Acme Mathematics 8

Acme Mathematics 8 251(c) Tetrahedron(d) Octahedron(e) Square Pyramid(f) Cone(g) CylinderTetrahedron Skeleton of Tetrahedron Net of TetrahedronOctahedron Skeleton of Octahedron Net of OctahedronSquare Pyramid Skeleton of Square Pyramid Net of Square PyramidCone Net of ConeCylinder Net of Cylinder

252 Acme Mathematics 8C. Prism and Pyramid (a) Prism: A prism is a solid whose base and top are identical polygons. Triangular prism Rectangular prism (Cuboid) Circular prism (Cylinder)Net of Triangular prismModel of Triangular prismSkeleton of Triangular prism(b) Pyramid: A pyramid is a solid figure whose base is any polygons and side face are triangles. The number of faces of a pyramid depends on the number of the sides of the base.Triangular pyramid Circle base pyramid Square base pyramidNet of triangular pyramid Model of triangular pyramid Skeleton of triangular pyramidNet of Square Pyramid Model of Square Pyramid Skeleton of Square Pyramid

Acme Mathematics 8 253Classwork1. Study the net of a cuboid with numbers 1 to 6 marked on its 6 faces. Answer the following questions. (a) Which face is opposite to 5? (b) Which face is opposite to 3? (c) Which face are adjacent to 6?(d) Which face are adjacent to 1? 2. Fill in the blanks. (a) A triangular prism has ....... vertices. (b) A rectangular prism has ........ faces. (c) A circular prism has ........... circular faces. (d) A square base pyramid has ......... vertices. (e) A triangular base pyramid has............... edges. Exercise 5.91. Construct the following solids. (a) Triangular prism (b) Square base pyramid (c) Triangular pyramid (d) Rectangular prism. 2. Complete the following table.Solid objects Number of vertices Number of edges Number of faceTriangular PrismTetrahedronSphereCone3. List the solid shape around you.4. (a) Name the solid of following net.(b) Draw the net of cube. c) Draw a neat Net of cone. 1 3 6 425

254 Acme Mathematics 85.5 CoordinatesA. Distance between two pointsIf we have the coordinates of two points, by using Pythagoras theorem, we can find thedistance between two points as follows. In figure, AB is a line segment such that the coordinates of A are (2, 3) and that of B are(7, 8). A right angled triangle ACB has been drawn, as shown, with AB as hypotenuse. AC and BC are base and perpendicular respectively. The coordinates of C are (7, 3). What are the length of AC and BC ? The length of AC = 7 – 2 = 5 units.[2 and 7 are the x-values] The length of BC = 8 – 3 = 5 units [3 and 8 are the y-values] Now, using Pythagoras theorem, we get AB2 = AC2 + CB2= 52 + 52 = 25 + 25 = 50 or, AB = 50 or, AB = 5 × 5 × 2 or, AB = 5 2 units. Hence, length of AB is 5 2 unitsB. Distance formula Let A (x1, y1) and B (x2, y2) be any two given points. To find the distance between A and B, we complete the right angled triangle ACB asshown in figure.Now, the coordinates of C are (x2, y1) The length of AC = x2 – x1The length of CB = y2 – y1By using Pythagoras theorem, AB2 = AC2 + CB2OA(2,3)B(7,8)C(7,3)X' XY'YX' XY'x2y2y1O x1A(x1,y1)B(x2,y2)C(x2,y1)Y

Acme Mathematics 8 255or, AB2 = (x2 – x1)2 + (y2 – y1)2or, AB = (x2 – x1)2 + (y2 – y1)2Hence, distance between A and B, AB is (x2 – x1)2 + (y2 – y1)2 units. This is called distance formula.Solved ExamplesExample 1: Find the length of the line segment joining the points A (– 3, 6) and B (– 10, – 13). Solution : Let, (x1, y1) = (– 3, 6)(x2, y2) = (– 10, – 13) Now, by using distance formula,AB = (x2 – x1)2 + (y2 – y1)2= [ – 10 – (– 3)]2 + (– 13 – 6)2= (– 10 + 3)2 + (– 19)2 = (– 7)2 + (– 19)2= 49 + 361 = 410 Hence required length is 410 units.Example 2: Show that the points (1, – 1), (– 1, 1) and (– 3 , – 3 ) are the vertices of an equilateral triangle.Solution : Let, A(1, – 1), B(– 1, 1) and C(– 3 , – 3 ) are the vertices of an equilateral triangle ABC. Then using distance formula,AB = (1 + 1)2 + (– 1 – 1)2= 4 + 4= 8 = 2 2 ............................. (i)BC = (– 3 + 1)2 + (– 3 – 1)2 = 3 – 2 3 + 1 + 3 + 2 3 + 1 ABC

256 Acme Mathematics 8= 8 = 2 2 ................................(ii)CA = (– 3 – 1)2 + (– 3 + 1)2= 3 + 2 3 + 1 + 3 – 2 3 + 1 = 8= 2 2 ...............................(iii)From (i), (ii) and (iii), AB = BC = CA Since, sides AB, BC and CA are equal.Hence, the points (1, – 1), (– 1, 1) and (– 3 – 3) are the vertices of an equilateral triangle.Example 3: Show that the points (– 2, 1), (0, 2), (4, 2) and (2, 1) are the vertices of a parallelogram.Solution : Let A(– 2, 1), B(0, 2), C(4, 2) and D(2, 1) are the vertices of a parallelogram ABCD. AB = (0 + 2)2 + (2 – 1)2= 4 + 1= 5................................(i) BC = (4 – 0)2 + (2 – 2)2= 16 + 0= 4 .................(ii) CD = (2 – 4)2 + (1 – 2)2= 4 + 1= 5..............(iii) AD = (2 + 2)2 + (1 – 1)2= 16 + 0 = 4 .................(iv)A(–2,1)D(2,1)B(0,2)C(4,2)

