GEOMETRY 245 all the sides and angles the same size, we know it as an equilateral triangle, or a regular triangle. A quadrilateral with all sides and angles the same is known as a square, or regular quadrilateral. A pentagon with all sides and angles the same is called a regular pentagon. A polygon with n sides and n angles the same is called a regular n-gon. III. Some special Types of Polygons Name Figure Properties Convex Polygon (i) All of its angles are less than 180°. (ii) All of the diagonals are internal. Concave Polygon (i) At least one angle measures more than 180°. (ii) At least one of the diagonals is outside the shape of the polygon. Equilateral Polygon All sides are equal. Equiangular Polygon All angles are equal. Regular Polygon They have equal angles and equal sides. Irregular Polygon They do not have equal angles and equal sides.
246 The Leading Maths - 8 IV. Angles of Regular Polygons Regular polygons also have two different angles related to them. They are interior angle and exterior angle. a) Interior Angle of a Polygon The interior angle of a polygon is an angle determined by its side which lies within the polygon. In the figure, angles w, x, y and z are interior angles of the quadrilateral ABCD. b) Exterior Angle of a Polygon An angle that forms a linear pair with one of the interior angle of the polygon is called the exterior angle of the polygon. In the figure, the angle blacken makes linear pair with the angle a and hence, is an exterior angle of the given polygon. V. Sum of Interior and Cyclic Exterior Angles of Polygon Draw a diagonal from one vertex in the quadrilateral. How many triangles do you get in it? What is the sum of interior angles of a triangle? Therefore, the sum of the interior angles of the quadrilateral is 2 times of the sum of the interior angles of the triangles formed in the quadrilateral. i.e., 2 ×180° = 360°. Similarly, what may be the sum of interior angles in pentagon, hexagon and so on? Now, we can show the sum of interior angles of the polygons as below: SN Name of Polygon No. of Sides No. of Triangles Sum of interior angles Pattern 1. Triangle 3 1 1 × 180o = 180o (3 – 2)180o 2. Quadrilateral 4 2 2 × 180o = 360o (4 – 2)180o 3. Pentagon 5 3 3 × 180o = 450o (5 – 2)180o 4. Hexagon .... .... .... .... 5. Heptagon .... .... .... .... 6. Octagon .... .... .... .... A B C w D x y z exterior angle exterior angle a a
GEOMETRY 247 7. Nonagon .... .... .... .... 8. Decagon .... .... .... .... ... .... .... .... .... .... n n-gon n n – 2 (n – 2) × 180o Since a n-gon has n sides and n angles, so the measure of each interior angle of the regular polygon is equal to (n – 2) n ×180°. Also, the sum of interior angle and exterior angle is 180°. Then the measure of each exterior angle of the regular polygon is 180° – Interior angle. i.e., 180° – (n – 2) n ×180° = n180° – n180° + 360° n = 360° n . Thus, the sum of exterior angles of the polygon in cyclic order is 360°. CLASSWORK EXAMPLES Example : 1 Find the interior angle of a regular octagon. Solution: Here, The number of sides in octagon (n) = 8 Now, we know that The measure of interior angle of the octagon = (n – 2) n × 180° = (8 – 2) 8 × 180° = 6 8 × 180° = 135° Example : 2 Solve for x in the given figure. Solution: The sum of the interior angles of a polygon is (n – 2) 180°, where n represents the number of sides. The sum of the angles of a hexagon (six sides) is equal to (6 – 2) 180o = 4 × 180° = 720o . Add the interior angles, the sum equal to 720° and solve for x: 120o + (8x – 8o ) + (4x + 14o ) + 7x + (5x – 6o ) + 6x = 720o or, 120o + 8x – 8o + 4x + 14o + 7x + 5x – 6o + 6x = 720o or, 30x + 120o = 720o 120° 5x – 6° 5x – 6° 7x 8x – 8° 6x
248 The Leading Maths - 8 or, 30x = 720o - 120o or, 30x = 600o or, x = 20o Example : 3 Solve for x in the given figure. Solution: The sum of the interior angles of a polygon is (n – 2) 180°, where n represents the number of sides. The sum of the angles of a pentagon (five sides) is equal to (5 – 2) 180o = 3 × 180° = 540o The pentagon is missing one interior angle, say y. Then, 100o + 120o + 90o + 108o + y = 540o or, 418o + y = 540o or, y = 540o – 418o = 122o The interior and exterior angles of a polygon are supplementary. Therefore, 122o + x = 180o or, x = 180o – 122o = 58o . EXERCISE 12.4 Your mastery depends on practice. Practice like you play. Read, Understand, Think and Do Keeping Skill Sharp 1. (a) What is polygon? Define it. (b) What is regular polygon? (c) Write the name of the interior angles and exterior angle from the given polygon. (d) How many triangles are formed in a hexagon from the same point? (e) What is the sum of the interior angles of a pentagon? 2. Circle ( ) the correct answers. (a) How many sides there are in a heptagon? i. 5 ii. 6 iii. 7 iv. 8 90° 120° 100° 108° x A B C D E F
GEOMETRY 249 (b) Regular polygon has ...................... i. equal all sides ii. equal all angles iii. unequal sides and angle iv. both (i) and (ii) (c) Which formula do we use to find the value of a regular polygon? i. (n – 2) 180° ii. n – 2 n × 180° iii. 360° n iv. n + 2 2 × 180° (d) Which formula is used to find each angle of regular polygon? i. n – 2 2 × 180° ii. 360° n iii. n 360° iv. n – 2 2 × 180° 3. Find the sum of interior angles of the following regular polygons: (a) Octagon (b) Heptagon (c) Nonagon (d) Decagon (e) Pentagon (f) Hexagon (g) Triangle (h) Dodecagon 4. Find the measure of the interior angle of the following regular polygons: (a) Trigon (b) Tetragon (c) Pentagon (d) Hexagon (e) Heptagon (f) Octagon (g) Nonagon (h) Decagon 5. Find the measure of the exterior angle of the following regular polygons: (a) Square (b) Tetragon (c) Pentagon (d) Hexagon (e) Heptagon (f) Octagon (g) Nonagon (h) Decagon 6. Find the name of the polygons having the following sum of interior angles: (a) 540o (b) 900o (c) 1080o (d) 1440o 7. Find the name of the polygons having the following interior angles: (a) 120o (b) 135o (c) 140o (d) 90o 8. Find the name of the polygons having the following exterior angles: (a) 120o (b) 72o (c) 45o (d) 36o
250 The Leading Maths - 8 9. Find the value of x from the following polygons: (a) 3x 2x 60° (b) 118° 80° 3x x 2x (c) 3x 120° 110°2x 115°130° (d) 3x 3x 2x 2x 4x 3x 3x (e) 120° 2x + 25° 135° 125° 2x + 10° 4x – 15° (f) 2x + 35° 4x –10° 95° 4x – 5° 135° 10. Find the values of x and y from the following polygons: (a) 210° 30° x 40° y A C B D E (b) y P Q R S T U 2x–5° 92° 100° 98° 95° (c) 120° 135° 4x – 5° 145° 120° 105° A B C E D F y ANSWERS 3. (a) 1080° (b) 900° (c) 1260° (d) 1440° (e) 540° (f) 720° (g) 180° (h) 1800° 4. (a) 60° (b) 90° (c) 108° (d) 1201 2 (e) 1284 7 ° (f) 135° (g) 140° (h) 144° 5. (a) 90° (b) 120° (c) 72° (d) 60° (e) 513 7 ° (f) 45° (g) 40° (h) 36° 6. (a) Pentagon (b) Heptagon (c) Octagon (d) Decagon 7. (a) Hexagon (b) Octagon (c) Nonagon (d) Square 8. (a) Square (b) Pentagon (c) Octagon (d) Pecagon 9. (a) 42° (b) 57° (c) 49° (d) 45° (e) 40° (f) 29° 10. (a) 80°, 100° (b) 80°, 25° (c) 25°, 85°
