel line. (c) Prove with statements and reasons that two lines perpendicular to the same line are parallel to each other. 13. (a) Prove that the corresponding angles are equal, if pair of parallel lines are intersected by a transversal. (b) Prove that if a pair of parallel lines are intersected by a transversal, each pair of alternate interior angles are equal. (c) Prove that if a pair of parallel lines are intersected by a transversal, each pair of co-interior angles are supplementary. a e d h b p t q f c g
GEOMETRY 217 ANSWERS 1. (a) ∠h, ∠a, ∠b, ∠e, ∠f (b) Yes (c) ∠f (d) ∠e (e) ∠b (f) ∠a, ∠h, ∠b, ∠g 2. (a) ∠a, ∠f (b) ∠a (c) (i) co-interior (ii) co-interior angles (iii) corresponding angle (iv) Alternate angles (v) Corresponding angles 3. (a) a = 10°, b = 50°, c = 20°, d = 170° (b) a = 60°, b = 120°, c = 60°, d = 20° 6. (a) x = z = p = r = 60°, y = q = s = 120° (b) x = y = a = c = p = r = 60°, z = b = d = q = s = 120° (c) p = x = z = 135°, q = r = s = y = 45° (d) b = e = f = i = j = 140°, a = c = d = g = h = k = 40° (e) w = y = d = 80°, x = z = a = c = 100° (f) p = r = w = y = a = 110°, q = x = s = z = b = d = 70° (g) a = c = w = y = 125°, b = d = x = 55° (h) a = d = p = w = y = 110°, b = c = q = r = x = z = 70° 7. (a) 70° (b) a = c = 70°, b = d = e = 110° (c) a = 50° (d) a = c = 120°, b = d = 60° (e) a = 110° (f) b = l = n = h = k = g = e = 115°, a = c = d = f = j = i = o = m = 65° (g) a = 120°, b = d = 60°, c = f = 80°, g = 100°, e = 120° (h) a = x = y = q = 60°, w = p = r = 120° Project Work 11 Take three flat wooden listy or sticks and put in screw or nail such that they are easily moved as shown in the adjoining figure. Name their ends and crossing points as you like. Now, write the pair of corresponding angles, co-interior angle, co-exterior angles, interior alternate angles and exterior alternate angles. Write their relations when two sticks or listy are parallel. Prepare a report and present it in the classroom.
218 The Leading Maths - 8 At the end of this topic, the student will be able to: ¾ test and prove the theorems related to angles of a triangle. Learning Objectives 12.1 Triangle and Sum of Its Angles I. Introduction to Polygon By a simple closed figure, we shall mean a curve with its ends joined but not cross itself. Simple closed figures. Closed figures but not simple. A polygon is a simple closed curve made up of finite number of line segments each end point of a segment is joined, with one and only one end point of other segment. A common end meet of sides is called a vertex. A line segment is called a side. A segment joining two non-adjacent vertexes of a polygon is called a diagonal. If all the diagonals lie on the interior of the polygon, it is called a concave polygon, otherwise convex polygon. Concave Polygons Convex Polygons Non- polygons CHAPTER 12 PLANE FIGURES
GEOMETRY 219 II. Triangle A triangle is a special kind of polygon having three sides and three angles. We shall discuss different kinds of triangles and theorems related to the triangles. A triangle is a three sided polygon figure. ABC is a triangle. Triangles are classified according to the length of sides. Name of Triangle Definition Figure Example Equilateral Triangle All three sides are equal. A B C AB = BC = CA Isosceles Triangle At least two sides are equal. A B C AB = AC Scalene Triangle No sides are equal. A B C AB ≠ BC ≠ CA Triangles are also classified according to the size of the angles in them. Name of Triangle Definition Figure Example Right Triangle One angle is 90° (right angle). A B C ∠B = 90° A B C Vertex Vertex Vertex Side Angle Side Side
220 The Leading Maths - 8 Acute Triangle All angles are less than 90° (acute). A B C 58° 59° 63° ∠A < 90° ∠B < 90° ∠C < 90° Obtuse Triangle One angle is greater than 90° (obtuse). A B C 30° 30° 120° ∠ABC > 90° III. Angle Properties of a Triangle Mathematical Modeling for Sum of Angles of Triangle From a sheet of paper cut out some triangles as shown below. (a) y z x B C A A (b) B A C y z x (c) A B C z x y Now, take triangle of figure (a) and tear its corners. Arrange the corners as shown below. x y z A A B B CC A B C Do the corners x, y, z or the angles ∠x, ∠y and ∠z make a straight angle? ∴ ∠x + ∠y + ∠z = 180° Repeat this activity with triangular pieces in figures (b) and (c). What did you notice? Can you write a valid statement about the sum of interior angles of a triangle?
