Oasis School Mathematics 8 New

Oasis School Mathematics-8 197 16.3 Similar Figures Let's compare the following pair of figures. A B C E F D A D B P S Q C R The above pair of figures are similar in shape but different in size. Hence, above pairs of figures are similar figures. Two geometrical figures having same shape are called similar figures. It is denoted by ~ sign. Condition of similarity of triangles In the adjoining figures; ∠A = ∠P = 600 ∠B = ∠Q = 700 and ∠C = ∠R = 500 Hence, ABC and PQR are equiangular triangles. Such equiangular triangles are similar triangles. It is written as ∆ABC ~ ∆PQR. Two triangles are similar if three angles of the first triangle are respectively equal to the three of the other. What happens when two triangles are similar? If two triangles are similar, then their corresponding sides are proportional. How to show corresponding sides proportional? Here, ∆ABC ~ ∆PQR. Take the ratio of opposite sides of equal angles Opposite side of ∠A Opposite side of ∠P BC QR = Opposite side of ∠B Opposite side of ∠Q AC PR = Opposite side of ∠C Opposite side of ∠R AB PQ = Note: Similar figures may not be of the same size. P 600 700 50 700 0 600 500 Q R B C A P Q R B C A

198 Oasis School Mathematics-8 ∴ BC QR = AC PR = AB PQ Note:Iftwoangelsofthe twotriangles are equal,the thirdangle is automatically equal. Therefore, two triangles are similar by AA Test. Worked Out Examples Example 1 In the given figure, ∆ABC ~ ∆DEF. Make their corresponding sides proportional. Solution: Here, ∆ABC ~ ∆DEF [Given] ∴ AC DF = BC EF = AB DE [Corresponding sides of similar triangles are proportional) Example 2 In the given figure, ∆ABC ~ ∆DEF, find the value of x. Solution: Here, ∆ABC ~ ∆DEF [Corresponding sides of similar triangles are proportional] ∴ AC DF = BC EF = AB DE or, 4 12 = 2 x = AB DE Taking first and second ratios, or, 4 12 = 2 x or, 1 3 = 2 x or, x = 6 ∴ x = 6 cm Example 3 In the given figure, PQ//BC, prove that ∆APQ ~ ∆ABC. Solution: Given : In ∆ABC, PQ//BC To prove : ∆APQ ~ ∆ABC D E F B C A E F A D B C 4 cm 2 cm x 12 cm B C P Q A

Oasis School Mathematics-8 199 Proof : Statements Reasons 1. In ∆APQ and ∆ABC, (i) ∠APQ = ∠ABC (A) (ii) ∠AQP = ∠ACB (A) (iii) ∠PAQ = ∠BAC (A) 1. (i) Corresponding angles (ii) Corresponding angles (iii) Common angles 2. ∆APQ ~ ∆ABC 2. By A.A.A. Proved. Example 4 In the adjoining figure DE //BC, find the value of x and y. Solution: Given : In ∆ABC, DE//BC To find: Value of x and y. Procedure: In ∆ABC and ∆ABC, (i) ∠DAE = ∠BAC (A) [Common angle] (ii) ∠DEA = ∠ACB (A) [Corresponding angles, DE//BC] (iii) ∠ADE = ∠ABC (A) [Corresponding angles, DE//BC] ∴ ∆ADE ~ ∆ABC [By A.A.A. test] Then, AE AC = DE BC = AD AB [Corresponding sides of similartriangles] or, x x +18 = 6 y = 10 10+20 or, x x+18 = 6 y = 10 30 or, x x+18 = 6 y = 1 3 Taking the first and the third ratios, x x+18 = 1 3 or, 3x = x + 18 or, 3x–x = 18 or, 2x = 18 x = 9 cm Again, taking the second and the third ratios. A E C B D 10cm 20cm 6cm 18cm y x

200 Oasis School Mathematics-8 6 y = 1 3 or, y = 18 cm ∴ x = 9 cm, y = 18 cm Exercise 16.3 1. In the given figure, ∆ABC~∆PQR. Make their corresponding sides proportional. (a) (c) (b) A A (d) P Q R A A B C R Q P 2. (a) In the given figure ∆ABC~∆DEF, find the value of x. (b) In the given figure ∆PQR~∆XYZ, find the value of x and y. (c) In the given figure ∆ABC~∆PQR, find the value of x and y. 3. Prove that following pair of triangles are similar and make their corresponding sides proportional. D E F A B x C 8cm 10cm 3cm y 5cm Q P R (a) A (b) C E D F 250 700 B 700 (c) X Y Z C A B 500 600 700 500 250 A B 350 650 C P Q R

Oasis School Mathematics-8 201 4. (a) In the given figure XY//BC. Prove that ∆ABC~∆AXY. (b) In the given figure AB//CD. Prove that ∆AOB~∆COD. 5. (a) In the given figure, prove that (i) ∆APQ~∆ABC. (ii) Find the value of 'x' and 'y'. (b) In the given figure, prove that ∆ABC~∆DCE and find the values of x and y. (c) In the adjoining figure, prove that ∆PQR ~ ∆PST. If PR = 30 cm, QS = 8 cm, PQ = 24 cm and ST = 24 cm, find RT and QR. (d) In the given figure, AB and CD are perpendiculars drawn on BE. AC=2cm, AE = 8 cm, CD = 3 cm, DE = 4.5 cm. (i) Prove that ∆ABE~∆CDE. (ii) Find the length of AB and BD. A B C D O A P Q 5cm B C 4cm 8cm 2cm x y A C x y B Answer 1. Consult your teacher. 2. (a) 14 cm (b) x = 12 cm, y = 7.5 cm (c) x = 6 cm, y = 4 cm 3. Consult your teacher. 4. Consult your teacher. 5. (a) x = 5 cm, y = 2cm (b) x = 5 cm, y = 3.3 cm (c) RT = 10 cm, QR = 18 cm (d) AB = 4 cm, BD = 1.5 cm

202 Oasis School Mathematics-8 Unit 17 Solid Objects Flash Back Identify the following object O B C P D A 17.1 Triangular Prism and Square Based Pyramid Triangular prism The given figure is a triangular prism. It has a triangular base. The base of this prism is ∆ABC or ∆A'B'C, which are two congruent triangles. It has three rectangular surfaces. Its rectangular surfaces are ABB'A', BCC'B' and ACC'A'. A' C' B' C A B Base Cross section You know! • Cube • Tetrahedron • Octrahedron • Dodecahedron • Icosahedron

Oasis School Mathematics-8 203 Surface area of triangular prism The given figure is a triangular prism. ∆ABC is its base. Three sides of ∆ABC are 'a', 'b' and 'c'. Let 'h' be the height of the prism. If we unfold figure (i), we will get a structure like figure (ii). The surface area of this prism is the sum of area of 2 triangular surfaces and the area of 3 rectangular surfaces. Square based pyramid Observe the given solid objects and answer the questions given below. • How many faces does it have? • How many square faces are there? • How many triangular faces are there? • Are the area of all triangular faces equal? Prepare the solid object like this and find its different feature. Prepare a net of a pyramid as shown in the figure. Fold and paste them to form a pyramid. Make three pyramids of the same size. Paste 3 Pyramids, then a cube is formed. Exercise 17.1 1. Name the following solid objects. b b a a h h h h h C' Fig (i) Fig (ii) B' B' C' A' A' A' A' B C B C A A A A c c O A D B C (a) (b)

204 Oasis School Mathematics-8 2. Study the given figure and answer the questions given below. (a) What is the name of given object? (b) What shape is there on the base of this object? (c) What are its edges? (d) What are its vertices? (e) What are its rectangular faces? 3. Study the given figure and answer the questions given below. (a) What is the name of given object? (b) What is the shape of the base of given object? (c) What are its triangular faces? (d) What are its vertices? (e) What are its edges? (f) What are its faces? 4. Study the given figure and answer the questions given below. (a) Identify the net of which object is given in the figure? (b) What are its rectangular faces? (c) What are its triangular faces? (d) Make the figure formed by this net. 17.2 Solids and their Nets We have already studied about solid figures and their nets in previous classes. These are made up with the combination of different planes. Let us review some solid figures such as cube, cuboid, cylinder, prism, pyramid, triangular prism, etc. and their nets. O A D B C Answer Consult your teacher. A B A' B' C C' b a h h h h B' C' A' A' A' B C A A c A