Acme Mathematics 8 257From (i), (ii), (iii) and (iv), we get,AB = CD and BC = ADSince opposite sides of the quadrilateral are equal. The points (– 2, 1), (0, 2), (4, 2) and (2, 1) are the vertices of the parallelogram.Example 4: Find the value of k. if the distance between the points (2, k) and (6, 7) is 42units.Solution : Here, given points are (2, k) and (6, 7) and distance = 4 2units.Now, Using distance formula:d = (x2 – x1)2 + (y2 – y1)24 2= (6 – 2)2 + (7 – k)2Squaring both sidesor, 16 × 2 = 16 + 49 – 14 k + k2or, k2 – 14k + 33 = 0or, (k – 3) (k – 11) = 0 Either, k – 3 = 0or, k – 11 = 0∴ k = 3 and k = 11Hence, the value of k = 11 or 3Classwork1. Calculate the Distance between the given points. (a) A (2, 3) and B (8, 7) (b) C (3, 6) and D (4, 12) (c) E (– 16, 5) and F (5, 10) (d) G (– 8, 0) and H (4, 7) (e) K (7, – 9) and L (8, – 1) (f) X (0, 0) and Y (9, 9) Exercise 5.101. Calculate the Distance between the given points. (a) P (0, – 3) and Q (– 4, 0) (b) R (– 3, – 6) and S (– 1, – 3)(c) A (– 10 – 9) and B (0, 0) (d) S (1 + 2, 1 – 2) and T (1 – 2, 1 + 2)(2, k) 4 2 units(6, 7)

258 Acme Mathematics 82. The coordinates of the vertices of triangle ABC are given alongside calculate. (a) The length of side AB. (b) The length of side BC. (c) The length of side CA.3. Find the distance between the origin O and the following points.(a) B (7, 3) (b) A (– 10, – 8) (c) C (0, 12) (d) E (– 10, 0) (e) R (9, 12) (f) Z ( 3, 5) 4. (2, – 4), (– 4, – 4) and (– 1, 5) are the three vertices of a triangle. Find the length of the side of this triangle. What type of triangle is this ?5. Calculate the length of radius of given circle. Check whether the point R(1, 5) lies on circumference or not?6. Calculate the length of AB where A and B are the points on y-axis and x-axis respectively.OABX' XY'Y7. Identify the following triangles according to their sides.(a)B(5, –2) C(1, 1)A(4, 5) (b)B(–4, 5) C(–5, –1)A(–2, –1)B(–5,5) C(3,20)A(3,5)A(2,3)B(4,7)

Acme Mathematics 8 259(c)B(2, 0) C(0, 3 3)A(–2,0) (d)B(3, 1) C(5, 5)A(–2,4)8. Show that the following points are the vertices of an equilateral triangle:(a) p(3, 3), Q(– 3, – 3) and R( –3 3 , 3 3 )(b ) A(–1, 1), B(1, – 1) and C( 3 , 3 )9. Show that the following points are the vertices of an isosceles triangle:(a) (0, 0), (8, 2) and (5, – 3)(b) (1, 2), (3, 5) and (4, 4)(c) A(0, – 2), B( – 2, – 2) and C(2, – 2)10. (a) Show that the points (2, 3), (5, 8), (0, 5) and (– 3, 0) are the vertices of a rhombus.(b) Show that the points ( – 3, 3), (– 2, 6), (– 5, 5) and ( – 6, 2) are the vertices of a rhombus.(c) Show that the points A(1, – 4), B(3, – 3), C(5, – 4) and D(3, – 5) are the vertices of a rhombus.11. (a) Find the value of k if the distance between the points (6, k) and (4, 3) is 20 units.(b) If the points A(12, 8), and B(6, a) are at a distance of 10 units, find the value of 'a'.(c) The distance between the points A(a, 2a), and B(4, 3) is 10units, find the value of 'a'.12. Prove that the following points lies on a same straight line: [Note: Use graph](a) A (2, 0), B (0, 3) and C (– 4, 9) (b) D (1, 2), E (3, 4) and F (– 2, – 1)(c) M (– 3, – 3), N (3, 3) and O (6,6) (d) P (1, 3), Q (3, 4) and R (5,5) 13. Answer the following questions.(a) Define the Pythagoras theorem.(b) A ladder 10 m long rests against a vertical wall 6 m above the ground. At what distance does the ladder touch the ground from the bottom of the wall?

260 Acme Mathematics 8A. Symmetrya. Line of SymmetryLook at the following figures.mmmIn the above figures. Line 'm' is called the line of symmetry. These figures are called symmetrical figures. The lines which divides the figures into two equal halves is called the axis or line of symmetry.ObjectLine of symmetry Two identical parts5.6 Symmetry and Tessellations

Acme Mathematics 8 261b. Order of SymmetryLook at the following figures carefullyl1ml2ml1Heart Butterfly Isosceles triangleThis figure fitted onto (completely) the original shape after the rotation of 360° it has 1 older of rotational symmetry.The following figures have 2 order of rotational symmetry:l2l1Sine of crossl2l1Rectanglel3l2l1Equilateral triangleEquilateral triangle has 3 order of rotational symmetryl4l2l3l1SquareSquare has 4 order of rotational symmetryl1l2l3l5l6A circle has infinite order of rotational symmetry.Parallelogram and scalene triangle has no rotational symmetrym

262 Acme Mathematics 8Exercise 5.111. Draw the line of symmetry of the following figures, if possible.(a) (b) (c) (d)(e) (f) (g) (h)2. Find the order of rotational symmetry of the following figures.(a) (b) (c) (d)(e) (f) (g) (h)

Acme Mathematics 8 263B. TessellationTessellation is covering of the surface with regular congruent geometrical shapes in a repeating without any gaps. It is used in the surface, walls, floor or carpets to make the area more attractive. We can make attractive designs by drawing many triangles, squares, pentagons and hexagons. Look at the following designs.A tessellation of triangles A tessellation of square A tessellation of hexagonsIn all the figures a tessellation is created when a shape is repeated over again covering a surface without any gaps. It is regular tessellation.The following tessellations are the semi tessellationTessellation of hexagon and dodecagonTessellation of octagon and squareTessellation of triangle and parallelogramThe tessellations given below are the irregular tessellation.