GEOMETRY 221 THEOREM - 4 The sum of interior angle of a triangle is 180°. (A) Experimental Verification Step: 1 Draw three triangles ABC of different sizes. Step: 2 Measure the three angles and complete the following table. A B C Fig (i) A B C Fig (ii) B B B Fig (iii) Figure ∠A ∠B ∠C Sum of three angles (i) ∠ A + ∠ B + ∠ C = ________ (ii) ∠ A + ∠ B + ∠ C = ________ (iii) ∠ A + ∠ B + ∠ C = ________ Conclusion: .................................... (B) Theoretical Proof Given: ∠ BAC, ∠ ABC and ∠ ACB are the angles of ∆ABC. To prove: ∠BAC + ∠ABC + ∠ACB = 180° Construction: Draw a line DE from A parallel to BC. Proof Statement Reason 1. ∠BAD + ∠BAC + ∠CAE = 180° 1. Straight angles 2. ∠BAD = ∠ABC, and ∠CAE = ∠ACB 2. Alternate interior angles (AIA) 3. ∠BAC + ∠ABC + ∠ACB = 180° i.e. ∠ABC + ∠BAC + ∠ACB = 180° 3. From statements (1 and 2) Hence, it is proved. B C y z D A a b E x
222 The Leading Maths - 8 Remarks: Theorem of the acute angles of a right triangle is 90°. Why? B A A C C CLASSWORK EXAMPLES Example : 1 Observe the following triangles. (i) A 60° 70° x B C (ii) 30° 2x x P Q R (a) Find the value of x. (b) Find the measure of the unknown angles. Solution: (i) (a) From the given figure, ∠A + ∠B + ∠C = 180° (Why?) or, 60° + 70° + x = 180° (Replacement) or, 130° + x = 180° (Adding) or, x = 180° – 130° (Transposition) ∴ x = 50° (Simplification) (b) Hence, the measure of unknown ∠C is 50°. (ii) (a) From the given figure, ∠P + ∠Q + ∠R = 180° (Why?) or, 2x + x + 30° = 180° (Why?) or, 3x + 30° = 180° (Why?) or, 3x = 180° – 30° = 150° (Why?) or, x = 150° 3 (Why?) ∴ x = 50° (b) The measures of the angles, ∠Q = 50° and ∠P = 2 × 50° = 100°.
GEOMETRY 223 Example : 2 In ∆ABC, find the size of angle marked x. Solution: From the given figure, i. ∠ACB + ∠ACD = 180° (why?) or, ∠ACB + 120° = 180° or, ∠ACB = 180° – 120° = 60° ii. BAC + ABC + ACB = 180° or, x + 40° + 60° = 180° or, x + 100° = 180° or, x = 180° – 100° ∴ x = 80° as required. EXERCISE 12.1 Your mastery depends on practice. Practice like you play. Read, Understand, Think and Do Keeping Skill Sharp 1. (a) Define triangle with illustration. (b) What is the sum of the interior angles of a triangle? (c) What is the sum of the acute angles of a right-angled triangle? (d) What is the measure of the third angle of a triangle whose other angles are 90° and 30°? (e) What is the value of x in the given triangle? 2. Circle ( ) the correct answers. (a) What is the name of the polygon made by three sides? i. trilateral ii. triangle iii. trigon iv. all of the above (b) What is the sum of the angles of a triangle? i. 0° ii. 90° iii. 180° iv. 360° (c) What is the sum of the acute angles of a right triangle? i. 0° ii. 90° iii. 180° iv. 360° B A 40° x 120° C D 45° 95° x
224 The Leading Maths - 8 (d) What is the value of the angle 'a' in the given triangle PQR? i. 20° ii. 30° iii. 40° iv. 50° (e) If one acute of a right-angled triangle is 50°, what is the measure of next acute angle? i. 90° ii. 30° iii. 40° iv. 50° (f) What is the value of the angle x in the given triangle? i. 110° ii. 120° iii. 130° iv. 50° 3. Observe the following triangles and answer the given questions. (i) A B 62° 85° C x (ii) P 110° x 32° Q R (iii) (iv) E D F x x 120° (v) J K x 3x 40° L (vi) X 3x 2x x Y Z (a) What is the sum of the interior angle of a triangle? (b) Find the sizes of the unknown angles in the given triangles. 4. Study the following figures and write the answers of the given questions. (i) x x A C D 120° B (ii) P 2x x 135° Q R S (iii) 70° 80° a P Q R 60° 70° x 35° 20° O N x M I K 100° L 20° y x