Oasis School Mathematics-8 205 Solid figures Net of solids Cube Cubiod Triangular prism Tetrahedron Cylinder Square pyramid

206 Oasis School Mathematics-8 Exercise 17.2 1. Draw the net of following solid figures: (a) Cube (b) Cuboid (c) Cylinder 2. Draw the net of : (a) Prism (b) Pyramid (c) Cone 3. Name the solid objects having the following nets: (a) (b) (c) Answer Consult your teacher. Project Work Take a chart paper and draw a net of cube, cuboid, cylinder, etc. Miscellaneous Exercise 1. Study the given figure and answer the questions. (a) Write the name of two pairs of parallel lines. (b) Calculate the value of a and b. (c) If AB is not parallel to CD, What would be the effect and the result? [Ans: (a) Consult your teacher (b) a = 70°, b = 70° (c) Consult your teacher] 2. Study the given figure and answer the questions given below. (a) What is the relation of ∠BCD? why? (b) Which angle is the corresponding angle of ∠CDE? (c) Using the relation of (b), find the value of x. [Ans: (a), (b) Consult your teacher (c) 100°] A B C D a b 700 E G F B A C D E 300 y x 1300

Oasis School Mathematics-8 207 A C D B E 950 350 3. Study the given figure and answer the questions given below. (a) Write the name of a pair of parallel line. (b) Produce ED to F of AB, then what is the value of ∠DFB? (c) Find the value of ∠DFA and ∠CDE [Ans: (a) Consult your teacher (b) 50° (c) 130°, 130°] 4. Study the given figure and answer the questions given below. (a) If AB = BC, what type of triangle is ∆ABC? Which two angles of ∆ABC are equal? (b) Draw a line PQ of any length and find the point Rsuch that PR = QR. Then ∆PQR is an isosceles triangle. Draw two such triangles and verify experimentally that ∠RPQ = ∠RQP. (c) If there is PQ = PR = QR, what type of triangle is this? What is the value of each angle of the triangle? [Ans: consult your teacher] 5. Study the given figure and answer the questions given below. (a) What type of triangle is shown in the figure? If XY = YZ then what is its type? (b) What is the value of ∠X and ∠Z if XY = YZ? (c) Draw two isosceles right angled triangle of different size and verify experimentally that value of both its acute angle is 45°. [Ans: Consult your teacher] 6. In the given figure, PM ⊥ QR and N is the mid point of QR (a) What do PN and PM represent? (b) Draw two isosceles triangle ABC of different size. Where AB = AC. From A draw AM⊥BC. Then verify by actual measurement that BM = CM (c) Is the median and altitude of an isosceles triangle can be same? Justify. [Ans: consult your teacher] A B C Y Z X Q N M R P

208 Oasis School Mathematics-8 7. (a) What type of triangles are called equilateral triangle? (b) Draw two equilateral triangle of different sizes and verify by actual measurement that value of each angle is 60°. (c) Is the median and the altitude of equilateral triangle same? Give reason. [Ans: consult your teacher] 8. Study the given figure and answer the questions given below. (a) What type of triangle is ∆ABC? (b) What is Am called? Is AM⊥BC? Justify your answer. (c) Find the values of x and y? [Ans: (a), (b) consult your teacher (c) x = 35°, y = 55°] 9. Study the given figure and answer the questions given below. (a) In the given figure, AB//DC and AD//BC. What is this figure called? (b) Which of the following relation is not true? (c) Construct a parallelogram ABCD where AB = 6.5cm, BC = 7.2cm, ∠B = 60°. [Ans: Consult your teacher] 10. Observe the given figure and answer the questions given below. (a) What is the name of given figure? IsAC = BD? Given reason. (b) Construct a square PQRS whose are side is 7cm. (c) Join PR and QS, then verify experimentally that PR = QS. [Ans: Consult your teacher] 11. Observe the given figure and answer the questions given below. (a) What is the name of given figure? What are its features? (b) Construct a rhombus ABCD, where AB = 6.8cm, ∠B = 60°. (c) Join AB and BD, then measure the length of AO, BO, CO, DO and ∠AOD then draw the conclusion. [Ans: Consult your teacher] A D B C A D B C O B A M C 550 y x A B O D C

Oasis School Mathematics-8 209 Choose the correct alternative questions. 1. The value of x in the given figure is (i) 45° (ii) 90° (ii) 135° 2. Which one of the following is not the property of an isosceles triangle. (i) Its base angles are equal (ii) Perpendicular drawn from its vertex bisects its base. (iii) Its all the altitudes are equal in length. 3. I am a quadrilateral, my opposite sides are parallel my both diagonals are equal then I am. (i) a parallelogram (ii) a rectangle (iii) a rhombus 4. Which one of the following statement is not true? (i) Size of similar figures may not be same. (ii) Shape and size of congruent figures are always the same. (iii) Two triangles be congruent by ASS axiom. 5. The value of x and y in the given figure is (i) x = 32°, y = 64° (ii) x = 64°, y = 32° (iii) x = 60°, y = 36° 6. In the given figure, ABCD is a square, the number of right angled triangle it has is (i) 4 (ii) 8 (iii) 12 7. The formula to calculate an interior angle of regular polygon is (i) (n – 2)180° (ii) (n – 2)180° n (iii) 360° n 8. The given figure is the net of: (i) Pyramid (ii) Cylinder (ii) Cone 9. Which axiom is not used to make two triangles congruent (i) AAA (ii) SAS (iii) AAS 10. The formula to calculate an interior angle of regular polygon is (i) Its diagonals bisect each other at right angle. (ii) Its all sides are equal. (iii) Its each angle is 90°. A E B C 96° x A D B O C Objective Questions x

210 Oasis School Mathematics-8 Assessment Test Paper Attempt all the questions. Full marks – 25 1. In the given figure AB//EF//CD. Study the given figure and answer the questions given below. (a) Calculate the value of 'b'. (1) (b) Write the alternate angle of DEF. Using this relation find the value of 'a'. (2) 2. In the given figure, P is the mid-point of BC. (a) What is the line AB called? (1) (b) If ABC is an isosceles triangle, where AB = AC then what is the relation of AP with BC? Why? (2) (c) Draw two isosceles triangles of different sizes and verify experimentally that its base angles are equal. (3) 3. (a) Identify the given figure and mentions its properties. (2) (b) If AC and BD intersect at the point 'O', what type of O triangle is ∆AOB? Give reason. (2) (c) If none of its angle is 900 , then what type of quadrilateral will it be? (1) (d) Construct a rhombus ABCD whose one angle is 600 and a side is 7 cm. (2) 4. (a) Which of the following statement is not true? (1) (i) Size and shape of two congruent figures are same. (ii) Similar figure are equal in size. (iii) Circles of any size are similar. (b) In the given figure ∆ABC is an isosceles triangle, M is the mid point of BC make ABM and ∆AMC congruent. (2) (c) Show that AM ⊥ BC. (1) 5. (a) Which formula is used to calculate the interior angle of the regular polygon? (1) (b) If the sum of the angle of a polygon is 3600 , find its number of sides. (2) (c) Draw a net of tetrahedron. (2) D F E A B C 1100 1400 b a A B C M A D B C A B C P

