Maths new edition

["13.2 Chapter 13 INTRODUCTION Mensuration is a branch of mathematics. It deals with computation of geometric magnitudes, such as the length of a line segment, area of a surface and volume of a solid. In this chapter, we will deal with computation of areas and perimeters of plane figures and surface areas and volume of solid figures. To be precise, we will learn to find the perimeters and areas of plane figures including triangles, quadrilaterals, etc. Further, we will study prisms, cubes, cuboids and cylinder. Then we will learn how to compute area and volume of pyramids, cones, cone frustums, tetrahedrons and octahedrons. Finally, we will discuss spheres, hemispheres, hollow-spheres, hollow-hemispheres and torus. PLANE FIGURES A figure lying in a plane is called a plane (or planar) figure. Triangles, rectangles, circles are some examples of plane figures. A plane figure is a closed figure if it has no free end. It is called a simple closed figure if it does not cross itself. A plane figure is made up of line segments or curves or both. A plane figure is called a rectilinear figure, if it is made up of only line segments. Triangles, squares, pentagons, etc. are some examples of rectilinear figures. A circle is not a rectilinear figure. The part of the plane which is enclosed by a simple closed figure is called a plane region. The magnitude of a plane region is called its area. A line segment has one dimension, i.e., length. Hence, its size is measured in terms of its length. A planar region has two dimensions, i.e., length and breadth. Hence, its size is measured in terms of its area. The perimeter of a closed planar figure is the total length A SR of the lines bounding the figure. In Fig. 13.1, ABC is a triangle and PQRS is a quadrilateral. The perimeter of the \u2206ABC is AB + BC + CA. The perimeter of the quadrilateral PQRS is PQ + QR B CP Q + RS + SP. Figure 13.1 The perimeter of a rectangle of length l and breadth b is 2(l + b). The perimeter of a square of side s is 4s. Units of Measurement The length of a line is measured in units, such as centimetres, inches, metres, feet. Area is measured in units, such as square centimetres, square inches, square metres, square feet. ESTIMATING AREAS The area of a plane figure can be estimated by finding the number of unit squares that can fit into a whole figure. Observe the square ABCD (Fig. 13.2), in the square ABCD, there are 16 unit squares. Therefore, its area is 16 sq. units.","Mensuration 13.3 In Fig. 13.3, PQR is a right triangle. All the small units are not squares. D C Just as the length of a line need not be a whole number of units, the area of a plane figure also need not be a whole number of units. In the figure, the triangle PQR does not cover 8 unit squares. But, its A B area is equal to that of 8 unit squares. This is because, the partial units, when added, are equal to 2 unit squares. Besides this, there are 6 unit squares. Hence, its area is equal to 8 unit squares. Area of a Rectangle Figure 13.2 P A rectangle is a four-sided closed figure in which both pairs of opposite sides are equal and each angle measures 90\u00ba. In Fig. 13.4, ABCD is a rectangle. In rectangle ABCD, AB = 5 cm and BC = 4 cm. Each of the small squares measures 1 cm by 1 cm and the area occupied by each small square is 1 cm2. There are 4 rows, each row consist of 5 unit squares. \u2234 The area occupied by the rectangle = (4)(5) = 20 cm2. QR The number of unit squares in the rectangle ABCD is equal to the Figure 13.3 product of the number of unit squares along the length and the number C of unit squares along the breadth. Thus, we observe that the area of a D rectangle = (length)(breadth). Area of a Parallelogram Consider parallelogram ABCD as provided in Fig. 13.5: Draw a perpendicular DE from D to AB. AB Draw a perpendicular CF from C to AX (AB extended). Figure 13.4 By the ASA congruency property, the area of D C \u2206AED is equal to the area of \u2206BFC. \u2234 