GEOMETRY 225 (iv) x y 72° 40° (v) 45° 60° y x (vi) x y 62° z 60° (a) Why is the sum of the angles of a triangle 180°? Give reason. (b) Find the angles marked x, y and z. 5. (a) The sides of a triangle are produced outside so that the exterior angles a, b and c are formed. Find the sum of a, b, c. (b) In the given figure, find the sizes of the angles marked a and b. ANSWERS 3. (i) 33° (ii) 38° (iii) 125° (iv) 30° (v) 35° (vi) 30° 4. (i) 60° (ii) 45° (iii) 80°, 80° (iv) 68°, 40° (v) 45°, 75° (vi) 60°, 58°, 62° 5. (a) 360° (b) 90°, 150° A D B E E C c b a b a x 2x 3x Q S R T
226 The Leading Maths - 8 At the end of this topic, the student will be able to: ¾ test and prove the properties of rhombus, trapezium and kite. Learning Objectives 12.2 Experimental Verifications I. Review on Quadrilateral A quadrilateral is four-sided polygon. It has four angles and two pairs of opposite sides. A. Type of quadrilaterals Name Definition Figure Properties Quadrilateral A closed plane figure made by four sides. D A C B Two pairs of opposite sides; AB and DC, AD and BC. Two pairs of opposite angles; ∠A and ∠C, ∠B and ∠D. Diagonals are intersected. Trapezium One pair of opposite sides is parallel. A B D C Parallel sides are called bases. Non-parallel sides are called legs. Median is parallel to bases. Parallelogram Opposite sides are parallel. A D C B i. The opposite sides of a parallelogram are equal. ii. The opposite angles of a parallelogram are equal. iii. The diagonals of a parallelogram bisect each other. Rectangle A rectangle is a parallelogram in which each angle is right angle. D O A B C i. All the properties of a parallelogram are also the properties of a rectangle. ii. Diagonals are equal. iii. All the angles are equal.
GEOMETRY 227 Rhombus A rhombus is a parallelogram whose adjacent sides are equal. A O B D C i. All the properties of a parallelogram are also representation of a rhombus. ii. The diagonals are perpendicularly bisected. iii. All sides are equal. iv. The diagonals bisect the opposite angles. Square A square is a rectangle in which adjacent sides are equal. In other words, square is a rhombus having one angle as right angle. A B C D O i. All the properties of rectangle and rhombus are also the properties of a square. ii. The diagonals bisect the opposite angles. Kite Opposite adjacent sides are equal in a quadrilateral. i. The diagonals are perpendicularly intersected. ii. The vertical or long diagonal bisects the horizontal diagonal. iii. The vertical diagonal bisects the angle of equal of adjacent sides. iv. One pair of opposite angles formed in non-equal adjacent side is equal. II. Properties of Rhombus THEOREM - 5 The diagonals of a rhombus are perpendicular bisectors. A. Observation 1. Draw two rhombuses ABCD using set-square compass and ruler. 2. Cut out the rhombuses as shown in the figure.
228 The Leading Maths - 8 3. Fold both of them diagonally. 4. Cut one of them diagonally and form 4 triangles. 5. Measure the angle at O in all four triangles. What did you get? 6. Measure the side of hypotenuse in all four triangles. What did you get? 7. Measure the side other than the side of rhombus in all four triangles. What did you get? B. Experimental Verification Step 1: Draw three rhombuses in different shape and size by using ruler, compass and set-squares, and label each of them as ABCD. A B D Fig. (ii) C O A B D C Fig. (iii) O A B D C Fig. (i) O Step 2: Draw the diagonals AC and BD in each rhombus that intersect at O. Measure OA, OB, OC and OD and ∠O in each figure. Step 3: Fill the result in the table below: Fig OA OB OC OD ∠O Remarks (i) (ii) (iii) Conclusion: From the table, we note that OA = OB = OC = OD and ∠O = 90°. Hence, the diagonals of a rhombus are perpendicular bisectors. THEOREM - 6 The diagonals of a rhombus bisect the angles. A. Observation Overlap the all four triangles obtained from the observation of theorems. They are congruent. Hence, the acute angles are separately equal. B. Experimental Verification Step 1: Draw three rhombuses in different shape and size by using ruler, compass and set-squares, and label each of them as ABCD.