GEOMETRY 251 CHAPTER 13 CONGRUENCY AND SIMILARITY Lesson Topics Pages 13.1 Congruency of Triangles 252 What are the shape and size of your both hands? Alike or different? What are the shape and size of your maths book? What are the shape and size of your pencil and erasers and your friends? Whose have the same and whose have different? Are the squares with the same and different sides equal? WARM-UP
252 The Leading Maths - 8 At the end of this topic, the student will be able to: ¾ identity the congruent shapes. ¾ test the conditions of the congruent triangles. ¾ solve the problems related to congruency of triangles. Learning Objectives 13.1 Congruency of Triangles I. Introduction Trace each of the pair of figures given below and place the tracing and one on the other. (a) 4 cm 5 cm 4 cm 4 cm 5 cm 4 cm (b) 4.5 cm 4 cm 2 cm 2 cm (c) 6 cm 1.5 cm 4cm 4cm Now answer the following questions : (i) Which pair of figures have the same shape and size? (ii) Which pair of figures have the same shape but different size? (iii) Which pair of figures have both shape and size different? ○ Geometrical figures having the same shape and size are called congruent figures. In the figure, the pair of triangles in (a) are congruent. ○ Geometrical shapes having the same shape are called similar. In the figure, the pair of squares in (b) are similar. From the definition, it should be cleared that all congruent shapes are similar but all similar shapes are not necessarily congruent. II. Similarity Two triangles are said to be similar of the corresponding angles are equal and the ratio of the corresponding sides are also equal. A B C Q P R
GEOMETRY 253 In similar triangles, the corresponding angles are equal; i.e. ∠A = ∠P, ∠B = ∠Q and ∠C = ∠R. The sides which are opposite to the equal angles in similar triangle are corresponding sides. In the figure ∆ABC ~ ∆PQR ∠A = ∠P ⇒ BC and QR are corresponding sides. ∠B = ∠Q ⇒ AC and PR are corresponding sides. ∠C = ∠R ⇒ AB and PQ are corresponding sides. For similarity of two triangles, their sides are proportion. i.e. BC QR = AC PR = AB PQ III. Properties of Congruent Triangles ACTIVITY : 1 Trace ∆ABC and place the tracing over ∆DEF and see, whether ∠A = ∠D AB = DE ∠B = ∠E and BC = EF ∠C = ∠F AC = DF Trace ∆DEF and place the tracing over the ∆ABC and see whether ∠D = ∠A DE = AB ∠E = ∠B and EF = BC ∠C = ∠F DF = AC As the sides and angles are separately equal; the two triangles are congruent we write ∆ABC @ ∆DEF or ∆DEF @ ∆ABC Note: ∠A = ∠D and BC = EF BC and EF are opposite to the equal angles A and D. Similarly, ∠B = ∠E and AC = DF ∠C = ∠F and AB = DE Here, ∠A and ∠D, ∠B and ∠E , ∠C and ∠F are corresponding angles and AB and DE, BC and EF, AC and DF are corresponding sides. A B C Q P R A B C E F D
254 The Leading Maths - 8 ○ In congruent triangles, sides opposite to equal angles are corresponding sides. ○ In congruent triangles, corresponding sides and angles are equal. ∆ABC @ ∆DEF ⇒ AB = DE, BC = EF, AC = DF and ∠A= ∠D, ∠B = ∠E, ∠C = ∠F These are corresponding sides and angles and commonly as corresponding parts. IV. Test for Congruency in a Triangle ACTIVITY : 2 PQR is a given triangle. Copy this ∆PQR in your note copy. In how many ways did you do it? Are your ideas similar to the ones given below? Method: 1 Draw a line segment XY and cut off Q'R' equal to QR on XY. At Q' and R' construct ∠P'Q'R' = ∠PQR and ∠P'R'Q' = ∠PRQ in size. Let the sides meet at P'. Compare the remaining parts of ∆PQR and ∆P'Q'R'. You will flow ∠P = ∠P', ∠P = ∠P', PQ = P'Q' and PR = P'R'. Here you have copied ∆PQR by constructing two angles in one side. This type of construction follows the step of Angle-Side-Angle (ASA). This is called triangles ASA test of congruency. In two triangles, if a triangle one side equal to the one side of another and the angles on them in are separately equal, then the two triangles are congruent. Two triangles, ∆PQR @ ∆PQR by ASA property. Method: 2 On XY construct Q'R' equal to QR. At Q construct ∠M'Q'R' = ∠PQR cut off QP = Q'P' Join P'R'. Compare the remaining parts of the triangles you will flow ∠P = ∠P', ∠R = ∠R', PR= P' R' . Here, you have copied the ∆PQR. This construction follows the steps Side-Angle-Side (SAS). This is called SAS test for congruency. A B C E F D P' X Q' R' Y P P Q R Q R P' M X Q' R' Y
GEOMETRY 255 In two triangles, two sides of a triangle equal to two sides of other separately and the angle included between the two equal sides are equal, then the two triangles are congruent. Two triangles ∆PQR @ ∆PQR by SAS test for congruency of triangles. Method: 3 On XY, construct Q'R' = QR . Draw an arc from Q' taking radius equal to PQ . Draw an arc from R' taking radius equal to PR. Let these arcs intersect at P'. Join P' with Q' and R'. Then P'Q' = PQ and P'R' = PR. Compare the remaining part of the triangles. Here you have copied the ∆PQR onto ∆P'Q'R' by copying the equal length of sides. This type of construction follows the step of Side-Side-Side (SSS). This is called SSS test of congruency of triangles. V. Test for Congruency in a Right-Angled Triangle ACTIVITY : 4 In the figure ∆PQR is right-angled at Q. Can you copy this in your note copy? Follow the following steps. In XY, construct Q'R' = QR. Draw an angle PQR = 90° Taking length of radius equal to PR draw an arc from R’ that intersects Q'M at P'. Join P'R'. Now ∆PQR is copied over the ∆P'Q'R'. Here you have constructed a right angle, Hypotenuse AngleSide (RHS) equal in magnitude to the RHS of the given triangle. This is known as the RHS test for congruency of right-angled triangles. In two right angled triangles if the hypotenuse and a side are separately equal then the two triangles are congruent by RHS test for congruency. In the figure ∆ABC @ ∆PQR by RHS property. P X Q R Y Z X Q' R' P' Y P Q R X Q' P' R' M Y A B C P Q R
256 The Leading Maths - 8 Note: Two triangles are congruent if two angles and any one of the sides are respectively equal to the corresponding angles and side of the other because their remaining angles are also equal and follow ASA test of congruency as shown in the figure alongside. This is known as AAS or SAA test of congruency of triangles. CLASSWORK EXAMPLES Example : 1 In the given figure, the pair of triangles PQR and XZY are congruent. (a) Find the length of angles in each of them. (b) Find the length of sides in each of them. Solution: (a) In ∆PQR, ∠Q = 180° – (42° + 95°) = 180° – 137° = 43° In ∆XYZ = ∠Z = 180° – (42° + 43°) = 180° – 85° = 95° Now, ∠P = ∠X = 42° ∴ QR = ZY = 5 cm ∠Q = ∠Y = 43°, ∠R = ∠Z = 95° (b) PR = XZ = 5.6 cm ∴ PQ = XY or, 3x 6 cm = x + 2 cm or, 3x – x = 2 cm + 6 cm or 2x = 8 cm or, x = 8 cm 2 or, x = 4 cm Hence, PQ = XY = x + 2 = 4 + 2 = 6 cm P Q Z Y X 5 cm R 5 cm ? 95° ? 42° (3x – 6cm) 5.6 cm 42° 43° x + 2 cm
GEOMETRY 257 Example : 2 State the test by which the following pairs of triangles are congruent to each other. (a) A D B EC F (b) P X Q YR Z (c) N M O P Q R (d) X Y Z N P M Solution: (a) AB = DE (S) ∠B = ∠E (A) BC = EF (S) ∆ABC @ ∆DEF by SAS test for congruency. (b) ∠Q = ∠Y (A) QR = YZ (S) ∠R = ∠Z (A) ∆PQR @ ∆XYZ by ASA test for congruency. (c) ∠Q = ∠N = 90° (R) PR = MO (H) PQ = MN (S) ∆PQR @ ∆MNO by RHS test for congruency. (d) XY = MN (S) YZ = NP (S) YZ = MP (S) ∆XYZ @ ∆MNP by SSS test for congruency.