264 Acme Mathematics 8Exercise 5.121. Copy the given tessellation and complete it in A4 size paper.2. Using the regular polygons given below make different tessellations taking at list to polygons at a time.(a) (b) (c) (d)3. Copy the given tessellation and complete it.(a) (b) (c)(d) (e) (f)

Acme Mathematics 8 2655.7 TransformationIn general, shifting the position of the object by moving each point to a different position by a specific procedure is called transformation.The basic transformations are: 1. Translation 2. Reflection 3. Rotation 4. Enlargement A. Translation Study the following movement of the objects.In the given figure the car in moving from place A to place B with certain distance anddirection,. This movement of the car is its translation.A BSimilarly, in the given figure a load is moved from point A to point B. This is also calledtranslation. The first position is called the object and the second position is called theimage of the object B.A Ba. Translation of a point:Suppose A is given point.Its movement is 5 units up from the original position. This movement is shown in the grid alongside. Point B is the new position of A after 5 units up. Point A is translated to its new position B.5 unitsBA

266 Acme Mathematics 8b. Translation of a line:Suppose AB is given line.Its movement is 5 units up from the original position. This movement is shown in the grid alongside. Line A'B' is the new position of line AB after 5 units up.c. Translation of a triangleSuppose that Rajiv pushes a triangle A, 4 units to the right. Triangle moves the place A to the place B. Again if he pushes the triangle A, 5 units up then triangle moves from the place B to place C.Triangles B and C are called the images of triangle A.Movement on the grid:(a) Right - up ......................(b) Right - down .................(c) Left - up ........................(d) Left - down ...................Solved ExamplesExample 1 : Study the grid and translate the point A : 4 units right.ASolution : Here point A shifts 4 square rooms right from the given position, which is shown in the figure.A A'BB'AA'A BC4 units5 unitsStarting point is the any given point of the object

Acme Mathematics 8 267Example 2 : Study the grid and translate the line towards 5 units right.ABSolution : Here Line AB shifts 5 square rooms right from the given position. which is shown in the figure.A A'B B'Example 3 : Study the grid and translate the triangle ABC towards 4 units right.ABCSolution : Here triangle ABC shifts 4 square rooms right from the given position, which is shown in the figure.A A'B B'C C'4 units4 units4 unitsExample 4 : Study the grid and translate the point M, 4 units right and 3 units up.M

268 Acme Mathematics 8Solution : Here point M shifts 4 square rooms right and 3 square rooms upward from the given position.MM'4 units3 unitsExample 5 : Study the grid and translate the triangle M, 5 units right and 3 units up.MSolution : Here point M shift 5 square rooms right and 3 units upward from the given position.5 unitsM 3 unitsM'Classwork1. Describe the movement of the points A. in different position.(a) A to B(b) A to C(c) A to D A CDB

Acme Mathematics 8 2692. Describe the movement of the line, CD in different position.BADCFEHG(a) CD to GH(b) CD to EF(c) EF to GH(d) CD to ABExercise 5.131. Describe the movement of the triangles in different positions.ABCD(a) A to B (b) A to C(c) C to D (d) D to A

270 Acme Mathematics 82. Translate the given figures 5 units right.(a) (b) (c)3. Translate the given figures 4 units right and 5 units up(a) (b) (c)4. In a square grid draw a point A and find its image after the given movement units.(a) 7 units up (b) 10 units right(c) 3 units right and 5 units up5. In a square grid draw a line segment MN and find its image after the given movement units:(a) 5 units right (b) 3 units up (c) 4 units right 6 units up6. In a square grid draw a right angled triangle PQR and find its image after the given movement units : (a) 6 units up (b) 7 units right(c) 4 units up and 4 units right (d) Colour all image triangles.7. Translate the given figures in the given direction and give magnitude. (Use the length of arrow and its direction)AABAA A BBK L(b) M ABCD(a) (c) K LP A

Acme Mathematics 8 271B. ReflectionActivityStudy the given graph and discus about the coordinates .∆ABC is given in the adjoining graph paper. Do the following activities. (a) Write the co-ordinate of vertices A, B and C. (b) Reflect the vertices A, B, and C in the line YY'(X-axis). Name the imagetriangle as A', B' and C'. (c) Write the co-ordinate of image A', B', and C'. O X' XYY'ABCNow, compare the co-coordinate of vertices of ∆ABC and vertices of image ∆A'B'C' and write conclusion. Here, in ∆ABC, vertices are A(1,1), B(6,2) and C(2,5). After reflection, image points are A'(−1,1), B'(−6,2) and C'(−2,5). While reflecting ∆ABC in y-axis, the y-coordinate of vertices A, B, and C remains same and sign of the x-coordinate is changed.

272 Acme Mathematics 8Introduction When we look into the mirror we see our face facing towards us. The shape that we see behind the mirror is our image. The mirror reflect the object. This is called reflection. Image inthe mirror is far behind the mirror as we are in front. Thus, reflection is a transformation in which object and image arefacing opposite to each other with the mirror line. Mirror line is called the axis of reflection.Solved ExamplesExample 1 : In the adjoining figure 'm' is the axis of reflection. Reflect ∆XYZ in the axis of 'm'.Solution : Method of reflecting the figure (i) Draw the perpendicular lines YA, XB and ZC to the mirror (ii) Produce YA to Y' such that YA = AY' (iii) Produce XB to X' such that XB = BX' (iv) Produce ZC to Z' such that ZC = CZ' (v) Join X', Y' and Z' respectively. Hence, ∆X'Y'Z' is the image of ∆XYZ.Reflection on the grida. Reflection of a pointA point X is given on the grid, 'm' is mirror line and B is the image of X.Point X and B are equidistant from the mirror 'm'. b. Reflection of a line:A line AB is given on the grid 'm' is mirror line. CD is the image of line AB.Y Z mXYY'A BX'ZZ'C mXX B2 units 2 unitsAD3 unitsB3 units C'm'Bm

Acme Mathematics 8 273c. Reflection of a triangleDraw ΔABC on the grid. Let 'l' be the mirror line.Now,Point A is 3 units left from the mirror line so its image point A' is 3 units right from the mirror line.Similarly,B' is the image of the point B. C' is the image of the point C.A', B' and C' are joined. Thus, ΔA'B'C' is an image of ΔABC, after reflection on line 'l'.Reflection using coordinates a. Reflection in x-axis i. Reflection of a point.Let, A(2, 3) be the point and x-axis (X'X) is the axis of reflection(mirror line). Now, draw AR⊥X'X and produce AR to A' so that AR = RA', where coordinates of A' is (2, – 3). ii. Reflection of a line:Let A (2, 3) and B(6, 3) are the end point of line AB. X-axis is the axis of reflection.Now, draw AR⊥X'X and produce AR to RA'. So, that AR = RA' and draw BS⊥X'X and produce BS to SB'. So, that BS = SB'. Join A' and B'. Line A'B' is the image of line AB.ABC C'B'A'Mirror line (l)3 units3 units3 units3 units5 units 5 unitsO RA(2, 3)A'(2, –3)YY'X' XO R SA(2, 3) B(6, 3)A'(2, –3) B'(6, –3)YY'X' X