Oasis School Mathematics-8 211 Unit 18 Co-ordinates Flash Back Rectangular coordinate axes We are familiar with the number line of zero, negative integer and positive integers. • Negative integers lie left to '0'. • Positive integer lie right to '0'. -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 If we draw similar vertical number line in which upper part contains positive numbers, lower part contains negative numbers and zero at the centre and combine them, we get the number lines as below. Here, the horizontal number line XOX' is called X–axis and vertical line YOY' is called Y–axis and the point of intersection O is called origin. In the figure, the plane is divided into four parts which are called quadrants. Here we discuss the nature of each quadrant separately. (a) First quadrant: The XOY plane which is determined by positive part of both X–axis and Y –axis is called the first quadrant. Any point in this quadrant contains both positive coordinates. (b) Second quadrant: The X'OY plane which is determined by negative part of X–axis and positive part of Y –axis is called the second quadrant. Any point in this quadrant has x – coordinate negative and y – coordinate positive. -7 -6 -5 -4 -3 -2 -1 1 -1 -1 -2 -3 -4 -5 -6 2 3 4 5 6 7 O 2 3 4 5 6 7 X' X Y' Y You know! • Co-ordinates axes • Quadrants • Cartesian co-ordinates

212 Oasis School Mathematics-8 (c) Third quadrant: The X'OY' plane which is determined by negative part of both X–axisandY–axisiscalledthethirdquadrant.Anypointinthisquadrantcontainsboth x– coordinate and y – coordinate negative. (d) Fourth quadrant: The XOY' plane which is determined by positive part of X–axis and negative part of Y –axis is called the fourth quadrant. Any point in this quadrant contains x– coordinate positive and y – coordinate negative. The origin is considered as central point which is at a distance of 0 units from both X–axis and Y–axis. So, its coordinates are given by (0, 0). In general, a point in any quadrant is represented by (x, y) where x is a perpendicular distance from the Y–axis and y is a perpendicular distance from the X–axis. The sign of each coordinate in different quadrants are summarized in following table: Quadrants x-coordinate y-coordinate First + ve + ve Second – ve + ve Third – ve – ve Fourth + ve – ve 18.1 Pythagoras Theorem The Right Angled Triangle In the given figure, ABC is right-angled triangle in which ∠C = 90° and ∠B be the angle of reference. The side opposite to 90° is the hypotenuse (h). The side opposite to the angle of reference is the perpendicular (p). The side adjacent to the angle of reference is the base (b). A hypotenuse base perpendicular B C • Side opposite to right angle is 'h'. • Side opposite to angle of reference is 'p'. • Side adjacent to the angle of reference is 'b'. Remember !

Oasis School Mathematics-8 213 The Pythagoras Theorem ABC is a right angled triangle, right-angled at ∠C. Here, AC = 4 units and BC = 3 units and AB = 5 units. Squares on side AB, BC and AC are ARSB, BXYC and APQC of ∆ABC respectively. We know that, Area of a square = (side)2 ∴ Area of square on side AB = AB² = 52 = 25 Area of square on side AC = AC2 = 42 = 16 Area of square on side BC = BC2 = 32 = 9 Now, 25 = 16 + 9 i.e. Area of square ARSB = Area of square APQC + Area of square BXYC or, AB² = AC2 + BC2 i.e. square of its hypotenuse = sum of the squares on two sides of a right angled triangle. Hence, Pythagoras theorem states that in a right-angled triangle, the area of the square described on the hypotenuse is equal to the sum of the areas of the squares described on the other two sides. The adjoining figure shows a right–angled triangle ABC in which ∠C = 90° and AB is the hypotenuse. By Pythagoras' theorem: AB2 = AC² + BC2 Experimental Verification The square of the hypotenuse of a right-angled triangle is equal to the sum of the squares of perpendicular and base. Draw three right angled triangles ABC with different shapes and sizes and right– angled at B as shown in figures. A B C B C A C B A Fig. (i) Fig.(ii) Fig. (iii) R A C 4 3 5 Q P Y X S B Note: The longest side of the right angled triangle is hypotenuse. The angle opposite to the longest side of a right- angle triangle is a right angle.

214 Oasis School Mathematics-8 To Verify : AC2 = AB² + BC2 Verification: Measure the lengths AB, AC and BC and tabulate them as follows: Figure AB BC AC AB2 BC2 AC2 AB2 +BC2 Results i. AC2 = AB2 + BC2 ii. AC2 = AB2 + BC2 iii. AC2 = AB2 + BC2 Conclusion: The square of the hypotenuse of a right-angled triangle is equal to the sum of the squares of perpendicular and base. Pythagorean Triplets If three numbers a, b and c are such that a² + b² = c2 , then the numbers a, b and c are called Pythagorean triplets. e.g. 3, 4 and 5 are Pythagorean triplets as 32 + 42 = 9 + 16 = 25 = 52 Some Pythagorean Triplets 5, 12, 13 7, 24, 25 14, 48, 50 9, 12, 15 9, 40, 41 8, 15, 17 20, 21, 29 12, 35, 37 3, 4, 5 Worked Out Examples Example 1 Find the value of 'x' in the given figure. Solution: In the given right angled triangle ABC, h = AC = 13cm p = AB = x b = BC = 12cm Using Pythagoras theorem, h2 = p2 + b2 AC2 = AB2 + BC2 or, (13)2 = x2 + (12)2 or, 169 = x2 + 144 or, x2 = 169 – 144 or, x2 = 25 or, x = 5cm A B C 12 cm x 13 cm

Oasis School Mathematics-8 215 Example 2 Identify whether 20cm, 16cm and 10 cm form the sides of a right angled triangle. Solution: Now, h2 = p2 + b2 (20)2 = (16)2 + (10)2 or, 400 = 256 + 100 or, 400 = 356 which is not true. ∴ 20cm, 16cm and 10 cm do not form a right-angled triangle. Example 3 Find the value of x in the given figure. Solution: ABC is a right angled triangle where, AB = 3cm, BC = 4 cm. Using Pythagoras theorem, AC2 = AB2 + BC2 or, AC2 = (3)2 + (4)2 or, AC2 = 9 + 16 or, AC = 25 ∴ AC = 5cm. Again, ADC is a right angled triangle where AD = 12cm, AC = 5 cm, CD = x cm Using Pythagoras theorem, CD2 = AD2 + AC2 or, x2 = (12)2 + (5)2 or, x2 = 144 + 25 or, x2 = 169 or, x = 169 ∴ x = 13 cm. Example 4 A ladder 25 m long rests against a vertical wall. If the foot of the ladder is 24m far from the foot of the wall, find the height of the wall. Solution: In the figure, AB is a vertical wall i.e. ∠ABC = 900 AC = length of ladder = 25 m BC = the distance from the foot of the wall = 24 m AB = height of the wall Longest side is the hypotenuse B D C A x 4cm 3cm 12 cm C 24m B Ladder=25m A

216 Oasis School Mathematics-8 In the right angled ∆ABC, AC2 = AB2 + BC2 [using Pythagoras theorem] or, (25)2 = AB2 + (24)2 or, 625 = AB2 + 576 or, AB2 = 625 – 576 or, AB2 = 49 or AB = 49 = 7m. ∴ Height of the wall = 7m Exercise 18.1 1. Identify perpendicular (p), base (b) and hypotenuse (h) in each of the following right angled triangles with reference to the blank marked angle. (a) (b) (c) (d) A P X D B Q Y F C R Z E 2. Find the value of 'x' in each of the following figures. (a) (b) (c) (d) . 3. (a) Identify whether the following triangles are right angled or not. (i) (ii) (iii) (b) Using Pythagoras theorem determine whether each of the following triangles with given sides is right angled triangle or not . (i) 12 cm, 5 cm, 10 cm (ii) 6 cm, 8 cm, 10cm C x x x x 3cm 4cm 12cm 13cm 24cm 15cm 12cm 25cm C C C B B B B A A A A 4cm 12cm 13cm 5cm 3cm 3cm 6cm 9cm 2cm