258 The Leading Maths - 8 Example : 3 In the figure ∆ABC ~ ∆PQR with ∠A = ∠P, ∠B = ∠Q and ∠C = ∠R as corresponding angles. Given that AB = 4.5 cm, BC = 5 cm, PR = 15 cm, QR = 25 cm, find the length of AC and PQ. A B C Q R P 15 cm 5 cm 25 cm 4.5 cm Solution: As ∆ABC ~ ∆PQR, the ratios of corresponding sides are proportional. ∴ AB PQ = BC QR = AC PR Now taking, AB PQ = BC QR We get, 4.5 PQ = 5 25 ⇒ PQ = 4.5 × 25 5 = 22.5 cm Taking, BC QR = AC PR We get, 5 25 = AC 15 ⇒ AC = 5 × 15 25 = 3 cm ∴ AC = 3cm and PQ = 22.5 cm as required. EXERCISE 13.1 Your mastery depends on practice. Practice like you play. Read, Understand, Think and Do Keeping Skill Sharp 1. (a) What are congruency and similarity? Define. (b) Write the conditions of congruent triangles. (c) Why are ∆PQR and ∆KLM congruent? Give reason. (d) Are ∆ABC and ∆DEF similar? Give reason. K L M P Q R E F D B C A
GEOMETRY 259 (e) Are ABC and PQR congruent? Why? Give reason. (f) If ∆LMN @ ∆CDE, write the value of x. 2. Circle ( ) the correct answers. (a) Which is not the condition of congruency of triangles? i. SAA ii. ASA iii. SSS iv. ASS (b) What is the full form of RHS in congruency of two triangles? i. Right angle-Handed-Side ii. Right-Hand-Side iii. Right angle-Hypotenuse-Side iv. Right-Height-Side (c) If the given triangles KLM and CDE are congruent by SAA congruency test of triangles, what is the value of x? i. 80° ii. 40° iii. 60° iv. 120° (d) If ∆RIG ≅ ∆APN, which is the value of x? i. 6 cm ii. 3 cm iii. 5 cm iv. 4 cm 3. If the pair of triangles ABC and DEF are congruent, find the length of each side and the size of missing angles in each of the following triangles. (a) A B C D E F 2.4 cm 2.3 cm 1.7 cm 1.7 cm 2.3 cm 2.4 cm 71° 71° 44° (b) A B C 64° 64° 30° P Q R 2.5cm 2.5cm 4.5 cm 4.5 cm 5 cm 5 cm (c) (d) A B C R Q P L M N 60° 5 cm 4 cm C D E 60° 5 cm x L M M x 40° 4 cm D E C 80° 40° 4 cm R I 2x G 7 cm A N P 7 cm 6 cm 2 cm 30° A 1 cm P 3cm A B C P Q R (3x–0.6) cm (x+3) cm 86° 40° 54° 54° 3.1 cm 3.9 cm
260 The Leading Maths - 8 (e) (f) 4. State the test for congruency for the pair of the following triangles. (a) A D B EC F (b) G H I K J L (c) Q P O M N O (d) S T U W X V (e) (f) (g) A B D C (h) A B D C (i) A B C D (j) A B D C 40° 40° Q R P 22° 123° 5.3 cm (1.9x + 1 ) cm A B C (1.9x+1)cm 3.4cm 123° 35° P Q B R A C 28° 62° 2.94 cm 5.8 cm 5 cm B E D C A Q T S R P
GEOMETRY 261 5. In each of the following, the pair of triangles are similar. Find the length of the sides marked x and y. (a) A B C P Q R 6 cm 9 cm 7 cm y x 12 cm (b) A B C y Q R P 5 cm 12 cm 39 cm x 36 cm (c) A B C Y Zy X x 25 cm 12 cm 20 cm 21 cm (d) A B C M N O 14 cm 21 cm 18 cm y x 24 cm 6. (a) In the figure, ∠B = ∠C. Prove that AB = AC. (b) In ∆ABC, AD^BC and BD = DC. Prove that AB = AC. (c) In the figure, AB = AC and ∠BAD = ∠CAD. Prove that BD = DC. 7. (a) In the figure, AB//DE and AB = DE. Prove that BC = CD and AC = EC. (b) In the figure, AB//CD and AC = CE. Prove that ∆ABC @ ∆CDE and BC = CD. 8. (a) In the figure, ∠ABC = ∠DCB = 90° and AC = BD. Prove that ∆ABC @ ∆BCD and ∆AEB @ ∆CED. C A B A B D C A B D C B E D C A A B E C D
262 The Leading Maths - 8 (b) In the figure, if AD//BC and AD = BC, prove that ∆ABD @ ∆DBC. 9. (a) In the figure, AB = BC and AD = DC. Prove that ∠BAD = ∠BCD. (b) In the figure, ∠P = ∠R = 90° and PQ = QR. Prove that PS = RS. ANSWERS 3. (a) ∠A = 65°, ∠B = 44°, ∠E = 65° (b) ∠C = 86°, ∠P = 86°, ∠R = 30° (c) AB = 3 cm, ∠C = 60°, ∠Q = 60°, ∠R = 30° (d) ∠C = 40°, ∠Q = 86°, BC = 3.9 cm, AC = 4.8 cm, PQ = 3.1 cm, PR = 4.8 cm (e) ∠A = 22° ∠P = 35°, AC = 5.3 cm, QR = 3.4 cm (f) ∠C = 62°, ∠P = 28°, BC = 5.8 cm, QR = 2.94 cm, PQ = 5 cm 4. (a) SAS (b) ASA (c) RHS (d) SSS (e) ASA (f) ASA or SAA (g) RHS or SAS or SAA (h) SAS or ASA or SSS or SAA (i) SSS (j) ASA or SAS or AAS or SSS 5. (a) 10.5 cm, 8 cm (b) 15 cm, 13 cm (c) 43.75 cm, 9.6 cm (d) 27 cm 16 cm Project Work 13.1 Take a rectangular paper and at fold along diagonally. i. What do you find? ii. Are they of the same shape and size ? Why? Justify your answer using the condition of congruency of triangles. Prepare a report and present it in your classroom. A B D C B C D A P Q R S
GEOMETRY 263 CHAPTER 14 SOLID OBJECTS Lesson Topics Pages 14.1 Prism and Pyramid 264 What type of geometric shape is called solid? What is the different between solid shape and plane shape? Does the solids shape represent in the plane surface? Are all solid in regular shape? Tell some examples of regular and irregular solids. What is cross-section of solid? WARM-UP
264 The Leading Maths - 8 14.1 Prism and Pyramid At the end of this topic, the student will be able to: ¾ introduce the triangular prism and pyramid. ¾ prepare the nets of cube, cuboid, tetrahedron, cone and cylinder. Learning Objectives I. Triangular Prism and Pyramid A solid with two parallel polygonal bases whose cross-section is congruent to the bases is called a prism. The name of prism depends on its bases. A triangular prism is a polyhedron made of triangular base. A right triangular prism has rectangular lateral faces. A cube has squared base and a cuboid has rectangular base. A solid with two polygonal base and sloping sides or faces that meet at a point at the top is called a pyramid. The name of polygons depends on its bases. A triangular pyramid is a pyramid having a triangular base. The tetrahedron is a triangular pyramid having congruent equilateral triangles for each of its faces. II. Net of Solids A set of plane shapes which folds by attaching the edges can be folded to form a solid shape, is called a net. A net is a pattern made when the surface of a three-dimensional figure is laid out flat showing each face of the figure. That is, the net is formed by opening or unfolding the hollow solid shape. For examples; when we fold six attaching squares, forms a cube. When we fold four attaching equilateral/isosceles triangles edge by edge, forms a tetrahedron (triangular pyramid). Cube Cuboid Triangular Prism Triangular pyramid Squared pyramid