274 Acme Mathematics 8iii. Reflection of a triangle.Let the vertices of ∆ABC are A(2, 3),B(4, 5) and C(10, 2). Reflect it in x-axis.Now, Draw AD⊥OX and produce AD to DA' so that AD = DA'.Draw BE⊥OX and produce BE to EB' so that BE = EB'. And draw CF⊥OX and produce CF to FC' So that CF = FC'Join the points A', B' and C'. Colour the triangle A'B'C'.Here, in the figure alongside ∆A'B'C' is an image of ∆ABC.Thus, while reflecting a point in x-axis, the x-coordinate is unchanged and the y-coordinatehas the same magnitude with opposite sign. In general, we have, A(x, y) Reflection in x-axis A'(x, – y). b. Reflection in y-axis i. Reflection of a point.In the adjoining figure let, A(2, 3) be the point and y-axis, the axis of reflection. Now draw AR perpendicular to YY' and produce AR to A' such that AR = RA', where coordinates of A' is (– 2, 3). Point A' is image of point A, after reflection in y-axis.ii. Reflection of line:In the adjoining figure let, A(2, 3) and B(4, 1) are the end of line AB. y-axis be the line of reflection. Now draw AR perpendicular to YY' and produce AR to A' such that AR = RA'. Similarly, draw BS perpendicular to YY' and product BS to B' such that BS = SB'. Where coordinates of A' is (– 2, 3) B' is (– 4, 1). Line A'B' is image of line AB, after reflection in y-axis.O D E FA(2, 3)B(4,5)C(10, 2)A'(2, –3)C'(10, –2)B'(4,–5)YY'X' XORA'(–2, 3) A(2,3)YY'X' XORSA(2, 3)B'(–4, 1) B(4, 1)YY'X' XA'(–2, 3)

Acme Mathematics 8 275iii. Reflection of triangle:The vertices of triangle ABC are A(1, 0), B(3, 2), and C(1, 4) . Reflect it in y-axisand write down the coordinates of the vertices of image triangle A'B'C'. Here, in the figure alongside triangle ABC is reflected along y-axis. Triangle A'B'C' is an image of triangle ABC. From the graph coordinates of vertices of triangle A'B'C' are A'(– 1, 0), B'(– 3, 2) and C'(– 1, 4).Thus, while reflecting, a point on y-axis, the y-coordinate is unchanged and the x coordinate has the same magnitude with opposite sign.In general, we have, A(x, y) Reflection in y-axis A'(– x, y)Classwork1. Reflect the objects on the given mirror line 'm'.(a) (b) (c)2. Reflect the object on x-axis and colour the image.(a) (b) (c)OC'(–1,4) C(1,4)A'(–1,0)B'(–3,2) B(3,2)A(1,0)YY'X' XmAmABmYY'X' XAOYY'X' XABOYY'X' XPQ RO

276 Acme Mathematics 8Exercise 5.141. Describe the reflection axis.YY'X' XOB AC D(a) A to B(b) B to C(c) C to D(d) A to D2. In the figure alongside 'S' is reflected in the y-axis to form its image as triangle 'T'.(a) Draw the image triangle 'T' and Colour it.(b) Write down the coordinates of the vertices of triangle 'T'.(c) Translate triangle 'T' by 4 unit right and 6 unit up and color the triangle.3. Reflect the point A(– 4, 5) on the graph when the axis of reflection is given below: (a) x-axis (b) y-axisYY'X' XSO

Acme Mathematics 8 2774. The line AB joining the points A(3, –3) and B(7, 5) is reflected on the given axis: (a) x-axis (b) y-axis Find the coordinates of the images of the line AB. A B5. The vertices of ∆TMP are T(1, 3), M(10, 3) and P(10, 10). Write down the coordinatesof the image of the vertices of ∆TMP when reflected on the following reflection axis:(a) x-axis (b) y-axis YY'X' XTM PO6. Points A(0, 5), B(9, 0) and C(3, 10) are the vertices of a triangle ABC. The triangle ABC is reflected in x-axis. Present both the figures on the same graph and find the coordinates of the vertices of the image of the triangle ABC . 7. Triangle PQR with vertices P(– 7, 2), Q(0, 5) and R( – 7, 9) is reflected in y-axis. Present both the figures on the same graph and find the coordinates of the vertices of the image of the triangle PQR .8. Reflect the point A(– 4, 5) on the graph when the axis of reflection is given below:(a) first in x-axis then y-axis(b) first in y-axis then x-axis9. Draw the ∆ABC on yours graph paper and answer the following questions. Based ona given graph.(a) Reflect the ∆ABC in Y axis andwrite cordinates of the verticals of ∆ABC after reflection.(b) Double the cordinates of the points A, B, and C and draw on the another graph. Reflect ∆ABCin Y axis measure the length of AA' and new AA'. Is new AA' is double of AB? Write the reason. O XYA BC

278 Acme Mathematics 8C. RotationLook at the given adjoining figures:The figure (i) describes the rotation of a point Aabout the point K by 50° in clockwise direction so that the image of A is A'.Similarly, B' is an image of B after 180°rotation.KB and KB' are related to each other by:1. Centre K.2. Angle of rotationIn the above figures AK = KA' and KB = KB'Thus, rotation is the transformation where every point of an object move about a fixed point through a given angle. The fixed point is called centre of rotation and the angle is called the angle of rotation.(a) Rotation through (– 90°) The rotation of an object A with an angle of (–90°) is given alongside.It is clockwise rotation. (– 90°) rotation is called negative rotation.In the figure A to A' is (–90°) rotation through point ‘K’. (b) (+ 90°) rotationThe rotation of an object with an angle of (+90°) is given along side. It is anticlockwise rotation. (+ 90°) rotation is called positive rotation.In the figure B to B' is (+90°) rotation. Ingeneral (+90°) is denoted by 90o rotation through point ‘R’..Note: The angle of rotation can be taken of any measurement.Rotation on the gridUse of coordinates is the easiest way to find the image of the given object after rotatingthem about some specific angle as 90o and its multiples such as 180o, 270o, 360o. When we use coordinates, we take (0, 0) as centre of rotation in this chapter.KFigure (i) Figure (ii)A A'B'B180°K50°90°A'AKBB'90°R