Oasis School Mathematics-8 217 4. Find the length of the diagonal of the following squares. (a) (b) 5. Find the length of the diagonal of the following rectangles. (a) (b) 6. Find the value of 'x' in each of the following figures. 7. (a) In the adjoining figure, find the length of the ladder that goes 8 ft up from the foot of the wall. (b) A ladder 5m long rests against a vertical wall. If the foot of the ladder is 3m far from the foot of the wall, find the height of the vertical wall. 4 cm A D B C 8 cm P Q S R 8 cm 6 cm E H F G 9 cm 12 cm P S Q R (a) (c) (b) (d) 6 cm x P S Q 4 2cm R 4 2cm 12 cm x A D B C 8 cm x A B C D 6cm 9cm 10 cm A D B 9cm C x 37cm 15cm 9 cm Answer 1. Consult your teacher. 2. (a) 5 cm (b) 5 cm (c) 7 cm (d) 9 cm 3. (a) (i) Yes (ii) No (iii) No (b) (i) No, (ii) Yes 4. (a) 5.6 cm (b) 11.31 cm 5. (a) 10cm (b) 15 cm 6. (a) 17 cm (b) 10 cm (c) 17 cm (d) 35 cm. 7. (a) 11.31 ft (b) 4m. R R R 8 ft 450

218 Oasis School Mathematics-8 • Co-ordinates of point A is (3, -3). • Co-ordinates of point B is (7, -3) • Distance between A and B is 4 units. (count horizontal steps) • The length of PQ is 6 units (count vertical steps) Distance between two points Observe the given graph. Let's try to find the length of PQ, QT, TR, QR, PS, ST. From the given graph it is clear that horizontal distance between P and Q is 7 units. [Q is 7 units right from P]. To find the length of QT, let's see how many units below from point Q, the point T is > From the graph it is clear that point T is 5 units below from Q. Hence, the length of QT is 5 units. Similarly, we can find the length of TR, ST and PS. TR = 3 units [R is 3 units below T] ST = 7 units [ T is 7 units right from S] PS = 5 units [S is 5 units below P] How to find the length of SR and PT? Let's observe the above figure, ∆SRT is a right angled triangle. Now, using Pythagoras theorem, h2 = p2 + b2 or, SR2 = ST2 + TR2 -7 -6 -5 -3 -2 -1 1 -1 -1 -2 -3 -4 -5 -6 2 3 4 5 6 7 O 2 3 4 5 6 7 R T P Q S X' X Y' Y -4 18.2 Distance between Two Points Observe the given figure and answer the questions given below. • What are the co-ordinates of the points A and B? • What is the distance between A and B? • What are the co-ordinates of points P and Q? • What is the length of PQ? -7 -6 -5 -3 -2 -1 1 A P Q B -1 -1 -2 -3 -4 -5 -6 2 3 4 5 6 7 O 2 3 4 5 6 7 X' X Y' Y -4

Oasis School Mathematics-8 219 = 72 + 32 = 49 + 9 = 58 SR = 58 units. Again, to find the length of PT. Take a right angled triangle PQT, PQ = 7 units, PT = 5 units Using Pythagoran theorem, h2 = p2 + b2 or, PT2 = PQ2 + QT2 = 72 + 52 = 49 + 25 = 58 SR = 74 units. • Find horizontal distance. • Find vertical distance. • Using Pythagoras theorem, find the required length. Worked Out Examples Example 1 Observe the given figure and find the length of AB, BC and AC. Solution: Here, point B is 6 units below from the point A. So, AB = 6 units Similarly, point C is 4 units right from B. So, BC = 4 units In right angled triangle, ABC h2 = p2 + b2 or, AC2 = AB2 + BC2 = 62 + 42 = 36 + 16 = 52 Hence, AC = 52 units. -7 -6 -5 -3 -2 -1 1 -1 -1 -2 -3 -4 -5 -6 2 3 4 5 6 7 O 2 3 4 5 6 7 A B C X' X Y' Y -4

220 Oasis School Mathematics-8 Example 2 From the given figure, find the length of PQ. Solution: Here, From the given figure, Co-ordinates of point P is (2, 6) Co-ordinates of point Q is (-4, 2) Draw a horizontal line from Q and draw a vertical line from P which meet at R. It is clear that point R is 6 units right from Q. So, QR = 6 units Similarly, point R is 8 units below from P. So, PR = 8 units Using Pythagoras theorem in right angled triangle PQR, PQ2 = PR2 + QR2 = 62 + 82 = 36 + 64 = 100 Hence, PQ = 100 units. Exercise 18.2 1. From the given figure, find the length of: (a) (b) -7 -6 -5 -3 -2 -1 1 -1 -1 -2 -3 -4 -5 -6 2 3 4 5 6 7 O 2 3 4 5 6 7 P Q R X' X Y' Y -4 -7 -6 -5 -3 -2 -1 1 -1 -1 -2 -3 -4 -5 -6 2 3 4 5 6 7 O 2 3 4 5 6 7 P Q X' X Y' Y -4 1 1 -1 -2 -3 -4 -5 -6 2 3 4 5 6 7 8 -7 -6 -5 -4 -3 -2 -1O 2 3 4 5 6 7 P • X • M Q X' X Y' (i) PQ (ii) MN (iii) XY (i) XY (ii) AB (iii) PQ (iv) MN Y 1 1 -1 -2 -3 -4 -5 -6 2 3 4 5 6 7 8 -7 -6 -5 -4 -3 -2 -1O 2 3 4 5 6 7 M P Q A N B X' X Y' Y X N

Oasis School Mathematics-8 221 2. Copy the given figure, in the graph paper, and find the length of given line. (a) (b) 3. Copy the given line in the graph, make the right angled triangle and find the distance between two given points, using Pythagoras theorem. (a) (b) 4. Plot the given points on the graph paper, draw a right angled triangle and using Pythagoras theorem, find the distance between two given points. (a) (1, 2 )and (5, 7) (b) (5, -2) and (-3, 4) (c) (5, -6) and (-3, 1) (d) (-2, -3) and (3, 4) (e) (5, 1) and (-1, -7) (f) (3, 7) and (-6, 2) 1 1 -1 -2 -3 -4 -5 -6 2 3 4 5 6 7 8 -7 -6 -5 -4 -3 -2 -1O 2 3 4 5 6 7 Q C D P X' X Y' Y 1 1 -1 -2 -3 -4 -5 -6 2 3 4 5 6 7 8 -7 -6 -5 -4 -3 -2 -1O 2 3 4 5 6 7 E X Y F X' X Y' Y 1 1 -1 -2 -3 -4 -5 -6 2 3 4 5 6 7 8 -7 -6 -5 -4 -3 -2 -1O 2 3 4 5 6 7 W Q X P X' X Y' Y 1 1 -1 -2 -3 -4 -5 -6 2 3 4 5 6 7 8 -7 -6 -5 -4 -3 -2 -1O 2 3 4 5 6 7 A N M X' X Y' Y B Answer 1. Consult your teacher. 2. (a) AB = 52 units, MN = 145 units (b) PQ = = 125 units, CD = 205 units 3. (a) EF = 74 units, XY = 65 units (b) PQ = 200 units, WX = 185 units 4. (a) 39 units (b) 10 units (c) 113 units, (d) 74 units, (e) 10 units, (f) 106 units,

222 Oasis School Mathematics-8 18.3 Distance between Two Points Using Formula Let A (x1 , y1 ) and B (x2 , y2 ) be any two points, and d be the distance between them. Draw AM and BN perpendicular to x-axis. Here, OM = x1 , AM = y1 ON = x2 , BN = y2 Draw AP⊥BN. Now, AP = MN = ON – OM = x2 – x1 Again, BP = BN – PN = y2 – y1 In right angled ∆ABP, Using Pythagoras theorem, AB2 = BP2 + AP2 d2 = (y2 – y1 )2 + (x2 – x1 )2 or, d = (x2 – x1 )2 + (y2 – y1 )2 ∴ Distance between two points. d = (x2 – x1 )2 + (y2 – y1 )2 Note: X - coordinate on Y-axis = 0 Y- coordinate on X-axis = 0 Any point on X-axis = (x, 0) Any point on Y-axis = (0, y) Important formula: Distance between two points (d) = (x2 – x2 )2 + (y2 – y1 )2 Equilateral triangle : All sides are equal. Isosceles triangle : Any two sides are equal. Scalene triangle : No sides of triangle are equal. Right angled triangle : Pythagoras theorem is satisfied. Parallelogram : Both pairs of opposite sides are equal or parallel. Rhombus : All sides are equal. Rectangle : Opposite sides and diagonals are equal. Square : All sides and diagonals are also equal. (x1 , y1 ) (x2 , y2 ) O M N X A P B Y