The area of parallelogram ABCD = The area of rectangle DEFC. \u21d2 (EF)(FC) = (AB)(DE) \u2234 Area of parallelogram = (Base)(Corresponding A EB FX height) Figure 13.5 Area of a Triangle Consider \u2206ABC as provided in Fig. 13.6: A D Draw a line AD parallel to and equal to BC. Join CD. ABCD is a parallelogram. By ASA axiom, the area of triangle ABC is equal to the area of triangle ADC. \uf8eb 1\uf8f6 B E C \uf8ed\uf8ec 2\uf8f7\uf8f8 Figure 13.6 \u2234 The area of triangle ABC = (Area of a parallelogram)","13.4 Chapter 13 (Where AE is the height of the triangle or parallelogram.) \u2234 The area of a triangle = \uf8eb 1\uf8f6 (Base)(Height) \uf8ec\uf8ed 2\uf8f8\uf8f7 The area of a triangle is also given by s(s \u2212 a)(s \u2212 b)(s \u2212 c ), where a, b and c are the lengths of the sides of the triangle and s is the semi-perimeter, i.e., s = (a + b + c). This relation is known 2 as Heron\u2019s formula. Area of a Trapezium Consider trapezium ABCD (Fig. 13.7), in which AD and BC are the parallel sides and AB and DC are the oblique sides (not parallel). Join BD. DE is the perpendicular distance between AD and A D BC. The area of the trapezium is equal to the sum of the areas of the \u2206ABD and \u2206BCD. = \uf8eb 1 \uf8f6 ( AD )(DE ) + \uf8eb 1 \uf8f6\uf8f7\uf8f8 (BC )(DE ) \uf8ec\uf8ed 2 \uf8f8\uf8f7 \uf8ed\uf8ec 2 \uf8eb 1 \uf8f6 B E C \uf8ed\uf8ec 2 \uf8f8\uf8f7 Figure 13.7 = ( AD + BC )(DE ) \u2234 Area of a trapezium = \uf8eb 1 \uf8f6 (Sum of the parallel sides)(Distance between them). \uf8ed\uf8ec 2 \uf8f8\uf8f7 Area of a Rhombus P S R A rhombus is a parallelogram in which all sides are equal. O Consider rhombus PQRS (Fig. 13.8), in which PR and SQ Q are the diagonals of the rhombus. Figure 13.8 In a rhombus, the diagonals bisect each other at right angles at the point O. \u2234 \u2220POS = 90\u00b0 and \u2220POQ = 90\u00b0. Area of the rhombus PQRS = Area of \u2206PRS + Area of \u2206PQR = \uf8eb 1 \uf8f6 (PR )(OS ) + \uf8ed\uf8eb\uf8ec 1 \uf8f7\uf8f8\uf8f6 (PR )(OQ ) \uf8ed\uf8ec 2 \uf8f7\uf8f8 2 = \uf8eb 1 \uf8f6 (PR )(OS + OQ ) = \uf8eb 1 \uf8f6 (PR )(SQ ) \uf8ec\uf8ed 2 \uf8f8\uf8f7 \uf8ed\uf8ec 2 \uf8f8\uf8f7 \u2234 Area of rhombus = \uf8eb 1 \uf8f6 (product of the diagonals). \uf8ed\uf8ec 2 \uf8f8\uf8f7","Mensuration 13.5 Area of a Square A square is a rectangle in which all the sides are equal. Area of a rectangle = (Length)(Breadth) In a square, Length = Breadth \u2234 Area of a square = (Side)(Side) = (Side)2 Perimeter of a square = 4(Side) Diagonal of a square = 2 (Side) \u2234 Area of a square = (Diagonal )2 . 2 POLYGON A polygon is a closed rectilinear figure that has three or more sides. The following is a list of different types of polygons: Name Number of Sides Triangle 3 Quadrilateral 4 Pentagon 5 Hexagon 6 Septagon 7 Octogon 8 Nonagon 9 Decagon 10 In a polygon, if all the sides are equal, and all the angles are equal, then it is a regular polygon. Area of a Polygon D In Fig. 13.9, ABCDE is a polygon. AE This polygon may be divided into triangles by joining EC points B and E, E and C. Then the areas of the triangles can Figure 13.9 be found. The sum of the areas of these triangles is the area of the polygon. Area of a Regular Polygon The regular polygon of n sides may be divided into n triangles of equal areas. Consider the regular octagon in Fig. 13.10. The octagon can be divided into 8 equal triangles. Let OP be the perpendicular distance from centre O to side AH. OP is the height of the triangle OHA. Area of triangle OHA = \uf8eb 1 \uf8f6 ( AH )(OP ) \uf8ec\uf8ed 2 \uf8f8\uf8f7","13.6 Chapter 13 Since the octagon consists of 8 equal triangles, the area of the ED O octagon = (8) \uf8eb 1 \uf8f6 ( AH )(OP ). F C \uf8ed\uf8ec 2 \uf8f8\uf8f7 G H PA B But, the perimeter of the octagon = 8(AH). \u2234 The area of a regular polygon = \uf8eb 1 \uf8f6 (Perimeter of the polygon) \uf8ec\uf8ed 2 \uf8f7\uf8f8 (Perpendicular distance from the centre to any side.) Area of an Equilateral Triangle Figure 13.10 A In Fig. 13.11, ABC is an equilateral triangle. Let AB = BC = CA = a Draw perpendicular AD from A to BC. aa AD is the height of the triangle. ADC is a right triangle, w=here DC 2a=and AC a. B C \uf8eb a \uf8f6 2 3 D \uf8ed\uf8ec 2 \uf8f8\uf8f7 2 a AD = (a)2 \u2212 = a. Figure 13.11 \u2234 The height of the triangle 3 a. 2 The area of the triangle = 1 (a) \uf8eb 3 a\uf8f8\uf8f6\uf8f7 \uf8eb 3 a2 \uf8f6 . 