Acme Mathematics 8 279Remember !(i) – 90o rotation about origin(0,0) is clockwise rotation through 90°(ii) + 90o rotation about origin(0,0) is anti-clockwise rotation through 90°(a) - 90o rotation about origin(0,0)Example 1 : Rotate the point B with O as center of rotation and -90° as angle of rotation.Solution: Steps:• Join O and B with dotted line.• Draw 90° clockwise at O with dotted line. • Taking O as center and OB as radius, draw an arc towards 90° line.• The intersecting point of 90° line and arc is the image of point B. Mark it as B'.Now,B' is an image of point B.(b) 90o rotation about origin (0,0)Example 2 : Rotate the point A with O as center of rotation and 90° as angle of rotation.Solution: Steps: • Join O and A with dotted line. • Draw 90° anticlockwise at O with dotted line. • Taking O as center and OA as radius, draw an arc towards 90° line. • The intersecting point of 90° line and arc is the image of point A. Mark it as A'.Now, A' is an image of point A.90°(0, 0)B'(5, – 4)B(4, 5)X' O XY'Y90°(0, 0)A(8, 4)A'(–4, 8)X' O XY'Y

280 Acme Mathematics 8Example 3 : The points A(5, 8) and R(7, 4) are the ends of the line AR. The line AR is rotated (+ 90o) about origin. Find the image on the graph.Solution: Steps:• Join O and A with dotted line.• Draw 90° anticlockwise at O with dotted line. • Taking O as center and OA as radius, draw an arc towards 90° line.• The intersecting point of 90° line and arc is the image of point A. Mark it as A'.Repeat the same process for the Point R and find R'. Join A’ and R’.The image of the line AR is A'R' which is shown on the graph.Example 4 : The points A(–2, 5), R(1, 9) and B(4, 7) are the vertices of a triangle ARB. The ∆ARB is rotated (–90o) about origin. Find the image on the graph.Solution: Steps:• Join O and A with dotted line.• Draw 90° clockwise at O with dotted line. • Taking O as center and OA as radius, draw an arc towards 90° line.• The intersecting point of 90° line and arc is the image of point A. Mark it as A'.Repeat the same process for the points B and C and find A' and C'. Join the points A', B' and C' and Color the triangle A’B’C’.Now, ΔA'B'C' is an image of ΔABC.x'R'A'AO90°R(7, 4)(5, 8)(– 4, 7)(– 8, 5)xy'yBA90°RB' (7, – 4)R'(9, –1)(4, 7)(– 2, – 5)A'(2, 5)(9, 1)X' O XY'Y

Acme Mathematics 8 281Exercise 5.151. Rotate the following figures through 90o taking K as centre of rotation.(a) (b) (c)2. Rotate the point A(2, 2) taking origin as centre of rotation and the following angles as angle or rotation.(a) + 90o (b) – 90o3. Rotate the line AB joining the point A(2, 2) and B(5, 5) taking origin as centre of rotation and the following angles as angle of rotation.(a) 90o (b) – 90o4. Plot on the graph and rotate the quadrilateral ABC with vertices A(6,1), B(10,–2) and C(4,7) taking (0, 0) as centre of rotation and following angles as angle of rotation:(a) positive quarter turn(+90o)(b) negative quarter turn(–90o)5. In the graph A (2, 6), B (2, 2) and C (6, 2) are the verticals of ∆ABC(a) Draw ∆ABC on your graph. Rotate ∆ABC through -900 taking (0, 0) as a centre of rotation and colour the image ∆ A' B' C.(b) Calculate the length of OA and OA'. Are OA and OA' equal?AKBAKBAB KBO XYCA

282 Acme Mathematics 8A. BearingACTIVITYStudy the given figure.” Name the figure.” What is represented by N, E, S, W?” Where is it used?In the adjoining figure the 4 directions are shown. N → North direction S → South direction E → East direction W → West directionHow for is E from N? A bearing is a direction. Bearings are measured clockwise from north direction. a Compass bearingThe compass indicates the direction. The four main points of the compass are North (N), South (S), East (E) and West (W). The direction halfway between North and West is described as North West (NW). Similarly we have North East (NE), South East (SE) and South West (SW). b. 3-figure bearing:Direction is measured from clockwise direction from north direction. It is angular measurement. It is expressed as '000° for 0° and 025° for 25° from the north direction. We express angles in 3 digits. So it is called 3-figure bearing.Comparing of compass bearing and 3-figurebearings are given below :Compass bearing 3-figure bearingNorth 000°North East (NE) 045°East 090°South East (SE) 135°W270°E090°N360°/000°S 180°NENSW ESWSEWN5.8 Bearing & Scale Drawing

Acme Mathematics 8 283Compass bearing 3-figure bearingSouth 180°South West (SW) 225°West 270°North West (NW) 315°In general it is used in navigation on sea, land.Example:Imagine that you are in a boat at sea. You are lost and you cannot see any land. You cry for help and are told to sail at a bearing of 125∘. What does this mean? A bearing is a direction. It is always measured clockwise from North. The bearing 125∘ is shown in the diagram alongside.Finding the bearing of A from BHere is an example. Imagine that you are at point B. You need to travel to point A. On which bearing should you travel? You will need to use a protractor.The bearing of A from B is 037∘.Solved ExamplesExample 1 : Find the bearing of A from B. Solution: We use a protractor to measure the angle SBA.It is 46∘The bearing of A from B is S46∘ Eor, (180° – 46°) = 134°Example 2 : Find the bearing of A from B.Solution: We use a protractor to measure the angle SBA.It is 30∘The bearing of A from B is S 30°W or (180°+ 30°) = 210°N125°Direction of travelNABNA37°BNW EBSA46°BASNW E

284 Acme Mathematics 8Example 3 : Draw the bearing of 312°.Solution : Here, Bearing of the point R is 312°Example 4 : Write the bearing of point O from point B.Solution : Here, bearing of ∠NOB is 70°bearing of point B is 070° bearing of O from B = 360° –∠OBN' (x) = 360° – 110° = 250°Hence, bearing of points O from B is 250°.Classwork1. Study the given figure given alongside and find the degree measure of: (a) NE (b) SE(c) SW (d) NW2. Find the bearing of the point R.(a) (b) (c)(d)ORN (e)O RN (f)ORNORNSW E312°NN'070°250°x°OBOSSW SEW ENW NENORNORNORN