Oasis School Mathematics-8 223 Worked Out Examples Example 1 Find the distance between the points A (1, 2) and B (–2, 3). Solution: Here, A (1, 2) and B (–2, 3) are given two points. Let A (1, 2) be (x1 , y1 ) and B (–2, 3) be (x2 , y2 ). By distance formula, d x x y y units = − + − = − − + − = − + = + ∴ = ( ) ( ) ( ) ( ) ( ) ( ) 2 1 2 2 1 2 2 2 2 2 2 1 3 2 3 1 9 1 10 Hence, distance between the points A(1, 2) and B(–2, 3) is 10 units. Example 2 Prove that the points P(3, – 7), Q( – 3, 3) and R(7, 9) are the vertices of an isosceles triangle. Solution: Here, P(3, – 7), Q( – 3, 3) and R(7, 9) are the three vertices of ∆PQR. For PQ, Let P(3, – 7) be (x1 , y1 ) and Q( – 3, 3) be (x2 , y2 ) So, by distance formula. PQ x x y y PQ = − + − = − − + + = − + = + ∴ = ( ) ( ) ( ) ( ) ( ) ( ) 2 1 2 2 1 2 2 2 2 2 3 3 3 7 6 10 36 100 136 units For QR, Let, Q( – 3, 3) be (x1 , y1 ) and R(7, 9) be (x2 , y2 ). So, by distance formula,

224 Oasis School Mathematics-8 QR x x y y QR = − + − = + + − = + = + ∴ = ( ) ( ) ( ) ( ) 2 1 2 2 1 2 2 2 2 2 7 3 9 3 10 6 100 36 136 For PR, Let P(3, – 7) be (x1 , y1 ) and R(7, 9) be (x2 , y2 ) So, by distance formula, ∴ ∆ PQR is an isosceles triangle. Example 3 Prove that: A(2, – 1), B(3, 4), C(–2, 3) and D( – 3, –2) are vertices of a rhombus. Solution: Here, A(2, – 1), B(3, 4), C(–2, 3) and D( – 3, –2) For AB, A(2, –1) = (x1 , y1 ), B (3, 4) = (x2 , y2 ) We have Similarly, Again, A(2, –1) = (x1 , y1 ), D (–3, –2) = (x2 , y2 ) We have, PR x x y y PR uni = − + − = − + + = + = + ∴ = ( ) ( ) ( ) ( ) 2 1 2 2 1 2 2 2 2 2 7 3 9 7 4 16 16 256 272 ts Wehave,PQ = = QR 136 units A(2, -1) B(3, 4) D(-3,-2) C(-2,3) AB x x y y AB units = − + − = − + + = + = + ∴ = ( ) ( ) ( ) ( ) 2 1 2 2 1 2 2 2 2 2 3 2 4 1 1 5 1 25 26 d x x y y uni = − + − = − − + − + = − + − = + = ( ) ( ) ( ) ( ) ( ) ( ) 2 1 2 2 1 2 2 2 2 2 3 2 2 1 5 1 25 1 26 ts

Oasis School Mathematics-8 225 Again, B(3, 4) = (x1 , y1 ), C (–2, 3) = (x2 , y2 ) d x x y y BC u = − + − = − − + − = − + − = + ∴ = ( ) ( ) ( ) ( ) ( ) ( ) 2 1 2 2 1 2 2 2 2 2 2 3 3 4 5 1 25 1 26 nits Again, C(–2, 3) = (x1 , y1 ), D (–3, –2) = (x2 , y2 ) d x x y y CD = − + − = − + + − − = − + − = + ∴ = ( ) ( ) ( ) ( ) ( ) ( ) 2 1 2 2 1 2 2 2 2 2 3 2 2 3 1 5 25 1 26 units Here, AB = BC = CD = AD = 26 units ∴ Given vertices are the vertices of a rhombus. Example 4 Find a point on X–axis whose distance from a point (5, 4) is 5 units. Solution: Here, Given point is (5, 4). Let, the point (5, 4) be (x1 , y1 ) and the point (x, 0) be (x2 , y2 ) [∵ point on the X-axis is (x, 0] Then, by distance formula, or, Squaring on both sides, we get 5² = (x – 5)2 + (0 – 4)2 or, 25 = x2 – 2x . 5 + 52 + 16 or, 25 = x2 – 10x + 25 + 16 or, 0 = x2 – 10x + 25 + 16 – 25 or, x2 – 10x + 16 = 0 or, x2 – (8 + 2)x + 16 = 0 or, x2 – 8x – 2x + 16 = 0 or, x(x – 8) – 2(x – 8) = 0 d x = − x y + − y = − + − ( ) ( ) (x ) ( ) 2 1 2 2 1 2 2 2 5 5 0 4

226 Oasis School Mathematics-8 or, (x – 8) (x – 2) = 0 Either, x – 8 = 0 or, x – 2 = 0 ∴ x = 8 x = 2 Hence, the required points on X–axis are (8, 0) or (2, 0). Exercise 18.3 1. Find the distance between the following points. (a) (3, –2) and (–7, 5) (b) ( – 2, 3) and (6, 1) (c) (0, – 3) and ( – 7, 0) (d) ( – 6, 7) and ( – 1, – 5) (e) (a, 2a) and (4a, 4a) (f) (–8, 7) and (–3, 4) 2. (a) Prove that the following points are the vertices of an isosceles triangle. (i) (4, –1), (2, –5), (8, 3) (ii) (1, 1), (–4, 4), (4, 6) (b) Prove that the points (9, 0), (3, 3) and (12, 21) are the vertices of a right angled triangle. (c) Prove that the triangle with vertices (2, 1), (3, 0) and (1, 0) is a right angled isosceles triangle. (d) Prove that the triangle with vertices (1, – 1), ( 3 , 3 ) and ( – 1, 1) is an equilateral triangle. 3. (a) Prove that the following points are the vertices of a rhombus. (i) (5, 0), (8, 5), (2, 3) and (0, – 3) (ii) (7, 5), (10, 11), (4, 8) and (1, 2) (b) Prove that the following points are the vertices of a parallelogram. (i) (0, 0), (3, 2), (7, 7) and (4, 5) (ii) (– 4, – 4), (2, –4), (4, 4) and ( – 2, 4) (c) Prove that the points ( – 7, 2), ( – 3, 2), ( – 3, – 1) and ( – 7, – 1) are the vertices of a rectangle. (d) Prove that the points (1, 5), (3, 7), (5, 5) and (3, 3) are the vertices of a square. 4. (a) Find the value of 'a' if the point (a, 0) is at a distance of 5units from the point (2, – 3). (b) Find a point on X–axis which is at a distance of 5 units from the point (5, 4). (c) Find a point on Y –axis which is at a distance of 17 units from the point ( – 8, 3). (d) Find the value of 'a' if the distance between (0, 6) and (a, 0) is 6 units. Answer 1. (a) 149 units (b) 2 17 units (c) 58 units (d) 13 units (e) a 13 units (f) 34 units 4. (a) 6 (b) (8, 0) (c) (0, 18) (d) 0

Oasis School Mathematics-8 227 19.1 Symmetry a. Line Symmetry: As we have already studied about line of symmetry, a line divides the given figure into two parts such that one part is the reflection of another part about that line. This line is called line of symmetry or axis of symmetry. b. Rotational symmetry: Figures which can be made to coincide with themselves by rotation through an angle about a point are called the rotational symmetry. Look at this figure: If the first figure is rotated through 180° about O, we will get the second figure. If the second figure is rotated through 180° about O, we will get the third figure which is the original position of the figure. Such symmetry is called rotational symmetry. Order of rotational symmetry: Let us look some figures. (i) This figure has only one line of symmetry. If we rotate it about a point we will get the same shape only after a rotation of 360°. Therefore, order of this rotational symmetry is 1. Line of symmetry Line of symmetry Line of symmetry Rotation through 180º about O. Rotation through 180º about O. A D A B D C B O O O A D C B C Fig. (i) Fig. (ii) Fig. (iii) Unit 19 Symmetry and Tessellation

228 Oasis School Mathematics-8 (ii) If we rotate the given figure through 180°, it can be fitted over the original figure. Again if it is rotated through another 180° we will get the original position of the figure. Therefore order of this rotational symmetry is 2. (iii) If we rotate the given figure through 120°, it can be fitted over the original figure. To bring in the original position it has to rotate three times. Therefore order of this rotational symmetry is 3. Similarly, order of rotational symmetry of a square is 4. Exercise 19.1 1. Draw the possible lines of symmetry in each of the following figures. (a) (b) (c) (d) (e) (f) Rotation through 180º about O. Rotation through 180º about O. O B A O A B O A B 120º C A B O 120º A B C O 120º C A B O 120º B C A O Rotation through 120º about O. Rotation through 120º about O. Rotation through 120º about O.