GEOMETRY 265 Below are the steps to determine whether a net forms a solid: 1. Make sure that the solid and the net have the same number of faces and that the shapes of the faces of the solid match the shapes of the corresponding faces in the net. 2. Visualize how the net is to be folded to form the solid and make sure that all the sides fit together properly. III. Construction of Net of Solid Shape (a) Steps of Construction of Net of Cube ○ Draw six adjoining squares in a piece of chart paper as shown in the figure. ○ Draw the dotted line segments at inner edges. ○ Mark flaps on the alternative edges of the net. ○ Cut down the net with flaps. ○ Fold the nets along the dotted line segments. ○ Paste the flaps by using glue. Hence, the shape of solid so formed is the required cube. (b) Steps of Construction of Net of Cuboid ○ Draw three sets of rectangles attaching alternatively in the piece of a chart paper as shown in the figure. ○ Draw the dotted line segments at inner edges. ○ Mark flaps on the alternative edges of the net. ○ Cut down the net with flaps. ○ Fold the nets along the dotted line segments. ○ Paste the flaps by using glue. Hence, the shape of solid so formed is the required cuboid or rectangular based prism. Cube
266 The Leading Maths - 8 (c) Steps of Construction of Net of Tetrahedron ○ Draw an equilateral triangle in the piece of chart paper as shown in the figure. ○ Find out the mid-points of each side. ○ Join the mid-points by dot line segments. ○ Mark flaps on the alternative sides of the net. ○ Cut down the net with flaps. ○ Fold the nets along the dotted line segments. ○ Paste the flaps by using glue. Hence, the shape of solid so formed is the required equilateral triangular pyramid or tetrahedron.. (d) Steps of Construction of Net of Pyramid ○ Draw a square with dot line segments in the center of the piece of chart paper as shown in the figure. ○ Draw four equilateral/isosceles triangles in the same shape and size on the base of the sides of the square. ○ Mark flaps on the alternative sides of the net. ○ Cut down the net with flaps. ○ Fold the nets along the dotted line segments. ○ Paste the flaps by using glue. Hence, the shape of solid so formed is the required squared base pyramid. (e) Steps of Construction of Net of Cylinder ○ Draw a rectangle with length equal to the circumference of a pair of circles with the same radius in the piece of chart paper as shown in the figure. ○ Draw the same two circles as mentioned above on the length of the rectangle and mark flaps. ○ Cut down the net with flaps.
GEOMETRY 267 ○ Paste the breadth of the rectangle by using glue as round/ circular shape and circles with its circular edges so formed. Hence, the shape of solid so formed is the required circular prism or cylinder. (f) Steps of Construction of Net of Cone ○ Draw a sector attaching circle whose circumference is equal to the arc length of the sector on the piece of chart paper as shown in the figure and mark flaps. ○ Cut down the net with flaps. ○ Paste the radii of the sector and form circular shape from arc length and paste the circle with it by using glue. Hence, the shape of solid so formed is the required circular pyramid or cone. EXERCISE 14.1 Your mastery depends on practice. Practice like you play. Read, Understand, Think and Do Keeping Skill Sharp 1. (a) What is solid? Why is the solid different from plane figure? (b) What is prism? Give some examples of prism. (c) What is triangular prism? Draw its shape. (d) Write any one difference between cube and cuboid. (e) What is pyramid? Give some examples of pyramid. (f) What is the net of solid shape?
268 The Leading Maths - 8 2. Circle ( ) the correct answers. (a) Which is not a solid? i. cube ii. cone iii. triangle iv. sphere (b) Which is a prism? i. cuboid ii. pyramid iii. sphere iv. cone (c) Which is an example of cone? i. wooden log ii. birthday cap iii. pencil box iv. dice 3. Identify the solid shape to be formed by the following nets: (a) (b) (c) (d) (e) (f) 4. Draw the solid shapes formed by the above question no. 3. 5. Draw the net of the following solid shapes: (a) Cube (b) Cuboid (c) Pyramid (d) Tetrahedron (f) Cylinder (f) Cone
GEOMETRY 269 6. Draw the net of the following physical shapes: (a) (b) (c) (d) (e) (f) (g) (h) ANSWERS Consult with your teacher. Project Work 14.1 Make a net of cube or cuboid or tetrahedron, or cylinder or cone or triangular prism by cutting a whole chart paper. Demonstrate the process of preparing that net in your classroom and show the solid from it to your friends.