Acme Mathematics 8 285Exercise 5.161. Write down the bearing of the aero plane from the Bharatpur airport to the given places. (a) Pokhara (b) Kathmandu (c) Biratnagar (d) SalyanDadeldhuraHumlaSalyanPokharaButwal KathmanduBirgung Bharatpur airportNamche BazarBiratnagarRolpaN100 km200 km2. Find the bearing of the place A from the place B.(a)AB 50°N1N(b)A B90°N1 N(c)A120°BN N1(d)ABNN1240°(e)AB300°N1 N (f)AB100°NN1

286 Acme Mathematics 83. Convert the following compass bearing to the angle. (a) NE (b) NNE (c) NW (d) NNW4. Express each of the following compass bearing as 3-figure bearing.(a) S34°E (b) N53°W(c) S29°W (d) N88°E5. Express each of the following 3-figure bearing as compass bearing.(a) 283° (b) 089°(c) 249° (d) 052°6. The bearing of point Q from R is 120°. State the bearing of the point R from Q. Give your answer in,(a) 3-figure bearing (b) compass bearing7. The bearing of point J from K is S51°W. State the bearing of the point K from J. Give your answer in,(a) 3-figure bearing (b) compass bearing8. The point C is on a bearing of 065° from point A and on a bearing of 310° from point B. Draw figure and locate the point C.

Acme Mathematics 8 287B. Scale DrawingACTIVITYConsider the given map of 5 districts. ” Is size of map is real?” How big is Chitwan?” Which district is largest?Here, the map is reduced to very small size. Size of chitwan is second largest among 5 districts. Makawanpur is the largest among all.This is the drawing only. A drawing that shows a real object either in large size or in small size is its scale drawing. Scale indicates the ratio between the size of real object and its figure (drawing).For example: The height of tree 30 meter can be represented by the given figure along side.Solved ExamplesExample 1 : The figure given alongside is the map of \"Ludo\"? Find its scale, if its actual length is 42 cm. Solution : Here, actual length of its side is 42 cm. But in the figure (map) its length = 3.5 cm Now, Scale = Length in figureActual length= 3.5 cm 42 cm= 112 = 1 cm : 12 cm Hence, the scale is 1 : 12ChitwanAEB CD100 km50 kmMakawanpurPrasaBaraRautahatN

288 Acme Mathematics 8Exercise 5.171. Taking scale 1 mm: 1 m, find the actual length and breadth of the following objects.(a) (b) (c)2. A rectangular ground is 500 m long and 200 m broad. Using 1 mm: 100 m as scale, draw the figure of the ground.3. Find the actual length of the following objects whose photo are given below. (a) (b) (c)4. Study the floor-plan map of a house and answer the following questions. [Scale 1 cm = 2 m](a) Find the length of bedroom 2. (b) Find the length of passage. (c) Find the length and the breadth of baranda. (d) Find the length and the breadth of bathroom. (e) Find the length and the breadth of the house.5. Study the map of a Narayani Zone (given above and answer the following questions.(a) Find the distance between A and E.(b) How far is D from B?(c) Find the distance between C and E.(d) Which place is far from E ?1 Passage 2Kitchen BarandaBedroom BedroomBathroomChitwanAEB CD100 km50 kmMakawanpurPrasaBaraRautahatN

Acme Mathematics 8 2891. (a) Write the formula to find the distance between two points.(a) If the distance between A(0, 4) and B(3, a) is 5 units then, find the value of a.2. The vertices of a triangle ABC are A(3, 2), B(4, 5) and C(–2, 3). (a) Draw the triangle ABC on a graph paper and reflect it in x-axis. (b) Find the coordinate of its image and draw its image on the same graph paper.3. (a) Write the formula to find distance between two points A(x1, y1) and B(x2, y2)(b) If P(0, b) and Q(a, 0) are the points and distance between them is 6 units then find the value of a.4. In the given diagram, the bearing of A from C is 060°.(a) Which is the base line direction to find the bearing of any place?(b) How many digits are used for denoting the bearing angle?(c) Find the bearing of C from A.5. (a) If P(2, 3) is a point on the line joining the points A(4, 5) and B(0, 1), show the PA = PB(b) M(2, – 5) and N(– 1, 2) are the ends of the line segment MN. Find the coordinates of image line M'N' under the rotation through +90° about origin. Also plot the object and image on the same graph paper.6. Do the followings:(a) Define tessellation.(b) Using the vertices (V), Face(F) and Edges (E) of any solid objectProve that V – E + F = 2.(c) Plot the vertices of triangle A(1, 2), B(5, 2) and C(3, 4) on a graph paper. Reflect it through x-axis and write the co-ordinate of image.7. The vertices of triangle PQR are P(2, 3), Q(5, 2) and R(3, – 2).(a) Draw triangle PQR on the graph and reflect in y-axis.CN N1A 060°Mixed Exercise

290 Acme Mathematics 8(b) Write down the coordinate of an image.(c) Draw a diagram to show the bearing of 080° angle.(d) A(3, – 4) and B(– 4, 3) are the ends of a line AB, find the image under the rotation about origin through +90°. Also plot the line AB and its image on its same graph.8. (a) Find the distance between two points (–a, 0) and (0, – b)(b) If the bearing of point B from point A is 060°, what is the bearing of A from B?(c) Find the vertices of the image ΔD'E'F' of ΔDEF with vertices D(4, 5),E(– 6, 1) and F(– 2, 7) under the rotation through 180° about the origin. Show both ΔDEF and ΔD'E'F' in a same graph.9. (a) By using regular pentagon, what type of tessellation can be made?(b) Write down in angle, the bearing of the place of B from K in the given figure.(c) Find the vertices of the image ΔA'B'C' of ΔABC withvertices A(– 4, 5), B(6, 3) and C(4, 0) after the rotation through positive 90° about the centre at origin. Represent ΔABC and ΔA'B'C' on the same graph.10. (a) What type of triangles can be used to make the regular tessellation?(b) The vertices ofΔPQR are P(4, 5), Q(7, 6) and R(–1, 9). RotateΔPQR through (–90°) about the origin and write down thecoordinates of the image triangle PQR.(c) Write down the bearing of point P from the point O.11. (a) Write the coordinate of the image when the point P(x, y) is displaced a unit to the right and b unit to down.(b) Construct a net of tetrahedron formed by four equilateral triangle.12. The vertices a triangle ABC are A(3, 6), B(–2, 4) and C(5, 1).(a) Find coordinates of its image when it is reflected on y-axis.(b) Draw the triangle ABC and its image on the same graph paper. 13. (a) The bearing of Kathmandu to Pokhara is 293°. Find the bearing of Pokhara to Kathmandu.65°KNB30°OP N