GEOMETRY 229 A B D Fig. (ii) C O A B D C Fig. (iii) O A B D C Fig. (i) O Step 2: Draw the diagonals AC and BD in each rhombus that intersect at O. Measure OA, OB, OC and OD and ∠O in each figure. Step 3: Measure the angles ∠BAO, ∠DAO, ∠BCO, ∠PCO, ∠ABO, ∠CBO, ∠ADO, ∠CDO. Fill the result in the table below: Fig ∠BAO ∠DAO ∠BCO ∠DCO ∠ABO ∠CBO ∠ADO ∠CDO (i) (ii) (iii) III. Properties of Trapezium THEOREM - 7 The median or mid-segment of a trapezium is parallel to its bases and half of the sum of bases. A. Observation 1. Cut out a trapezium along midsegment. 2. Arrange these two pieces as shown in the figure. The bases are coincident. 3. Again, join the base edges and midsegment horizontally. What do you find their sum? 4. Conclusion: ................. B. Experimental Verification Step 1: Draw three trapeziums of different shape and size by using ruler and set-squares, and label each of them as ABCD. D M N A B C D M N A C B D M N A B C N
230 The Leading Maths - 8 A E F B D C Fig. (i) A B D C E F Fig. (iii) A B D C Fig. (ii) E F Step 2: Draw median or mid-segment EF and join AF and DF in each figure. Step 3: In each figure, the angles ∠BAF, ∠AFE, ∠EFD and ∠FDC as well as the segments AB, DC and EF measured and the observations are tabulated as shown below: Fig ∠BAF ∠AFE ∠EFD ∠FDC AB DE EF Remarks (i) (ii) (iii) Conclusion: From the table, we note that the alternate angles are equal. So AB||EF and EF||DC. Hence, ............. IV. Properties of Kite THEOREM - 8 One pair of opposite angles formed in non-equal adjacent sides of a kite is equal. A. Observation 1. Draw a vertical line segment AB. 2. Construct two arcs from A or down ward. 3. Also, construct the arcs from B that intersect the previous arcs. Say the intersection points as C and D. We get the kite ACBD. 4. Cut down the kite ACBD. 5. Fold it vertically. What do you get? The angles at C and D are overlapped. Hence, ..................... B. Experimental Verification Step 1: Three different kites of different size are drawn with the help of ruler and compass and each is labeled ABCD. A C D B A C D B
GEOMETRY 231 A B D C A B D C A B D C Fig. (i) Fig. (ii) Fig. (iii) Step 3: Measure ∠ABC and ∠ADC in each figure. The results are tabulated below: Fig ∠ABC ∠ADC Remarks (i) (ii) (iii) Conclusion: .......................... THEOREM - 9 The longer diagonal of a kite bisects the shorter diagonal at right angle. A. Observation 1. Draw a kite in a paper and cutout. 2. Fold it diagonally vertical and horizontal. It forms a triangle as shown in the figure. 3. Measure the angle at the point of intersection diagonals. What do you find? Hence, .................. B. Experimental Verification Step 1: Three different kites of different size are drawn with the help of ruler and setsquares and each is labeled ABCD. Step 2: Diagonals AC and BD are drawn so as to intersect at O. Fig. (i) Fig. (ii) Fig. (iii) A B D O C A B D C O A B D C O
232 The Leading Maths - 8 Step 3: Measure the angle AOB and the length of BO and OD. The results are tabulated in the following table: Fig ∠AOB BO OD Remarks (i) (ii) (iii) Conclusion: From the table, it is cleared that .............. CLASSWORK EXAMPLES Example : 1 ABCD is a rhombus with BC produced to E and ∠DCE = 60°. (a) What is a rhombus? Define it. (b) Find the sizes of angles A, B and D. (c) Write the relation between ∠A and ∠C. Solution: (a) A parallelogram having equal adjacent sides is called a rhombus. (b) In the given