270 The Leading Maths - 8 CHAPTER 15 COORDINATES Lesson Topics Pages 15.1 Distance Between Two Points 271 What is number line ? Can you draw a number line ? Read and write numbers of number line. What are perpendicular lines ? Can you draw perpendicular lines ? Where are the negative and positive numbers on the graph paper ? Can you count the square grids on the graph paper ? WARM-UP South Pole North Pole Equator Prime Meridian Latitude Longitude 90° 60° 30° 0° -30° -60° -90° 0° 30° 60° 90° 120° 150° 180° -150° -120° -90° -60° -30°
GEOMETRY 271 At the end of this topic, the student will be able to: ¾ find the distance between the given two points by using formula. Learning Objectives 15.1 Distance Between Two Points I. Introduction From the very beginning, we considered points on a straight line. We further assumed that there are points in between any two points on a line. Suppose we could assign a positive number (or a numerical measure) to the set of all points between two given points P and Q by some rule. We call such a number the length or distance between them. It is denoted by d or PQ In our study, we shall be considering: (a) Distance between two points on a line along or parallel to the x-axis, (b) Distance between two points on a line along or parallel to the y-axis, and (c) Distance between any two points on a line inclined to the x-axis or y-axis. II. Distance between two points on a line along or parallel to the x-axis Consider two points M(x1 , 0) and N(x2 , 0) on the x-axis, then OM = x1 and ON = x2 . Thus, the distance between M and N is given by d = MN = ON – OM = (x2 – x1 ) units. Similarly, the line segment joining points P (x1 , y1 ) and Q (x2 , y1 ) is parallel to the x-axis, then d = PQ = (x2 – x1 ) units. Note: If x1 > x2 then d = (x1 – x2 ) units. O M N O N M Two sub-cases arise: i) The point N is to the right of the point M, i.e., x2 > x1 . Then, the length of the line segment MN is, d = MN = ON – OM = x2 – x1 P Q d O X' X Y' Y M(x1 , 0) P(x1 , y1 ) Q(x2 , y1 ) N(x2 , 0)
272 The Leading Maths - 8 ii) The point N is to the left of the point M, i.e., x2 < x1 Then, the length of the line segment MN is, d = MN = – (ON – OM) = – (x2 – x1 ) = x1 – x2 In short, the two together are denoted by |x2 – x1 | and is read as modulus of x2 – x1 . In other words, we have the following definition: |x2 – x1 | = |x1 – x2 | = + (x2 – x1 ) 2 = x2 – x1 if x2 > x1 x1 – x2 if x1 > x2 Note: The length of line segment calculated above remains the same even if the two points lie on a straight line parallel to the x-axis. The two points considered need not lie on the positive side only. They may lie on the negative side or one each on either side. III. Distance between two points on a line along or parallel to the y-axis Consider two points K(0, y1 ) and L(0, y2 ) on the y-axis or on a line parallel to the y-axis. Then, OK = y1 and OL = y2 . Thus, the distance between the points and L is given by, d = KL = OL – OK = (x2 – x1 ) units. Similarly, the lines segment joining the points P(x1 , y1 ) and Q(x1 , y2 ) is parallel to the y axis. Then d = PQ = (x2 – x1 ) units. Remark: If y1 > y2 then d = (x1 – x2 ) units. ∴ d = y2 – y1 if y2 > y1 y1 – y2 if y1 > y2 Note: The length of the line segment is the same as found above if the two points lie on a straight line parallel to the y-axis. O X' X Y' Y K(0, y1 ) L(0, y2 ) P(x1 , y1 ) Q(x1 , y2 )
GEOMETRY 273 IV. Distance between two points on a line inclined at an angle to the x-axis Consider two points P(x1 , y1 ) and Q(x2 , y2 ) on a line inclined at a certain angle with the x-axis. From P and Q, draw PM and QN perpendiculars on the x-axis. Also, draw PR perpendicular to QN. Then, (i) OM = x1 , ON = x2 . Since PMNR is a rectangle, so PR = MN = ON – OM = x2 – x1 (ii) Since PMNR is a rectangle, NR = MP = y1 and NQ = y2 , RQ = NQ – NR = y2 – y1 . Now applying to the Pythagorean theorem for the right-angled triangle PRQ, we have PQ2 = PR2 + RQ2 . If we denote the distance between the points P(x1 , y1 ) and Q(x2 , y2 ) by d, then, we have d2 = (x2 – x1 )2 + (y2 – y1 )2 Hence, d = (x2 – x1 )2 + (y2 – y1 )2 units. This is one of the most important formula of coordinate geometry and is known as the Distance Formula between two points P(x1 , y1 ) and Q(x2 , y2 ). CLASSWORK EXAMPLES Example : 1 (a) Find the distance between the points (– 4, 6) and (5, 6) on a line parallel to the x-axis. (b) Find the distance between the two points (1, 10) and (1, 5) on a line parallel to the y-axis. (c) Find the distance between the two points P(3, 5) and Q(6, 9). Solution: (a) Since the y-components of the points (– 4, 6) and (5, 6) are the same, so the distance between two points on the line parallel to the x-axis is, d = x2 – x1 = 5 – (– 4) = 5 + 4 = 9 units x1 x2 M P(x R 1 , y1 ) Q(x2 , y2 ) O Y' Y X' X y2 – y1 y1 y1 y2 x N 2 – x1
274 The Leading Maths - 8 (b) Since the x-components of the points (1, 10) and (1, 5) are the same, the distance between two points on the line parallel to the y-axis is d = y2 – y1 = 10 – 5 = 5 units. (c) The distance d between the points P(x1 , y1 ) and Q(x2 , y2 ) is given by d = (x2 – x1 )2 + (y2 – y1 )2 Here, P(3, 5) = P(x1 , y1 ) and Q(6, 9) = Q(x2 , y2 ), then d = (6 – 3)2 + (9 – 5)2 = 32 + 42 = 9 + 16 = 25 = 52 = 5 units Example : 2 (a) Find the value of x if the distance between the two points (3, 5) and (x, 5) is 10 units. (b) Find the value of y if the distance between the two points (5, – y) and (5, 4) is 2 units. Solution: (a) Given, (3, 5) = (x1 , y1 ) and (x, 5) = (x2 , y2 ) Distance between two points, d = 10 units. Now, we have d = (x2 – x1 )2 + (y2 – y1 )2 or, 10 = (x – 3)2 + (5 – 5)2 or, 10 = (x – 3)2 + 0 ∴ ± 10 = x – 3 Either, x – 3 = + 10 or x – 3 = – 10. Hence, x = + 13 or x = –7. (b) Here, suppose, (5, -y) = (x1 , y1 ) and (5, 4) = (x2 , y2 ) Distance between two points, d = 2 units Now, we know that d = (x2 – x1 )2 + (y2 – y1 )2 or, 2 = (5 – 5)2 + (4 + y)2 or, 2 = (4 + y)2 ∴ ± 2 = 4 + y So, y + 4 = + 2 or y + 4 = – 2. Hence, y = – 2 or x = – 6.