Oasis School Mathematics-8 229 2. Find the order of symmetry of the following figures. 3. Indicate the point of rotation and order of rotational symmetry of the following figures. Answer Consult your teacher. 19.2 Tessellation A tessellation is a method which is used to cover surface with congruent geometrical shapes in a repeating pattern without leaving any gaps and without overlapping each other. Tessellation is done on the surface of floor or wall or carpet to make them more attractive. Types of Tessellations Depending upon the types used in tessellation, there are three types of tessellations. They are: (a) Regular tessellation (b) Semi regular tessellation (c) Non regular tessellation O O O

230 Oasis School Mathematics-8 a. Regular tessellations: Regular tessellations are those tessellations in which regular polygons of only one type are used. Some examples of regular tessellations are given below. b. Semi-regular tessellations: Semi-regulartessellations are those tessellations in which two or more than two-regular polygons are used. Some examples of semi-regular polygons are given below. c. Non-regular tessellations: Non-regular tessellations are those tessellations in which non-regular polygons are used. Some examples of non-regular polygons are given below. Tessellation having square, triangle and hexagon Tessellation having square, triangle and hexagon

Oasis School Mathematics-8 231 Exercise 19.2 1. Complete the following regular tessellations and expand them in full page. 2. Complete the following semi regular tessellations and expand them in full page. 3. Complete the following non–regular tessellations and expand them in full page. 4. Complete the following tessellations and also mention their types. 5. Take the following figures and tessellate them. (a) Triangle (b) Regular hexagon Answer Consult your teacher.

232 Oasis School Mathematics-8 19.3 Designs Using Circles and Polygons Different designs can be made of circles, polygons, arc, etc. They can be used to decorate walls or floors of rooms or halls. Look at the following designs, study them and try to draw them. (a) Design using circle: (b) Design using circle and arc. (c) Design using squares: (d) Design using circles and triangles: (e) Design using pentagon: Exercise 19.3 1. Draw a circle of radius 4 cm and design as in (a). 2. Construct a square of side 4 cm and make a design as in (c). 3. Draw a regular pentagon of 5 cm and design it as in (e). Answer Consult your teacher.

Oasis School Mathematics-8 233 Flash back An object (or a geometrical figure) changes its position or size (or both) by definite rule which is known as the 'transformation'. There are many kinds of geometrical transformations. They are (i) Reflection, (ii) Rotation (iii) Translation (or displacement) (iv) Enlargement. etc. 20.1 Reflection Look at the mirror. You will get your image on the mirror. Now investigate the following: (i) Distance of the image. (ii) Nature of the image. Distance of the image from the mirror is equal to the distance of the object from the mirror. The image so formed is inverted. This phenomenon is the reflection. I. Properties of reflection The following are the properties of reflection. (i) The object and its image are equidistant from the axis of reflection. Here, in the figure, PA = P'A and QB = Q'B. (ii) The areas of object and its image are equal. Here in the figure, area of ∆ABC = area of ∆ A'B'C'. O Object Image Mirror P I Unit 20 Transformation You know! • Reflection of a point on co-ordinate axes • Displacement of a line horizontally and vertically. P Q P' A A A A' B B' M A' B M Q'

234 Oasis School Mathematics-8 (iii) Object and image are opposite to each other. (iv) The points on the axis of reflection are invariant points i.e., their images are the same points. Image of point C is point C itself. Image of point D is point D itself. II. The Method of Finding the Image Under Reflection In the given figure, ABC is a triangle and M is the axis of reflection. Now, the reflection image of ∆ABC can be obtained by the following procedures: • Draw AX perpendicular to the axis of reflection 'M' and produce it to A' such that AX = XA'. • Draw BY perpendicular to the axis of reflection 'M' and produce it to B' such that BY = YB' • Draw CZ perpendicular to the axis of reflection 'M' and produce it to C' such that CZ = ZC'. • Join A'B', A'C' and B'C' respectively. Hence, ∆A'B'C' is the image of the ∆ABC after reflection in the line 'M'. III. Reflection of Geometrical Shapes Using Co-ordinates a. Reflection on x–axis (y = 0): Take a point A(3, 4) and reflect it on x–axis. From A, draw AM perpendicular to x–axis and produce it to A' such that AM = MA'. Then the co-ordinates of A' will be A' (3, – 4). ∴ A(3, 4) Reflection x-axis A'(3, –4) ∴ Reflection of P(x, y) on x–axis is P(x, y) Reflection x-axis P'(x, –y) The points which are not changed in transformations are called invariant points. B A C M Y Y' X' X A B Y X Z C B' C' A' Y Y' X' X M A'(3, –4) A(3, 4) M M C D

Oasis School Mathematics-8 235 b. Reflection on y–axis (i.e. x = 0): Take a point P(3, 1) Plot the point P(3, 1) on graph paper. From P, draw PM perpendicular to y–axis and produce it to P' such that PM = MP'. Then, the co-ordinates of P' be (–3, 1) i.e. P(3, 1) Reflection y-axis P'(– 3, 1) i.e. P(x, y) Reflection y-axis P'(– x, y) In symbol, the reflection on y–axis is denoted by P(x, y) Reflection y-axis P'(–x, y). For example, (i) P(1, 3) Reflection y-axis P'(–1, 3) (ii) Q(–3, –1) Reflection y-axis Q'(3, –1) Worked Out Examples Example 1 Find the image of the ∆PQR under the reflection on the line M. Solution: Here, Example 2 Find the images of the following points under the reflection on x–axis (i.e. y = 0). (i) A(5, 2)) (ii) B(4, –3) (iii) C(–7, 8) (iv) D(–4, –5) Solution: We have, P(x, y) Reflection x-axis P'(x, –y) Y Y' X' X P'(-3,1) M P(3,1) M R Q P

236 Oasis School Mathematics-8 (i) A(5, 2) Reflection x-axis A'(5, –2) (ii) B(4, –3) Reflection x-axis B'(4, 3) (iii) C(–7, 8) Reflection x-axis C'(–7, –8) (iv) D(–4, –5) Reflection x-axis D'(–4, 5) Example 3 Find the images of the following points under the reflection on y–axis (i.e. x = 0). (i) N(–3, 6) (ii) Q(4, –5) (iii) R(–9, –3) (iv) M(4, 7) Solution: We have, P(x, y) Reflection y-axis P'(–x, y). (i) N(–3, 6) Reflection x-axis N'(3, 6) (ii) Q(4, –5) Reflection y-axis Q'(–4, –5) (iii) R(–9, –3) Reflection y-axis R'(9, –3) (iv) M(4, 7) Reflection y-axis M'(–4, 7) Example 4 Find the reflection image of ∆ABC with vertices A(3, 3), B(–3, –2) and C(5, 1) on (i) x–axis (i.e. y = 0) (ii) y–axis (i.e. x = 0). Also, show ∆ABC and ∆A'B'C' on the same graph paper. Solution: (i) x–axis (.i.e. y = 0) We have, P(x, y) Reflection x-axis P'(x, –y) ∴ A(3, 3) Reflection x-axis A'(3, –3) ∴ B(–3, –2) Reflection x-axis B'(–3, 2) ∴ C(5, 1) Reflection x-axis C'(5, –1) Hence, ∆ABC and its image ∆A'B'C' are shown on the graph. (ii) Y–axis (i.e. x = 0) We have, P(x, y) Reflection y-axis P'(–x, y) ∴ A(3, 3) Reflection y-axis A'(–3, 3) ∴ B(–3, –2) Reflection y-axis B'(3, –2) ∴ C(5, 1) Reflection y-axis C'(–5, 1) Hence, ∆ABC and its image ∆A'B'C' are shown on the graph. Y Y' X' X B A C C' A' B' Y Y' X' X B A C B' C' A'