2 \uf8ec\uf8ed 2 = \uf8ed\uf8ec 4 \uf8f7\uf8f8 \u2234 The area of an equilateral triangle = 3 (side)2 sq. units. 4 Area of an Isosceles Triangle An isosceles triangle has two equal sides. Generally, the base is considered to be the unequal side. Consider \u2206ABC (Fig. 13.12), in which AB = AC. Let AB = AC = a and BC = b. Draw perpendicular AD from A to D on BC. In an isosceles A aa triangle, AD is the altitude as well as the median. As D is the mid-point of BC, DC = b . 2 ADC is a right triangle, \uf8eb b \uf8f6 2 4a2 \u2212 b2 \uf8ed\uf8ec 2 \uf8f7\uf8f8 2 AD = a2 \u2212 = . The area of triangle ABC = \uf8eb 1 \uf8f6 (BC )( AD ) BD C \uf8ec\uf8ed 2 \uf8f7\uf8f8 b \uf8eb 1\uf8f6 ) \uf8eb\uf8ec 4a2 \u2212 b2 \uf8f6 b Figure 13.12 \uf8ed\uf8ec 2\uf8f7\uf8f8 \uf8ed 2 \uf8f7 4 = (b \uf8f8 = 4a2 \u2212 b2 sq. units.","Mensuration 13.7 Area of Right Triangle A In Fig. 13.13, ABC is a right triangle and right angled at B. Since AB is perpendicular to the base BC of the triangle, AB is the a altitude of the triangle. \u2234 The area of right \u2206ABC = \uf8eb 1 \uf8f6 (b )(a) = ab sq. units. \uf8ec\uf8ed 2 \uf8f8\uf8f7 2 Isosceles Right Triangle Bb C Figure 13.13 A right triangle in which the perpendicular sides are equal is an isosceles right triangle. A In Fig. 13.14, ABC is an isosceles right triangle, where AB = BC a = a units. The area of the triangle=ABC 21=(a)(a) a2 sq. units. 2 From Pythagorean theorem, we have AB2 + BC2 = AC2 Ba C \u21d2 a2 + a2 = AC2 \u21d2 AC = 2a units. Figure 13.14 \u2234 AB : BC : CA = a : a : 2a = 1 : 1 : 2. The hypotenuse is the longest side of a right triangle. The hypotenuse is 2 times of each of the equal sides. Notes\u2002 1.\t In-radius (r) of a triangle is given by the formula, r = \u2206 . s 2.\t Circum-radius (R) of a triangle is given by the formula, R = abc . P 4\u2206 30\u00b0 Where \u2206 is area of the triangle with sides a, b and c, and s is the semi- perimeter. Special Right Triangle with Angles 30\u00ba, 60\u00ba and 90\u00ba\u2002 In the right triangle (Fig. 60\u00b0 90\u00b0 13.15), \u2220PSQ = 90\u00ba. Q S We have to determine the ratio of the sides PQ, QS and PS. Figure 13.15 Construction: Produce QS to the point R, such that SR = SQ (see Fig. 13.16). \u2234 \u2206PSQ \u2245 \u2206PSR (SAS Congruence property) P Let PQ = QR = PR = a. Since PS is also the median to QR, QS = a 2 . 30\u00b0 30\u00b0 In an equilateral triangle, altitude is 3 times its side. 90\u00b0 2 S \u2234 PS = 3 a. 60\u00b0 60\u00b0 2 Q R In triangle PSQ, Figure 13.16","13.8 Chapter 13 SQ : SP : PQ = a : 3 a : a 2 2 = 1 : 3 : 2. \u2234 In a right triangle with angles 30\u00ba, 60\u00ba and 90\u00ba, the ratio of the sides opposite to these angles is 1 : 3 : 2. Example 13.1 From each corner of a square sheet of side 10 cm, a square of side s cm is cut, where s is an integer. The remaining sheet is folded into a cuboid of volume C cubic cm. Which of the following cannot be a value of C? (a) 64\t (b) 72\t\t (c) 48\t\t (d) 30 Solution s cm (10 \u2212 2s) cm s cm s cm s cm (10 \u2212 2s) cm (10 \u2212 2s) cm s cm s cm s cm (10 \u2212 2s) cm s cm The dimensions of cuboid are l = b = (10 \u2212 2s) cm and h = s cm. Volume of cuboid = (10 \u2212 2s) (10 \u2212 2s)s i.e., (10 \u2212 2s)2s cubic cm. 10 \u2212 2s > 0, i.e., s < 5 and s is an integer. \u2234 s = 1, 2, 3 or 4. When s = 1, C = 64 When s = 2, C = 72 When s = 3, C = 48 When s = 4, C = 16 \u2234 Only option (d) cannot be a possible value of C. CIRCLES Circumference of a Circle Since, a circle is a closed arc, its perimeter is known