GEOMETRY 275 Example : 3 (a) In which condition do three points lie on the same line? (b) Show that the three points P(5, 0), Q(9, 0) and R(11, 0) lie on a straight line. Solution: (a) Three points lie on the same line when the some of any two line segments is equal to the third line segment. (b) The three points P, Q and R will lie in a straight line, if the sum of the distances between two pairs of points is equal to the distance between the third pair. Here, P (5, 0) = (x1 , y1 ), Q (9, 0) = (x2 , y2 ) and R (11, 0) = (x3 , y3 ) Now, since the given points are on the line parallel to x-axis so, PQ = x2 – x1 = 9 – 5 = 4, QR = x3 – x2 = 11 – 9 = 2 and PR = x3 – x1 = 11 – 5 = 6. Since PQ + QR = 4 + 2 = 6 = PR, the three points P, Q and R lie in a straight line. i.e. they are collinear. Example : 4 Show that the side PR is the hypotenuse of the right-angled triangle PQR having the vertices P (3, 0), Q (6, 0) and R (6, 4). Solution: A triangle is right-angled, if the sum of the squares on two of its sides is equal to the square on the third side. Here, P (3, 0), Q (6, 0) and R (6, 4) are three vertices of the triangle. So, PQ2 = (6 – 3)2 + (0 – 0)2 = 32 = 9 QR2 = (6 – 6)2 + (4 – 0)2 = 42 = 16 PR2 = (6 – 3)2 + (4 – 0)2 = 32 + 42 = 25. Clearly, PQ2 + QR2 = PR2 . Hence, PQR is right-angled at Q. That is, PR is the hypotenuse of the triangle PQR. Example : 5 Prove that the four points A(0, 0), B(1, 0), C(1, 1) and D(0, 1) are the vertices of a square. Solution: A square has all sides equal and has a right angle. A (0, 0) = (x1 , y1 ), B (1, 0) = (x2 , y2 ), C (1, 1) = (x3 , y3 ), D (0, 1) = (x4 , y4 ) Here, AB = x2 – x1 = 1 – 0 = 1 unit, BC = y3 – y2 = 1 – 0 = 1 unit, CD = x3 – x4 = 1 – 0 = 1 unit,
276 The Leading Maths - 8 and DA = y4 – y1 = 1 – 0 = 1 unit. Also, AB2 + BC2 = 12 + 12 = 2 and AC2 = (x3 – x1 )2 + (y3 – y2 )2 = (1 – 0)2 + (1 – 0)2 = 2. Thus, AB2 + BC2 = AC2 . Thus, the four sides of ABCD are equal and one of the angles is a right angle. So, ABCD is a square. Example : 6 Show that the origin is equidistant from the three points A(2, 0), B( 2, 2) and C(0, 2). Solution: The origin O has the coordinates O(0, 0), A (2, 0), B ( 2, 2) and C(0, 2) are three points. Then, OA = (2 – 0)2 + (0 – 0)2 = 2 = 2 units OB = ( 2 – 0)2 + ( 2 – 0)2 = 2 + 2 = 4 = 2units and OC = (0 – 0)2 + (2 – 0)2 = 22 = 2 units Thus, OA = OB = OC. Hence, it is proved. Example : 7 In the following figure, show that AG:GL is 2:1. Solution: Since A (2, 1), G(4, 3) and L (5, 4) are three given points. We have AG = (4 – 2)2 + (3 – 1)2 = 22 + 22 = 4 + 4 = 2 2 units GL = (5 – 4)2 + (4 – 3)2 = 12 + 12 = 2 units So, AG:GL = 2 2 : 2 = 2 : 1. Proved. A(2, 1) C(6, 2) B(4, 6) G(4, 3) L(5, 4)
GEOMETRY 277 EXERCISE 15.1 Your mastery depends on practice. Practice like you play. Read, Understand, Think and Do Keeping Skill Sharp 1. (a) What is the distance in coordinate plane? (b) What is the distance between (x1 , 0) and (x2 , 0)? (c) Write the distance formula joining two points (x1 , y1 ) and (x2 , y2 ). (d) What is the distance between (2, 4) ans (2, –3)? 2. Circle ( ) the correct answers. (a) Which is the distance between (x1 , y1 ) and x2 , y2 )? i. (x2 – x1 )2 ii. y1 – y2 iii. y2 – y1 iv. |y2 – y1 | (b) What is the distance between (a, b) and (c, d)? i. (a – b)2 + (c – d)2 ii. (a – d)2 + (b – c)2 iii. (c – a)2 + (d – b)2 iv. (a – c)2 + (b – d)2 (c) What is the distance between (2, 0) and (–2, 0)? i. 0 ii. 1 iii. 2 iv. 4 (d) What is the distance between (1, 2) and (5, 5)? i. 3 ii. 4 iii. 5 iv. 7 (e) What is the distance between (–2, 0) and (0, –2)? i. 0 ii. 1 iii. 4 2 iv. 2 2 3. Find the distance between the following two points on a line parallel to the y-axis: (a) (3, 5) and (5, 5) (b) ( –4, 4) and (5, 4) (c) (3, 5) and (3, 10) (d) (5, –4) and (5, 4) 4. Find the distance between the following two points on a line inclined to the axes: (a) (0, 0) and (3, 0) (b) (3, 0) and (0, 4) (c) (– 4, 0) and (– 4, 4) (d) (–1, 5) and (4, – 7) (e) (– 3, 4) and (3, – 4) (f) (a, 0) and (0, b) (g) (a + b, a – b) and (a – b, a + b) (h) (2a, b - a) and (a + b, 2b)
278 The Leading Maths - 8 5. Find the value of x or y if the distance between two points; (a) (3, 5) and (x, 5) is 2. (b) ( –x, 4) and (5, 4) is 10. (c) (3, 5) and (3, –y) is 2. (d) ( 5, –y) and (5, 4) is 10. (e) (0, 5) and (x, 9) is 5. (f) ( 2, –y) and (8, 0) is 10. 6. In each case given below, show that the three points lie on a straight line. (a) (5, 0), (7, 0) and (11, 0) (b) (3, 5), (3, 6), (3, 15) 7. Show that the following points are collinear: (a) (–1, –1), (1, 1) and (5, 5) (b) (–4, –3), (4, 3) and (8, 6) 8. (a) If P(2, 3) is a point on the line joining A(4, 5) and B(0, 1), show that PA = PB. (b) Show the points R(4, 2) and S(8, 6) are equidistant from the point T(6, 4). 9. Show that : (a) the point E(3, 3) is equidistant from the points A(0, 4), B(4, 6) and C(6, 2). (b) the point E(3, 2) is equidistant from the points A(0, 6), B(– 1, 5) and C(3, 7). 10. Show that the side PR is the hypotenuse of the right-angled triangle PRQ having the given the vertices: (a) P(5, 0), Q(8, 0), R(8,4) (b) P(0, 5), Q(0, 8), R(4, 8) 11. Use distance formula to show that each of the following sets of points are the vertices of a right-angled triangle: (a) (0, 0), (6, 0) and (6, 8) (b) (2, 4), (6, 4) and (6, 7) 12. Show that each of the triangles whose vertices are given below are isosceles: (a) A(0, 6), B(–5, 3), C(3, 1) (b) P(10, 8), Q(4, 0), R(0, 8) 13. Show that the following points are the vertices of an isosceles right-angled triangle: (a) (1, 0), (2, 1) and (3, 0) (b) (2, 0), (3, 1) and (4, 0)
GEOMETRY 279 14. (a) Show that the points A(0, 0), B(3, 3 ) and C(3, – 3) are the vertices of an equilateral triangle ABC. (b) Show that the points P(0, 0), Q(4, 0) and R(2, 2 3) form an equilateral triangle. 15. Prove that the following points are the vertices of a parallelogram taken in order: (a) (– 4, – 1), (2, –1), (0, – 4) and (– 6, – 4) (b) (– 3, 6), (2, 6), (0, 3) and (– 5, 3) 16. Prove that the following points taken in order are the points of a square. (a) (2, 2), (2, 6), (– 2, 6) and (– 2, 2) (b) (– 6, – 2), (– 2, – 2), (– 2, – 5) and (– 6, – 5) 17. Prove that the following points taken in order are the vertices of a rectangle: (a) (1, 3), (4, 3), (4, –3), and (1, – 3) (b) (2, 1), (2, –2), (– 4, – 2) and (– 4, 1) 18. (a) Prove that the quadrilateral having the vertices (2, – 1), (3, 4), (– 2, 3) and (– 3, – 2) is a rhombus. (b) Show that the points R(2, – 2), O(4, 1), M (2, 4) and B (0, 1) form a rhombus. 19. (a) P is a point on x-axis with its abscissa as 4 and Q is a point on y-axis with its ordinate as 3, show that the length of PQ is 5 units. (b) A is a point on x-axis with its abscissa as 6 and B is a point on y-axis with its ordinate as 8, show that the length of AB is 10 units. 20. (a) Find the point P in the x-axis which is 5 units from the point Q(0, 4). (b) Find the coordinates of A in the y-axis which is 41 units from the point B (4, –3) 21. (a) Show that the points (0, 3), (3, 0), (– 3, 0), (0, – 3), and (2 2, 1) lie on the circumference of a circle whose centre is at the origin. (b) Prove that the origin is the center of a circle circumscribing a triangle whose vertices are (0, 5), (– 1, – 2) and (1, 2). 22. (a) If O (3, 1) is the centre of a circle and P(8, 13) is any point on the circumference, find the length of its diameter of the circle. (b) If C (0, 2) is the centre of a circle and A (7, 5) is any point on its circumference, find the length of its circumference.