Oasis School Mathematics-8 237 Exercise 20.1 1. Copy the following figures in your copy and find the images of the following figures under the reflection on the line 'M'. 2. Find the images of the following points under the reflection on x–axis (i.e. on y = 0). (a) A(5, 3) (b) B(7, –4) (c) C(–2, –3) (d) D(–5, 4) 3. Find the images of the following points under the reflection on y–axis i.e. on x = 0). (a) N(3, 4) (b) Q(–4, 5) (c) R(–4, –9) (d) S(6, –5) 4. Find the axis of reflection in each of the following cases: (a) A(4, 6) → A'(4, –6) (b) M(6, 5) → M'(–6, 5) 5. Find the reflection image of ∆ABC with vertices A(3, 2), B(5, 6) and C(8, 1) on (a) x–axis (i.e. y = 0) (b) y–axis (i.e. x = 0). Also show ∆ABC and ∆A'B'C' on the same graph paper. 6. Find the reflection image of ∆PQR with vertices P(4, –2), Q(8, –2) and R(8, 2) on (a) y–axis (i.e. x = 0) (b) x–axis (i.e y = 0) Also show ∆PQR and ∆P'Q'R' on the same graph paper. 7. Draw a parallelogram on graph paper where A(–2, 3), B(–1, 1), C(4, 1) and D(3,3). Find the image of the parallelogram and plot on graph paper after reflecting on (a) x–axis and (b) y–axis. 8. Find the image of the quadrilateral ABCD whose vertices are A(–5, 4), B(–3, 1), C(–1, 3) and D(–2, 6) under reflection on y–axis. Present the quadrilateral ABCD and its image on the graph. (a) (d) (b) (e) (c) Answer 1. Consult your teacher. 2. (a) (5, –3) (b) (7, 4) (c) (–2, 3) (d) (–5, –4) 3. (a) (–3, 4) (b) (4, 5) (c) (4, –9) (d) (–6, –5) 4. (a) X-axis (b) Y-axis 5. (a) A'(3, -2), B' (5, -6) C' (8, -1) (b) A'(-3, 2), B' (-5, 6) C' (-8, 1) 6. (a) P'(-4, -2), Q' (-8, -2) R' (-8, 2) (b) P'(4, 2), Q' (8, 2) R' (8, -2) 7. (a) A'(-2, -3), B' (-1, -1) C' (4, -1), D' (3, -3) (b) A'(2, 3), B' (1, 1) C' (-4, 1), D' (-3, 3) 8. A'(5, 4), B' (3, 1) C' (1, 3), D' (2, 6)

238 Oasis School Mathematics-8 20.2 Rotation Study the following illustrations and get the idea of rotation of a point. Now, we take a point P, that rotates about a point O as shown in figures. In figure (i), ∠POP'= 90° and OP = OP' i.e. the point P rotates about O through angle 90° in the anticlockwise direction. In figure (ii), ∠POP' = 90° and OP = OP' i.e. the point P rotates about O through an angle 90° in the clockwise direction. In figure (iii), OP = OP' and ∠POP'= 180° i.e. the point P rotates about O through a 180° in both direction producing the same image. In figure, (iv), OP = OP' and ∠POP' = 270° i.e. the point P rotates about O through an angle 270° in anti-clockwise direction. In figure (v), OP = OP', ∠POP' = 360°, the point P rotates about O through an angle of 360° in both direction producing the same image. Thus, a rotation is defined when its centre, the angle of rotation and the direction of the rotation are given. The direction of rotation can be clockwise or anticlockwise direction. Rotation is a rule which shifts each point of an object in the same direction through a certain angle about a fixed point.

Oasis School Mathematics-8 239 I. Rotation of a Line PQ, about the Point O Through 90° in Anticlockwise Direction II. Rotation of a ∆ABC about the Point O Through 60° in Clockwise Direction Let, ABC be a triangle and O is the point. Steps: • Join OP and OQ. • Taking O as centre and OP as radius, draw an arc in anticlockwise direction. • At O, draw ∠P'OP = 90° and OP = OP' meeting the arc through P at P'. • Similarly, taking O as centre and OQ as radius draw an arc in anticlockwise direction. • At O, draw ∠Q'OQ = 90° and OQ = OQ', meeting the arc drawn through Q at Q'. • Join P'Q'. Hence, P'Q' is the image of PQ. Steps: • Join AO, BO and CO. • Taking O as centre and OA as radius, draw an arc in clockwise direction. • At O, draw ∠A'OA = 60° and OA = OA' meeting the arc drawn through A at A'. • Similarly, taking O as centre and OB and OC as radius, draw arcs in clockwise direction respectively. • At O, draw ∠B'OB = 60° and OB = OB' and ∠C'OC = 60° and OC = OC' meeting the arcs drawn through B and C at B' and C' respectively. • Join A'B', B'C' and A'C'. Hence, ∆A'B'C' is the image of ∆ABC.

240 Oasis School Mathematics-8 Properties of rotation The following are the properties of rotation. (i) Area of object and its image are equal. (ii) The centre of rotation is the invariant point. (iii) Each point of an object turns through an equal angular displacement in the same direction. (iv) The perpendicular bisector of the line segment joining a point of the object and its corresponding image passes through the centre of rotation. Rotation Using Co-ordinates Just as in reflection, the images of the objects can easily be obtained by the rotation through a given angle using co-ordinates. Look and learn the following illustrations. (c) Rotation through 180° in anti-clockwise and clockwise direction. Clockwise Anticlockwise P(3, 4) R(O,±1800 ) P'(–3, –4) Symbolically, P(x, y) R(O,±1800 ) P'(–x, –y) (a) Rotation through 90° (or –270°) in anticlockwise direction about origin. P(3,4) +900 P'(–4, 3) Symbolically, (b) Rotation through 90° in clockwise direction about origin. P(3, 4) – 900 P' (4, –3) Symbolically, P(x, y) R(O, –900 ) P' (y, –x) P(x, y) R(O,+900 ) P'(–y, x) Anticlockwise Clockwise

Oasis School Mathematics-8 241 • R(O, + 90°) is equivalent to R (O, – 270°) • R(O, –90°) is equivalent to R(O, +270°) • R (O, + 180°) is equivalent to R(O, – 180°) Remember ! Worked Out Examples Example 1 Find the image of the point A(–4, 6) under the rotation about the origin through the angle (i) +90° (ii) –90° (iii) 180° (iv) 270° Solution: We have, (i) P(x, y) R(O,+900 ) P' (–y, x) ∴ A(–4, 6) R(O,+900 ) A'(–6,–4) (ii) P(x, y) R(O,–900 ) P'(y , –x) ∴ A(–4, 6) R(O,–900 ) A'(6, 4) (iii) P(x, y) R(0,1800 ) P'(–x, –y) ∴ A (–4, 6) R(O, 1800 ) A'(4, –6) (iv) P(x, y) R(O,2700 ) P'(y, –x) ∴ A(–4, 6) R(O,2700 ) A(6, 4) Example 2 A(1, 3), B(6, 4), C(7, 2) are the vertices of a triangle ABC. Rotate the ∆ABC about origin through 90° in anticlockwise direction. Present the ∆ABC and its image in the graph. Solution: We have, R(O, + 90°) : P(x, y) → P'(–y, x) ∴ A(1, 3) R(O,+900 ) A'(–3, 1) B(6, 4) R(O,+900 ) B'(–4, 6) and C(7, 2) R(O,+900 ) C'(–2, 7) Hence, ∆ABC and ∆A'B'C' are presented on the graph.