figure, ∠BCD + ∠DCE = 180° [ Linear pair of angles] or, ∠BCD + 60° = 180° [ Substitution] or, ∠BCD = 180° – 60° [ Transposition] ∴ ∠BCD = 120° Now, ∠BAD = ∠BCD = 120° [ Opposite angles of a rhombus] ∠ABC = ∠DCE = 60° [ Corresponding angles] ∠ADC = ∠ABC = 60° [ Opposite angles of a rhombus] (c) ∠A = ∠C. Example : 2 ABCD is a trapezium in which CP = AP. PQ = BQ, AB = 3x + 5m PQ = 4x – 5 and CD = x + 1. (a) What is the median of a trapezium? Define it. (b) Find the value of x. (c) Find the actual length of AB, CD and PQ. (d) Justify CD + AB = ZCD. D E A B C 60°
GEOMETRY 233 Solution: (a) A line segment joint the mid-points of the legs of a trapezium is called a median. (b) From the given trapezium, AB + CD = 2 × PQ [ Relation between bases and mid-segment ] or, 3x + 5 + x + 1 = 2 (4x – 5) or, 4x + 6 = 8x – 10 or, 10 + 6 = 8x – 4x or, 16 = 4x or, x = 16 4 = 4 (c) AB = 3x + 5 = 3 × 4 + 5 = 17 units CD = x + 1 = 4 + 1 = 5 units PQ = 4x – 5 = 4 × 4 – 5 = 11 units (d) CD + AB = 5 + 17 = 22 units 2PQ = 2 × 11 = 22 units Hence, CD + AB = 2PQ. EXERCISE 12.2 Your mastery depends on practice. Practice like you play. Read, Understand, Think and Do Keeping Skill Sharp 1. (a) What is rhombus? Define it. Write any two prosperities of rhombus. (b) Write any one difference between rhombus and parallelogram. (c) Define trapezium with two properties. (d) Why is parallelogram different from trapezium? (e) What is kite? Why is the kite different from rhombus? (f) What is the perimeter of a rhombus whose one side is 7 cm long? (g) Write the length of median of a trapezium if its bases are 7 cm and 12 cm long. (h) In the given kite ABCD, if the measure of ∠ABC is 120°, write the measure of ∠ADC. A B C D P 4x – 5 Q x + 1 3x + 5 A C B 120° x D
234 The Leading Maths - 8 2. Circle ( ) the correct answers. (a) A parallelogram is a rhombus if .................. i. its adjacent sides are not equal. ii. its adjacent sides are intersected. iii. its adjacent sides are perpendicular. iv. its adjacent sides are equal. (b) A trapezium has .............................. i. Two pairs opposite sides parallel. ii. One pair opposite sides parallel. iii. all sides parallel. iv. No sides parallel. (c) If the lengths of the diagonal of a rhombus are 8 cm and 6 cm, what is the length of its side? i. 10 cm ii. 14 cm iii. 5 cm iv. 20 cm (d) If the length of mid-segment of a trapezium is 12 cm, what is the sum of its bases? i. 12 cm ii. 18 cm iii. 24 cm iv. 30 cm (e) In the given kite KITE, what is the value of x? i. 4 ii. 9 iii. 16 iv. 2 3. Observe the following plane shapes and find their perimeters. (a) 4 cm C D B A (b) Q 6 cm 3 cm R P S (c) Q 8 cm R P S 5 cm 7 cm K T I E 105° (2x+5)cm 13 cm 105°
GEOMETRY 235 4. Study the following quadrilaterals. (i) 3x+2 A B C D 5x–2 3y+2 (ii) 2y + 7 P Q R S 5y + 1 3x + 2 (iii) L M K N 3y+2 2y+5 7x–6 (iv) X 2x 4x – 3 5x – 4 Y A B W Z (v) I 7x – 2 3x + 3 3x + 4 R C D H M (vi) y – 3 4y + 5 9y + 5 M E D A B P (vii) y + 5 2y 4x – 1 3x + 5 A B D C (viii) T A S R 5x + 4 y + 7 7x – 2 3y + 1 (ix) x+3 2x+4 7x–1 3x–1 A B D C (a) Write the relation between the sides of the quadrilaterals. (b) Find the values of x and y. 5. Observe the following rhombuses and answer the given questions below. (i) 60° A B C D b a c (ii) 58° a b c B A C D (iii) 28° a b c A B C D (iv) 48° 72° a b c C B A D (a) What type of parallelogram is called a rhombus? (b) Find the size of angles marked a, b and c in the following rhombus ABCD.