280 The Leading Maths - 8 23. In the given figure, show that: (a) BG:GM is 2:1 (b) AL:AG is 3:2 (c) CN:GN is 3:1 ANSWERS 3. (a) 2 units (b) 9 units (c) 5 units (d) 8 units 4. (a) 3 units (b) 5 units (c) 4 units (d) 13 units (e) 10 units (f) a2 + b2 units (g) 2 2 b units (h) 2 (a2 + b2 ) units 5. (a) 1, 5 (b) 5, – 15 (c) – 3, – 7 (d) 6, –14 (e) 3, – 3 (f) 8, – 8 20. (a) (0, 3), (0, –3) (b) (0, 2), (0, – 8) 22. (a) 26 units (b) 44 2 or 62.23 units Project Work 15.1 Make a square grid paper from an A4 paper with 1 cm by 1 cm square. Draw the x-axis and y-axis at the middle. Take any four points in four quadrants and find the distance between any 2 – 2 points. Compare their length. Prepare a report and present it in your classroom. A(2, 1) M(4, 1.5) N(3, 3.5) G(4, 3) C(6, 2) L(5, 4) B(4, 6)
GEOMETRY 281 CHAPTER 16 SYMMETRY AND TESSELLATION Lesson Topics Pages 16.1 Tessellation 282 What are plane geometric shapes? Differentiate between them. When more same kind of triangles are joined side by side, what do you find? What is tessellation? Tell some examples of the tessellation. Discuss. WARM-UP
282 The Leading Maths - 8 At the end of this topic, the student will be able to: ¾ search regular and irregular tessellation and prepare the tessellation. Learning Objectives 16.1 Tessellation I. Introduction Atessellation of a flat surface is the tiling of a plane by using one or more geometric shapes, called tiles, with no overlaps and no gaps. In mathematics, tessellations can be generalized to higher dimensions and a variety of geometries. A periodic tiling has a repeating pattern. Some common tessellations are given below which are used on wallpapers or carpets: II. Types of Tessellation There are three types of tessellation. Regular tessellations are made up entirely of congruent regular polygons all meeting vertex to vertex. There are only three regular tessellations which use a network of equilateral triangles, squares and hexagons. Name Polygon Pattern Tessellation or Tile a tessellation of triangles a tessellation of squares a tessellation of hexagons
GEOMETRY 283 III. Regular Tessellation A regular tessellation means a tessellation made up of congruent regular polygons. [Remember: Regular means that the sides and angles of the polygon are all equivalent (i.e., the polygon is both equiangular and equilateral). Congruent means that the polygons that you put together are all the same size and shape. Some regular tessellations are given below: Regular Tessellation of Equilateral triangles Regular Tessellation of Squares Regular Tessellation of Regular Hexagons IV. Semi-regular Tessellation Regular tessellations of the plane by two or more convex regular polygons such that the same polygons in the same order surround each polygon vertex are called semi-regular tessellations. Some semi-regular tessellations are given below: Semi-regular tessellation of equilateral triangles and squares Semi-regular tessellation of squares and regular Hexagons Semi-regular tessellation of equilateral triangles and regular hexagons
284 The Leading Maths - 8 Semi-regular tessellation of equilateral triangles and squares Semi-regular tessellation of equilateral triangles, rhombus and squares Semi-regular tessellation of equilateral triangles, squares and hexagons V. Irregular Tessellation Irregular tessellations encompass all other tessellations, including the tiling in the main image. Many other shapes, including ones made up of complex curves can tessellate. The image below is an example of an irregular tessellation. Some irregular tessellations are given below: EXERCISE 16.1 Your mastery depends on practice. Practice like you play. Read, Understand, Think and Do Keeping Skill Sharp 1. (a) What is tessellation ? Define it. (b) Write the types of tessellations. (c) Write the name of the given tessellation.
GEOMETRY 285 2. Trace out the following tessellations on your note book and expand twice with colour: (a) (b) (c) (d) (e) (f) 3. Design the following tessellations with suitable different colours: (a) (b)
286 The Leading Maths - 8 (c) (d) (e) (f) 4. Make tessellations from the following polygons and colour them: (a) Equilateral Triangle (b) Square (c) Regular Hexagon (d) Regular Pentagon ANSWERS Consult with your teacher. Project Work 16.1 Take a square paper and fold it 4 times along horizontal and vertical. Unfold it . i. What do you see? ii. Which type of tessellation do you get? Regular or irregular? iii. Design it by using different colours you like. Prepare a report and demonstrate it in your classroom.
GEOMETRY 287 CHAPTER 17 TRANSFORMATION Lesson Topics Pages 17.1 Reflection 288 17.2 Translation 293 17.3 Rotation 299 What is the meaning of transformation? Define the transformation. Tell the types of transformation. Tell some examples of reflection, translation and rotation. X X' l Y' Z Y B B WARM-UP
288 The Leading Maths - 8 At the end of this topic, the student will be able to: ¾ reflect the plane geometric shapes in the graph. Learning Objectives 17.1 Reflection I. Transformation By a geometric figure, we shall mean points, line segments, lines, polygons, circles, etc. We use to trace the figures from one sheet of paper to another. In geometry turn, we transform the figure from one place to another. A transformation is a one to one mapping whose domain and range are the set of points in the plane. The figure in the domain is called pre-image or object and the figure in the range is called image. II. Introduction Copy the given figure. Fold the paper along the dotted line and see whether ∆XYZ, lets into ∆X'Y'Z'. This type of transformation is called reflection. The dotted line is called line of reflection or reflecting line. Notice that XX', YY', ZZ' are perpendicular to the dotted line l. A reflection in line l is a transformation that maps each point P onto a point P' such that (1) If P is on l, the image of P is P itself. (2) If P is not one, then l is perpendicular bisector of PP'. Reflections Using Coordinates (a) Reflection in x-axis ∆A'B'C' is the image of ∆ABC after reflection in x-axis. Complete the following coordinates from the given graph. Pre-image Image A(3, 3) R; x-axis A'( , ) B( , ) R; x-axis B'( , ) C( , ) R; x-axis C'( , ) ∴ P(a, b) R; x-axis P'( , ) X' X Y' Y O A B C B' A' C'