Acme Mathematics 8 291(b) A(2, 3), B(5, 4) and C(3, –2) are the vertices ofΔABC. Find the image ofΔABCunder the translation T–23 . Represent the object and its image by graph.14. A(1, 1), B(3, 1) and C(3, 4) are the vertices of a triangle.(a) Find the coordinate of the image of ∆ABC under the reflection on y-axis.(b) Plot the image in graph paper. (c) What will be the distance between the point C and its reflection image C'? 15. (a) Find the value of x if the distance between two points (3, 5) and (x, 5) is 10 units. (b) P(3, 4), Q(7, 3), and R(4, 3) are vertices of ∆PQR. Find the image of ∆PQR reflected by y-axis. (c) Plot the co-ordinates of vertices of ∆PQR and its image on the same graph. 16. (a) In the figure if the bearing of A from B is 060°, find the bearing of B form A. (240°)(b) The vertices of a triangle ABC are A(– 5, – 2), B(–1, – 4) and C(–3, 3). Reflect the triangle ABC about x-axis and present both the triangles on the same graph. (c) If 3x0, 4x0 and 5x0 are the angles of a triangle, find these angles. NBA060°N'

292 Acme Mathematics 8EvaluationTime: 96 minutes Full Marks: 401. (a) Write down the formula to find the distance between A and B. from the given graph. [1] O X' XYY'A(x1,y1)B(x2,y2) (b) Draw Δ??? ?? ????h ????? ??? ??????? ?? ?? ?−????.Then write its coordinate after reflection and also plot the image triangle. [3]CO X' XYY'AB (c) Write down any two properties of Δ??? ??? Δ?′?′?′. 2]2. The vertices of a ΔABC are A(1,1), B(3,0) and C(4,2). Answer the followingquestions. (a) Find the image of ΔABC under the translation '6 units right and 4unitd up' inand draw the graph paper. [3] (b) Find the image of ΔABC under the rotation of 900+ve about the origin. [3] (c) Write down the bearing of the point P from O. [1]N60oPO

Acme Mathematics 8 2933. (a) What is regular tessellation? [1](b) The bearings of B from A is 0650, find the bearings of A from B. [2] (c) Plot a ΔPQR with vertices P(2, 2), Q(4, 6) and R(6, 3) in a graph and reflect ΔPQR in Y-axis. Also find co-ordinates of its image ΔP'Q'R'. [3]4. (a) Write the formula to find the distance between two points. [1](b) Distance between A(0, 4) and B(3, a) is 5 units. Find the value of 'a'. [2]5. (a) Find the distance between two points (–a, 0) and (0, –b). [1](b) If the bearing of point B from point A is 060o, what is the bearing of A from B ? [2](c) Find the vertices of the image ΔD'E'F' of ΔDEF with vertices D(4, 5), E(–6, 1)and F(–2,7) under the rotation through 180o about the origin. Show both ΔDEF and ΔD'E'F' in a graph. [3]6. (a) By using regular pentagon, what type of tessellation can be made ? [1](b) Determine the scale of the given figure. [2](c) Find the vertices of the image ΔA'B'C' of ΔABC with vertices A(–4, 5), B(6, 3) and C (4, 0) after the rotation through positive 90o about the centre at origin. Represent ΔABC and ΔA'B'C' on the same graph. [3] 7. (a) Define Tessellation. [1](b) Using the vertices (V), Faces (F) and Edges (E) of any solid object, prove that V – E + F = 2. [2](c) Plot the vertices of triangle A(1, 2), B(5, 2) and C(3, 4) on a graph paper. translate it through '5units left and 5 units down' and write the co-ordinate of image. [3]

294 Acme Mathematics 86UNIT Statistics6.1 Revision Everyday we come across numerical data in the newspapers, magazines, television and financial review. The data may relate to pollution content of different towns, cost of fuel efficient cars, profit of an organization or certain other facts. Numerical data is used to draw differences. This entire process of collection and interpretation of data is statistics. Thus statistics is a branch of mathematics in which we deal with collection and classification of data, analyses and interpret them and make useful inferences. Collection of Data Let us suppose that we are interested in knowing about the performance of students in Mathematics in the half yearly examination. There are 24 students in the class. We ask each of the 24 students his/her score in Mathematics and note down the marks. The marks obtained by each student are as follows: 85 57 53 42 97 49 93 89 84 85 57 64 57 49 88 95 63 57 85 53 63 57 85 63 The maximum marks allocated to the subject are 100. Here each entry is a numerical fact and is called an observation. A collection of observations collected initially is called raw data. The data, as it is, does not help us in coming to any conclusion till we arrange the data in ascending or descending order. Let us arrange the data in ascending order:42 49 49 53 53 57 57 57 57 57 63 63 63 64 84 85 85 85 85 88 89 93 95 97 A glance at the above data shows that: the highest marks obtained are 97 and the lowest marks are 42. Data collected as numerical fact is called observation.The data obtained in the original form is called raw data or ungrouped data. The data arranged in an order – descending or ascending order is called arrayed data. The above data can be written asMarks 42 49 53 57 63 64 84Number of Students 1 2 2 5 3 1 1Marks 85 88 89 93 95 97 TotalNumber of Students 4 1 1 1 1 1 24The above table shows that 5 students got 57 marks each while 4 students scored 85 marks each. The quantity (marks) is called the variable (Variate) and the number of students who have scored a particular mark is called the frequency.