236 The Leading Maths - 8 6. Find the interior angles of a rhombus by finding the value of x. (a) (2x+5)° (3x–20)° A B D C (b) A B C D (5x–15)° (3x+15)° (c) A B C E D (2x–20)° (4x–40)° (d) A B C D E (15x–10)° (20x+15)° 7. Find the values of p, q and r in following trapeziums. (a) (b) Q 3r C J R M 60° 2r–10° 2p+10° 3q (c) p+40° 3q+10° 3+10° C T N U R P (d) 3r 105° 3q–30° q–10° 2p+20° A C E M F P 8. Find the values of a, b and c in the following kites. (a) M C P X 3a+20° 5a+10° (b) 9a–20° 7a+10° P S R Q (c) b 25° c a N K 105° M L (d) 2b 60° 3a+45° 30° 2c–10° 6a B C D A P r L K W 3p+20° 2p+50° 4q+15° 3p+5°
GEOMETRY 237 9. Complete the following steps. (a) Prove with statements and reasons that the diagonals of a rectangle are equal. Given: In the figure, ABCD is a rectangle with diagonals AC and BD. To prove: AC = BD Proof Statements Reasons 1. In ∆ ABC and ∆DCB 1. (i) AB = CD (S) (i) ___________________________. (ii) ∠ABC = ∠CDB (A) (ii) ___________________________. (iii) BC = BC (S) (iii) ___________________________. 2. ∆ABC @ DDCB 2. SAS test for congruency 3. AC = BD 3. ___________________________. (b) The diagonals of a rhombus bisect at right angles. Given: In the given, ABCD is a rhombus with diagonals intersecting at O. To prove: AO = CO, BO = DO and AC ^ BD Proof. Statement Reasons 1. In ∆AOB and ∆COD (i) ∠ABO = ∠CDO (A) (i) ___________________________. (ii) AB = CD (S) (ii) ___________________________. (iii) ∠BAO = ∠DCO (A) (iii) ___________________________. 2. ∆AOB @ ∆COD 2. ASA test for congruency triangles 3. AO = CO, BO = DO 3. ___________________________. 4. In ∆AOB and ∆AOD (i) AB = AD (S) (i) Adjacent sides of a rhombus (ii) BO = DO (S) (ii) From statement 3. (iii) AO = AO (S) (iii) ___________________________. A D B C A D O B C
238 The Leading Maths - 8 5. ∆AOB @ ∆AOD 5. SSS test for congruency triangles 6. ∠AOB = ∠AOD 6. ___________________________. 7. ∠AOB = 90° i.e. AO^AD 7. Linear pair of angles being equal 8. AC^BA 8. ___________________________. This completes the proof. 10. (a) Prove with statements and reasons that the diagonals of a square are equal and bisect at 90°. (b) If PQ||RS and PQ = RS prove with statements and reasons that PS and QR bisect at T. (c) MO and PN bisect each other at Q. If M, N, O, P are joined in order, prove that the resulting figure is a parallelogram. ANSWERS 3. (a) 20 cm (b) 18 cm (c) 20 cm (d) 17.4 cm (e) 18 cm (f) 18.4 cm 4. (i) 2 units, 1 unit, 28 units (ii) 1 unit, 2 units, 22 units (iii) 1 unit, 2 units, 34 units (iv) 2 units, 3 units, 26 units (v) 3 units, 2 units, 34 units (vi) 2 units, 3 units, 26 units 5. (i) 60°, 120°, 120° (ii) 122°, 58°, 122° (iii) 120°, 28°, 120° (iv) 60°, 72°, 60° 6. (a) 55°, 125°, 55°, 125° (b) 60°, 120°, 60°, 120° (c) 60°, 120°, 120°, 120° (d) 65°, 115°, 65°, 115° 7. (a) 22°, 20°, 105° (b) 25°, 20°, 38° (c) 30°, 25° (d) 35°, 20°, 50° 8. (a) 5° (b) 15° (c) 105°, 25° (d) 15°, 30°, 20° Project Work 12.2 Prepare a chart of the types of quadrilaterals with their names and properties and demonstrate it in your classroom. P R S Q T P M Q N O
GEOMETRY 239 At the end of this topic, the student will be able to: ¾ construct rectangle, square and parallelogram from the given measures. Learning Objectives 12.3 Construction of Quadrilaterals I. Constructing Rectangle By definition of a rectangle each angle is a right angle and opposite sides are equal. (i) When two adjacent sides are given. Construct a rectangle ABCD in which AB = 4 cm and BC = 2.5 cm. Steps i. Make a rough sketch of a rectangle on the basis of the given measures. ii. Draw a line segment AB of length 4 cm. At A, draw AE ⊥ AB by using compass. iii. With A as centre