242 Oasis School Mathematics-8 Exercise 20.2 1. Copy the figures given below in your exercise book. Rotate them about O through the given angle in the given direction. Draw the image formed by rotation. (a) (b) (c) 2. Draw a ∆ABC in which AB = BC = AC = 4.5 cm. Take any point O outside of ∆ABC and rotate it about O through 60° in anticlockwise direction. 3. (a) Find the image of the following points underthe rotation through +900 about origin. (i) (4, 3) (ii) (2, –3) (iii) (-5, -4) (iv) (-3, 2) (b) Find the image of the following points under the rotation through –90º about the origin. (i) (–2, 4) (ii) (1, 3) (iii) (5, –3) (iv) (-1, -4) (c) Find the image of the following points under the rotation through 180° about the origin. (i) (1, –5) (ii) (–3, –2) (iii) (2, 6) (iv) (–1, 2) 4. (a) Rotate the ∆ABC with verities. A(1, 2), B(4, 5) and C(6, 1) through +90° about origin. Draw both the figure on the graph paper. (b) ABC is a triangle with vertices A(–1, 4), B(5, 2) and C(3, 7). Find the images of the vertices A, B and C of the triangle ABC under the rotation about the origin through (i) 180° (ii) 270°. Plot both the triangles on the same graph paper. (c) ABCD is a quadrilateral with vertices A(2, 4), B(–1, 2), C(3, –1) and D(4, 1). Find the images of the vertices A, B, C and D of the quadrilateral ABCD under the rotation about the origin through (i) +90° (ii) –90°. Present the quadrilateralsABCD andA'B'C'D' on the same graph paper. A B C O • Rotation through +90 O • Rotation through –600 P Q R 3 O • Rotation through 1800 A B C 5 Answer 1. Consult your teacher. 2. Consult your teacher. 3. (a) (i) (–3, 4) (ii) (3, 2) (iii) (4, –5) (iv) (–2, –3) (b) (i) (4, 2) (ii) (3, –1) (iii) (–3 –5) (iv) (–4, 1) (c) (i) (–1, 5) (ii) (3, 2) (iii) (–2, –6) (iv) (1, –2) 4. (a) A'(–2, 1), B'(–5, 4) C'(–1, 6) (b) (i) A' (1, -4) B' (-5, -2), C' (–3, –7) (ii) A' (4,1), B' (2, -5), C'(7, -3) (c) (i) A' (–4, 2), B' (–2, –1), C' (1, 3), D' (–1, 4) (ii) A' (4, -2), B' (2,1), C' (-1,-3), D' (1,–4)

Oasis School Mathematics-8 243 20.3Translation or Displacement Each point of an object (or a geometrical figure) shifted along the same distance and same direction is called translation or displacement. The displacement of an object (or geometrical figure) has magnitude as well as direction. Hence it is a vector quantity. Example: Let AB be given line and a → be translation vector. Now, A and B are shifted to A' and B' such that AA' = a → .ThenABis saidtobe translatedtoA'B'bytranslation vector a → . Translation vector: The fixed value by which the geometrical figure will be displaced in definite direction of translation is called translation vector. Properties (i) Object and image are congruent. (ii) The line segments joining the object points and its corresponding image points are parallel and equal in length. Translation using co–ordinates: Any point P(x, y) moves a units parallel to x–axis and b units parallel to y–axis, then new co-ordinates of P will be (x + a, y + b). We mark the point P' on the plane whose co–ordinates are (x + a, y + b). Thus, if the translation vector T = ( a b), the point P(x, y) is translated to P'(x + a, y + b) Symbolically, P(x, y) T ( ( a b P' (x + a, y + b) Example 1 Translate ∆ABC given in the figure by the given vector. Here, ∆ABC is displaced by given vector a → . From A, draw AA' || a → such that AA'= a → From C, draw BB' || a → such that BB' = a → From C, draw CC' || a → such that CC' = a → Join A', B' and C' Hence, ∆A'B'C' is the displaced image of ∆ABC. a → A B A' B' Translation vector Image P'(x + a, y + b) P(x, y) b a Worked Out Examples

244 Oasis School Mathematics-8 Example 2 ABC is a triangle with vertices A(–2, 3), B(–2, 5) and C(2, 3). Translate the ∆ABC by translation vector T = ( ( 4 3 . Present the ∆ABC and its image on the graph. Solution: We have, P(x, y) T ( ( a b P' (x + a, y + b) A(–2, 5) T ( ( 4 3 A'(–2 + 4, 5 + 3) = A' (2, 8) B(–2, 3) T ( ( 4 3 B'(–2 + 4, 3 + 3) = B' (2, 6) and C(2, 3) T ( ( 4 3 C'(2 + 4, 3 + 3) = C' (6, 6) Hence, ∆A'B'C' is the translated image of ∆ABC. Exercise 20.3 1. Copy the figures given below in your copy and transform them by the translation vector in the direction and magnitude of it. Shade the image so formed. 2. Find the image of the following points (i) A(2, 3), (ii) B(6, –3), (iii) C(3, 0), (iv) D(–4, –6) when they are translated by the vectors (a) T = ( ( 4 3 (b) T =( ( -3 5 . 3. A point P(4, 5) is translated to the point P'(6, 8). Find the translation vector by which P is translated to P'. 4. ABC is a triangle with vertices A(3, 7), B(2, 2) and C(6, 1). Translate the ∆ABC by the vector (i) T = ( ( 2 3 (ii) T = ( ( -3 -4 . Present the ∆ABC and its image on the graph. Y Y' X' X O A A' B' C' B C (c) S P R O a → (b) A B C a → (a) A B a → Answer 1. Consult your teacher. 2. a) (i) A' (6,6) (ii) B' (10,0) (iii) C' (7, 3) (iv) D' (0,–3) b) A' (-1, 8), B' (3, 2), C' (0, 5), D' (-7, -1) 3. ( 2 3 ) 4. (i) A'(5,10), B' (4,5) C' (8,4) (ii) A' (0,3), B' (-1,-2) C' (3, –3)

Oasis School Mathematics-8 245 21.1Bearing Bearing is the method of finding the distance between the two places in degree. There are two types of bearing. I. The compass bearing II. The three figure bearing I. The compass bearing The compass bearing is the old method of locating the direction. In the given figure NOS and EOW represent north-south and east west direction respectively. O is the point of reference for the direction. We use line ON as the base line. Now, we have simple compass card having 8 directions as shown in the figure alongside. Where, NE - North East direction SE - South East direction SW - South West direction NW - North West direction Look at these two examples: The direction of P is N 30° E. The direction of Q is S 20° E. Note : The bearings are always measured from N or S but not from E and W. O NE E SE S N NW W SW S Q 200 W O E N N P W E S 30 O Unit 21 Bearing and Scale Drawing

246 Oasis School Mathematics-8 II. The three digits bearing The direction of any point in terms of an angle expressed in three digits as measured in clockwise direction with reference to the north line is called the three digit bearing. Here, the bearing of P from N is 060°. The bearing of Q from N is 150° Worked Out Examples Example 1 From the given figure, write down the compass bearing of P, NE and SW. Solution: Bearing of point P is N 15° E. Bearing of NE is N 45°E. Bearing of SW is S 45° W. Example 2 Write down the three digit bearing of the given points from O. (a) (b) (c) Solution: Here, (a) The bearing of A from O is 055°. (b) The bearing of B from O is 140°. (c) The bearing of C from O is (360° - 25°) = 335°. P N E Q O 300 600 S W Note : ∠NAB is the bearing of B from A. Reflex ∠N'BA is the bearing of A from B. N A N' B N P W E S 15º O SW SE NW NE O N A 550 N O B 1400 N O C 250