Acme Mathematics 8 295The data presented in the form is called Frequency Distribution Table for Ungrouped Data. Grouped Data If the variable extends over a wide range and the number of observations becomes very large, arranging the data in ascending or descending order becomes time consuming. In such cases, it becomes necessary to convert frequency into class group that is called class intervals. When the data is arranged in groups, it is called grouped data.Taking equal classes, one of them being 40-50, we can present the above data as below:Marks 40 - 50 50 - 60 60 - 70 70 - 80 80 - 90 90 - 100 TotalNumber of Students 3 7 4 - 7 3 24This is called Frequency Distribution Table.The first row indicates that the whole data has been distributed in six classes: 40 – 50, 50 – 60, …… 90 – 100. The second row indicates the number of students between the two limits of each class. The lower limit of the class 40-50 is 40 and the upper limit of the class is 50. Formation of ClassesRaw data is to be represented by grouped data by taking the following steps: 1. Find the range by finding the maximum and minimum of raw data. 2. Depending upon the range, decide about the number of classes to be formed. 3. Divide the range by the number of classes to be formed. 4. Write the classes.5. Use tally method to find frequency. For the sake of convenience, tally marks are put in bunches of five, the fifth one crossing the other four diagonally ( ) which represents Five. Let us illustrate it by taking examples.Solved ExamplesExample 1 : The weight of 25 students of a class (in Kg.) are as given below: 35, 40, 38, 39, 50, 54, 59, 37, 49, 51, 53, 42, 45, 46, 37, 48, 54, 58, 52, 38, 42, 41, 49, 55, 38Present the above data in the form of frequency distribution using the same class size, one such class being 40 - 45 (45 not included).

296 Acme Mathematics 8Solution : Here, the minimum and maximum observations are 35 and 59 respectively. The range of data is (59 – 35) = 24 As one of the suggested classes is 40 - 45, therefore the classes are: 35 -4 0, 40 -45, 45-50, 50-55, 55-60The frequency distribution is as below:Marks obtained Tally Bars Frequency35 – 40 || 740 – 45 |||| 445 – 50 550 – 55 | 655 – 60 ||| 3Total 25Cumulative frequency and cumulative frequency table The sum of frequencies of all the previous classes and that particular class is called the cumulative frequency of the class. Consider an example Cumulative frequency for the following data:Classes Frequency Cumulative Frequency0 – 10 7 710 – 20 8 (7 + 8) = 1520 – 30 12 (15 + 12) = 2730 – 40 9 (27 + 9) = 3640 – 50 4 (36 + 4) = 4050 – 60 3 (40 + 3) = 4343

Acme Mathematics 8 2976.2 Pie Chart ActivityThe details of yearly expenditure of Krishna's family is presented in the given pie chart.Discuss the following questions based on the above pie-chart. (a) Which headings have minimum expenditure?(b) What is the yearly expenditure of Krishna's family in education?(c) What is their expenditure on health, clothes and food?(d) What is the expenditure percentage in each heading?(e) What is the monthly expenditure on education if they are doing equal expenditure in each month of that year? (f) What is the yearly income of the Krishna's family if the family saved Rs 2,40,000 yearly in a bank? Why we use pie chart ? Discuss in the class.

298 Acme Mathematics 8IntroductionIn a pie chart, the values of different components of a frequency distribution are represented by the sectors of a circle. These sectors are constructed in such a way that the area of each sector is proportional to the corresponding value of the component. From geometry, we know that the area of a sector of a circle is proportional to the angle made by its arc at the centre of the circle. Thus, the central angle of each sector must be proportional to the corresponding value of the component.Since the sum of all the central angles is 360°, we have central angle of a component = value of the component total value × 360 °a. Construction of a pie- chart for a given data In order to construct a pie chart for a given frequency distribution, we proceed according to the following steps.Steps of Construction 1. Calculate the central angle for each component, given by central angle of a component = value of the component total value × 360 °2. Draw a circle of any convenient radius.3. Within this circle draw a horizontal radius.4. Starting with the horizontal radius drawn in step 3 ,draw the radii making central angles corresponding to the values of the respective components, till all the components are exhausted. Thus, the circle has been divided into various sectors. 5. Shade each sector with a different design. This will be the required pie-chart for the given data. Now, study the solved examples carefully.Solved ExamplesExample 1 : A man with monthly salary of Rs 18000, plans his budget for a month as given below:Item Food Rent Education Saving MiscellaneousAmount (In Rs) 5250 3500 3250 4000 2000Represent the above data in a pie-chart. Solution : Here, total value = 18000 Central angle of a component = value of the component total value × 360 °

Acme Mathematics 8 299Calculation of central anglesItem Amount (in Rs) Central angleFood 5250 5250 18000 × 360 ° = 105°Rent 3500 3500 18000 × 360 ° = 70°Education 3250 3250 18000 × 360 ° = 65°Saving 4000 4000 18000 × 360 ° = 80°Miscellaneous 2000 2000 18000 × 360 ° = 40°Construction of the pie-chart Steps of construction:1. Draw a circle of any convenient radius. 2. Within this circle draw a horizontal radius. 3. Starting with the horizontal radius drawn in step 2, draw sectors with central angles of 105°, 70°, 65°, 80° and 40° 4. Shade the sectors so obtained differently and label them.Thus, we obtain the required pie-chart as shown in the adjoining figure.Example 2 : Draw a pie-diagram to represent the following data on money spent (in cores of rupees) on various item during the tenth five year plan.Item Money spent (in crore of rupees)Agriculture 12500Energy 15000Industry 10000Rural development 20000Social Service 21250Miscellaneous 11250Total 90000Solution : Here, total value = 90000 Central angle of a component = value of the component total value × 360 °FoodMisc.SavingEducationRent

300 Acme Mathematics 8Calculation of central anglesItem Money spent (in cores of rupees) Central angleAgriculture 12500 12500 90000 × 360 ° = 50°Energy 15000 15000 90000 × 360 ° = 60°Industry 10000 10000 90000 × 360 ° = 40°Rural development 20000 20000 90000 × 360 ° = 80°Social Service 21250 21250 90000 × 360 ° = 85°Miscellaneous 11250 11250 90000 × 360 ° = 45°Construction of the pie-chart Steps of construction: 1. Draw a circle of any convenient radius.2. Within this circle draw horizontal radius.3. Starting with the horizontal radius, draw sectors with central angles of 50°, 60°, 40°, 80°, 85° and 45°. 4. Shade the different sectors differently and label them. Thus, we obtain the required pie-chart, shown in the figure.b. Relative frequencies Frequencies of components which are expressed as a percentage of the total frequency are known as relative frequencies. Thus, relative frequency of a component = frequency of the component total frequency × 100 % The relative frequencies can be easily represented by pie-charts, as shown in the following example.Solved ExamplesExample 1 : The following table shows the expenditure incurred in the construction of a house in Kathmandu.AgricultureEnergyIndustryRural DevelopmentSocialServiceMisc.