position and radius 2.5 cm, draw an arc cutting AE at D. iv. With B as centre position and radius 2.5 cm, draw an arc. v. With D as centre position and radius 4 cm, draw another arc cutting the previous arc at C. vi. Join BC and DC. Then, ABCD is the required rectangle. (ii) When one diagonal and one side are given. Construct a rectangle ABCD in the given sides AB = 4 cm and diagonal AC = 5 cm. Steps i. Make a rough sketch of a rectangle on the basis of the given measures. ii. Draw a line segment AB of length 4 cm. At B, draw BE AB. Rough Sketch D A C 4 cm B 90° 4 cm 2.5 cm 2.5 cm D A C 4 cm B 4 cm 2.5 cm 2.5 cm 4 cm 4 cm 5 cm D A C B 90° Rough Sketch
240 The Leading Maths - 8 iii. With A as centre position and radius 5 cm, draw an arc cutting BE at C. With centre B and radius equal to AC = 5 cm, draw an arc. iv. With centre position C and radius equal to AB = 4 cm, draw another arc, cutting the previous arc at D. v. Join AD and CD. Then, ABCD is the required rectangle. Steps i. Make a rough sketch of PQRS on the basis of the given measures. ii. Draw the line segment PR = 6 cm iii. Find the mid point using a compass. iv. Construct ROS = 45° using a compass at O, and produce the arm up to X and Y. v. Cut the line OX at S and OY at Q by taking a compass radius 3 cm. vi. Joint the points P, Q, R, S and P. Hence, PQRS is the required rectangle. II. Constructing Square A square is a rectangle with equal adjacent sides or is rhombus with one right angle. (i) When one side is given: Construct a square ABCD in which AB = 5 cm. Steps i. Make a rough sketch of a square on the basis of the given measures. ii. Draw the line segment AB = 5 cm. iii. Construct ∠BAD = ∠ABC = 90° iv. With A as the centre and radius equal to 5 cm, draw an arc on ray AX. v. In the same process on BY as iv. vi. Join points B, C, D and A. Hence, ABCD is the required square. 4.3 cm 3.6 cm B C A D E Rough Sketch 6 cm 45° P R y R O 6 cm 3 cm 3 cm x S P Q Rough Sketch A D C 5 cm B 90° A B D C x y
GEOMETRY 241 (ii) When one diagonal is given: Construct a square ABCD in which AC = 6 cm. Steps i. Make a rough sketch of the square ABCD on the basis of the given measures. ii. Draw a line segment Ac = 6 cm. iii. Construct perpendicular bisector xy of AC using a compass, say intersecting point O. iv. Cut the lines Ox ans Oy using compass radius 3 cm from O, say D and C respectively. v. Join A, B, C, D and A. Hence, ABCD is the required square. III. Constructing Parallelograms (i) When two adjacent sides and the angle between them are given: Construct a parallelogram PQRS having PQ = 3.2 cm, ∠Q = 60° and QR = 4 cm. Steps i. Make a rough sketch of a parallelogram on the basis of the given measures. ii. Construct QR = 4 cm iii. At Q construct ∠XQR = 60° by using a compass. iv. Locate P on QX taking QP = 3.2 cm. v. Fix the compass needle at P, open the compass to the length of 4 cm and take an arc. vi. Fix the compass needle at R and open the compass to the length 3.2 cm and draw another arc intersecting previous arc at S. vii. Join PS and RS. Hence, PQRS is the required parallelogram. Rough Sketch A D C B 6 cm 3 33 3 O 90° A D C B y 6 cm 3 3 3 3 x O Rough Sketch P Q S 4 cm R 60° 3.2 cm Q R X P S 4 cm 4 cm 3.2 cm 3.2 cm 60°