GEOMETRY 289 The point C(5, 0) R; x-axis C'(5, 0) is invariant point. In general, under reflection in x-axis, (x, y) R; x-axis (x, –y) (b) Reflection in y-axis DA'B'C' is the image of DABC after reflection in y-axis. Complete the following coordinates. Pre-image Image A(2, 2) R; y-axis A'( , ) B( , ) R; y-axis B'( , ) C( , ) R; y-axis C'( , ) ∴ P(a, b) R; y-axis P'( , ) The point B(0, 1) R; y-axis B'(0, 1) is invariant point. In general, under reflection in y-axis, (x, y) R; y – axis (–x, y) CLASSWORK EXAMPLES Example : 1 Find the image of ∆XYZ under the reflection over the line m. Solution: Construct perpendiculars from X, Y, Z to the line m. Plot the points X', Y' and Z' such that m is perpendicular bisector of XX', YY' and ZZ'. The reflection of ∆XYZ is ∆X'Y'Z'. We write ∆XYZ Ref. ∆X'Y'Z'. Example : 2 A(2, 4), B(6, 2) and C(3, 2) are the vertices of ∆ABC. Find the coordinates of the image of ∆ABC under reflection in the y-axis. Solution: Reflecting the verices A(2, 4), B(6, 2) and C(3, 2) of ∆ABC in the y-axis, we get (x, y) R; y-axis (– x, y) A(2, 4) R; y-axis A'(– 2, 4) B(6, 2) R; y-axis B'(– 6, 2) C(3, – 2) R; y-axis C'(– 3, – 2) X Y' X' Y O A B C A' C' X X' Y' Z Z' m O P Q Y
290 The Leading Maths - 8 Example : 3 A(– 2, 1), B(5, 2) and C(3, – 2). Find the coordinates of the image of ∆ABC under reflection in the x-axis. Graph the object and the image. Solution: Reflecting the vertrices A(– 2, 1), B(5, 2) and C(3, – 2) in the x-axis, we get (x, y) R; x-axis (x, – y). ∴ A(– 2, 1) R; x-axis A'(– 2, – 1) B( 5, 2) R; x-axis B'(5, – 2) C(3, – 2) R; x-axis C'(3, 2) The object and image so obtained are graphed in the figure. EXERCISE 17.1 Your mastery depends on practice. Practice like you play. Read, Understand, Think and Do Keeping Skill Sharp 1. (a) What is transformation? Give some examples of transformation. (b) Define reflection. Give some examples of reflection. (c) What are the images of a point P(x, y) when it is reflected about the x-axis and y-axis? (d) Write the image point of C(2, –3) reflection on the line y = 0. 2. Circle ( ) the correct answers. (a) Which is the image of Q(a, b) reflecting on the x-axis? i. (a, – b) ii. (–a, b) iii. (a, b) iv. (–a, –b) (b) Which is the image of P(p, q) reflecting on the line x = 0? i. (a, – b) ii. (–a, b) iii. (a, b) iv. (–a, –b) (c) Which is the image of the point (2, –3) reflection on the line y = 0? i. (–3, 2) ii. (–2, 3) iii. (2, 3) iv. (3, 2) X Y Y' X' O A(–2, 1) C(3, –2) B(5, 2) A'(–2, –1) B'(5, –2) C'(3, 2)
GEOMETRY 291 3. Find the image of the following shapes after reflection in the line m. (a) m (b) m (c) m 4. In figure (a) and (b), ∆ABC is the image of ∆A'B'C' after reflection in x-axis and in y-axis respectively. Complete the following table. (a) (b) (a) (b) Pre-image Image Pre-image Image A(1, 3) R; x-axis A'( , ) A(1, 1) R;y-axis A'( , ) B( , ) R; x-axis B'( , ) B( , ) R;y-axis B'( , ) C( , ) R; x-axis C'( , ) C( , ) R;y-axis C'( , ) ∴ P(x, y) R; x-axis P'( , ) ∴ P(x, y) R;y-axis P'( , ) 5. Repeat the question no. 2 for the following reflection: X' Y' X Y O A C B' A' C' B X' X Y Y' O A B C B' A' C' Are there any invariant points in reflection in (a) x-axis and (b) y-axis ? X Y' X' Y O A B C A' B' C' Y' X' X Y O A B C B' A' C'
292 The Leading Maths - 8 6. A(2, 1), B(5, 2) and C(3, 4) are vertices of ∆ABC. Find the coordinates of the image of ∆ABC under the following reflections: (a) In x-axis (b) In y-axis Graph the object and its image. 7. P (0, 1), Q (–2, –1), R (2, –3) and S (4, 0) are vertices of a quadrilateral ABCD. (a) R;x-axis (Quad. PQRS) (b) R;y-axis (Quad. PQRS) 8. Are the invariant points in reflection of (a) x-axis and (b) y-axis. A (–1, 1), B (2, –1), C (1, –1) and D (2, 1) is the vertices of a parallelogram ABCD? Find, (a) R;x-axis ( ABCD) (b) R;y-axis ( ABCD) 9. C (– 1, – 1), D (2, 1), E (1, –1) and F (2, –3) are the vertices of an overhead CDEF. Find, (a) R:x-axis (∆CDEF) (b) R;y-axis (∆CDEF) ANSWERS 4. (a) A(1, 3), B(– 1, 1), C(3,2); A'(1, – 3), B'(– 1, – 1), C'(3, – 2) (b) A(1, 1), B(2, – 2), C(3, 2); A'(– 1, 1), B'(– 2, – 2), C'(– 3, 2) 5. (a) A(3, 2), B(0, – 1), C(2, – 2); A'(3, – 2), B'(0, – 1), C'(2, 2) (b) A(1, 3), B(– 1, 1), C(2, – 1); A'(– 1, 3), B'(1, 1), C'(–2, – 1) 6. (a) A'(2, – 1), B'(5, – 2), C'(3, – 4) (b) A'(– 2, 1), B'(– 5, 2), C'(– 3, 4) 7. (a) P'(0, – 1), Q'(– 2, 1), R'(2, 3), S'(4, 0); (b) P'(0, 1), Q'(2, – 1), R'(– 2, – 3), S'(– 4, 0) 8. (a) A'(– 1, – 1), B'(2, 1), C'(1, 1), D'(2, – 1), No (b) A'(1, 1), B'(– 2, – 1), C'(– 1, – 1), D'(– 2, 1), No 9. (a) C'(– 1, 1), D'(2, – 1), E'(1, 1), F'(2, 3) (b) C'(1, – 1), D'(– 2, 1), E'(– 1, – 1), F'(– 2, – 3) Project Work 17.1 Draw a triangle in a graph. Reflect it in the x-axis and then reflect tis image in the y-axis. Write the all co-ordinates of the object triangle and its images. What do you notice the coordinates of the object triangle and final image triangle? Write your conclusion.
GEOMETRY 293 At the end of this topic, the student will be able to: ¾ find the coordinate of the vertices of the image of an object under translation. Learning Objectives 17.2 Translation I. Introduction There are different ways of transforming a figure from one place to another. Look at the figure B. Trace it on a sheet of paper and cut it out. Place the triangle B on another sheet of paper and trace it. Slide the cut out triangular piece and reach at the new position and join the corresponding vertices. The figure B is slide across the line l. This movement of figure B to B' is called a translation. A translation is a transformation that maps every point in the place to its image by moving each point the same distance in the same direction. II. Translation by Using Coordinates In a coordinate plane, each point if identified by its coordinates. When a point changes its position, its coordinates are also changed. A translation of 5 units in the direction of x-axis only is shown alongside. The coordinates of some points of the pre-images are given. The coordinates of the corresponding image points are in the table below. Coordinates of Preimage Movement of Point along x-axis Coordinates of Images A(3, 4) 5 A' (3 + 5, 4) = A'(8, 4) B(1, 1) 5 B'( , ) = ................... C(4, 0) 5 C'( , ) = ................... P(a, b) 5 P'( , ) = ................... B B' B l A B C C' A' B' X Y' O Y X'
294 The Leading Maths - 8 In general, if we translate a point (x, y) in the direction of ‘a’ units for the coordinates of the image is (x + a, y). i.e. A(x, y) Tx A'(x + a, y) Similarly, if we translate a point A(x, y) in the direction of y-axis only, then A(x, y) Ty A (x, y + b). Verify this by translating the line segment AB in the given figure 5 units in the direction of y-axis, and complete the table. Pre-image Movement of Point along y-axis Image A (– 3, 4) B (5, 2) P (a, b) 5 5 5 A' (– 3, 5 + 2) = A' (–3, 7) B' ................................ P' ................................. Again, the line segment A'B' is the image of the line segment AB after a translation 3 units along x-axis and 2 units along y-axis. Read the coordinates of the pre-image and corresponding image and fill in the table. Pre - image image O A B C Y Y' X' X A(2, 3) 3 2 T A' (...........) B(1, 1) 3 2 T B' (...........) Translating by a translation vector t = a b In general; a translation a b T maps (x, y) to (x + a, y + b). i.e. (x, y) a b T (x + a, y + b) A A' B' C' B C Y' X' X Y O A B B' A' Y Y' X' X