242 The Leading Maths - 8 (ii) When the base, one diagonal and the angle between them are given: Construct a parallelogram PQRS if QR = 4 cm, QS = 5.5 cm and ∠SQR = 30°. Steps i. Make a rough sketch of a parallelogram on the basis of the given measures. ii. Construct QR = 4 cm in length. iii. Construct ∠RQS = 30° by using a compass. iv. Locate S on QX taking QS = 5.5 cm. Join RS v. Fix the compass needle at S, open it to a length of 4 cm and draw an arc. vi. Fix the compass needle at Q, open it to a length equal to RS and draw an arc. Let these arcs intersect at P. vii. Join PQ and PS. Hence, PQRS is a required parallelogram. (iii) When two diagonals and angle between them are given. Construct a parallelogram ABCD in which AC = 8 c, BD = 6 cm and angle between AC and BD = 60° Steps i. Make a routh sketch of ABCD on the basis of the given measures. ii. Drew a line segment AC = 8 cm. iii. Find the midpoint of AC using a compass, say O. iv. Construct the angle 60° at O using a compass and produce the arm up to x and y. v. Cut the lines Ox and Oy from O with the compass. Radius 3 cm, say D and B respectively. vi. Join A, B, C, D and A. Hence, ABCD is the required parallelogram. Rough Sketch P Q S 4 cm R 30° 5.5 cm Q R X P S 4 cm 5.5 cm 30° A C B x 4 cm 4 cm y 8 cm O D Rough Sketch A C B 60° 4 4 3 3 O D
GEOMETRY 243 EXERCISE 12.3 Your mastery depends on practice. Practice like you play. Read, Understand, Think and Do Keeping Skill Sharp 1. Construct the following rectangles: (a) ABCD with the sides AB = 6 cm and BC = 4.5 cm (b) PQRS with the sides RS = 7.4 cm and QR = 3.5 cm (c) ABCD with the side AB = 5.6 cm and the diagonal AC = 7 cm (d) KLMN with the side KL = 5.2 cm and the diagonal LN = 6.4 cm (e) CDEF with the diagonal CE = 6 cm and angle between CE and DF = 45° (f) STUV with SU = 9 cm and angle between SU and TU is 60° 2. Construct the following squares: (a) ABCD with the side AB = 4.3 (b) PQRS with the side PQ = 5 cm (c) ABCD with the diagonal AC = 6.6 cm (d) KLMN with the diagonal LN = 7.4 cm 3. Construct the following parallelograms: (a) ABCD with AB = 5 cm, BC = 6.5 cm and ∠B = 45° (b) PQRS with PQ = 6.2 cm, QR = 7.5 cm and ∠Q = 60° (c) ABCD if BC = 6.5 cm, diagonal BD = 8 cm and ∠DBC = 30° (d) PQRS if QR = 7.2 cm, diagonal QS = 6.5 cm and ∠SQR = 45° (e) KLMN if KM = 6 cm, LN = 9 cm and angle between them is 30° (f) DFGH if DG = 8 cm, FH = 5 cm and angle between them is 75°
244 The Leading Maths - 8 At the end of this topic, the student will be able to: ¾ differentiate regular and irregular polygons. ¾ solve the problems related to interior and exterior angles of regular polygon. Learning Objectives 12.4 Polygons I. Introduction A polygon is any closed plane shape formed by three or more line segments. Triangles, quadrilaterals, pentagons and hexagons are all examples of polygons. The name tells you how many sides the shape has. For example, a triangle has three sides, and a quadrilateral has four sides. So, any shape that can be drawn by connecting three line segments is called a triangle, and any shape that can be drawn by connecting four line segments is called a quadrilateral. All of these shapes given in the figure are polygons. Notice how all the shapes are drawn with only straight lines. This is what makes a polygon. If the shape has curves or don’t fully connect, then it can’t be called a polygon. The rectangle with arrow shape is still a polygon even if it looks like it has an arrow. All the sides are straight, and they all connect. The rectangle with arrow shape has 11 sides. Polygons are not limited to the common ones we know but can get pretty complex and have as many sides as needed. They can have 4 sides, 44 sides, or even 444 sides. The names would be 4-gon, or quadrilateral, 44-gon, and 444-gon, respectively. An 11-sided shape can be called an 11-gon. II. Regular Polygons A special class of polygon exists; it happens for polygons whose sides are all the same length and whose angles are all the same size. When this happens, the polygons are called regular polygons. When a triangle has Geometric figures