as circumference. The circumference of a circle is \u03c0 (read as pie) times its diameter. \u03c0 is an irrational number. For numerical work, we use 22 the approximate value of \u03c0 as 3.14 or 7 . The circumference of a circle = \u03c0d = 2\u03c0r. The area of a circle = \u03c0r2.","Mensuration 13.9 Area of a Ring A B O\u2022 \u2022 The region enclosed between two concentric circles is known as a ring. Figure 13.17 In Fig. 13.17, the shaded region represents a ring. Let the radius of the smaller circle, OA, be r and the radius of the bigger circle, OB, be R. The area of the ring = (Area of the bigger circle) \u2212 (Area of the smaller circle) = \u03c0R2 \u2212 \u03c0r2 = \u03c0 (R2 \u2212 r2). Sector of a Circle A sector is formed when the endpoints of an arc of a circle are joined with two radii of the circle. In the circle with centre O, AB is an arc of the circle (see Fig. 13.18). O The endpoints of the arc, A and B are joined with two radii OA and \u03b8B OB of the circle. AOB is a sector of the circle. The arc makes an angle \u03b8 at the centre of the circle. This is also known as central angle or A sector angle. Figure 13.18 The length of an arc of the sector of a circle is proportional to the central angle of the sector of the circle. \u2234 The length of an arc of the sector of a circle (l) = \u03b8 \u00d7 2\u03c0 r, where r is the radius of the circle. 360\u00b0 The perimeter of the sector of the circle = l + 2r. The area of the sector of a circle is proportional to the central angle of the sector of the circle. \u2234 The area of the sector of a circle = \u03b8 (\u03c0 r 2 ) or lr . 360\u00b0 2 Example 13.2 A circle is placed in a rectangle such that it touches both the lengths of the rectangle. If the radius of the circle is one-fourth of the length of the rectangle, then find the ratio of the area of the region in the rectangle that is not covered by the circle to the area of the circle \uf8eb take \u03c0 = 22 \uf8f6 . \uf8ed\uf8ec 7 \uf8f8\uf8f7 (a) 21 : 17\t (b) 14 : 19\t (c) 17 : 11\t (d) 15 : 11 Solution Let l be the length of the rectangle. Radius (r) of the circle = \uf8eb 1 \uf8f6 (the length of the rectangle) \uf8ec\uf8ed 4 \uf8f7\uf8f8 = l 4 therefore, breadth (b) = 2 \u00d7 radius of the circle = 2 \u00d7 \uf8eb l \uf8f6 = \uf8eb l \uf8f6 \uf8ec\uf8ed 4 \uf8f8\uf8f7 \uf8ec\uf8ed 2 \uf8f8\uf8f7","13.10 Chapter 13 required ratio = Area of the region in the rectangle that is not covered by the circle : Area of the circle = (l ) \uf8ec\uf8ed\uf8eb l \uf8f6 \u2212 \u03c0 r 2 : \u03c0 r 2 2 \uf8f8\uf8f7 = \uf8eb l2 \uf8f6 \u2212 \uf8eb 22 \uf8f6\uf8eb l \uf8f62 : \uf8eb 22 \uf8f6\uf8eb l \uf8f62 \uf8ed\uf8ec 2 \uf8f7\uf8f8 \uf8ed\uf8ec 7 \uf8f8\uf8f7 \uf8ec\uf8ed 4 \uf8f7\uf8f8 \uf8ec\uf8ed 7 \uf8f8\uf8f7 \uf8ed\uf8ec 4 \uf8f7\uf8f8 = 17 : 11. Example 13.3 If a regular hexagon is inscribed in a circle of radius 4 cm, then find the area of the polygon in cm2. (a) 6 3 \t (b) 24 3 \t (c) 4 3 \t (d) 48 3 Hint Area of hexagon = 6(Area of equilateral triangle). SOLIDS A plane figure may have one dimension or two dimensions. Triangles and quadrilaterals have two dimensions. For two-dimensional figures, the dimensions are length and breadth or width or height. But many objects such as a brick, a match box, a pencil, a marble, a tank, an ice cream, a cone have a third dimension. Thus, a solid is a three-dimensional object. In general, any object occupying space can be called a solid. Some solids, like prisms, cubes and cuboids have plane or flat surfaces, while some solids, like cone, cylinder have curved surfaces as well as flat surfaces. Spheres have only curved surfaces. Lateral surface area of a solid having a curved surface is referred to as curved surface area. Volume of a Solid The amount of space enclosed by the bounding surface or surfaces of a solid is called the volume of the solid. It is measured in cubic units. Prisms A prism is a solid in which two congruent and parallel polygons form the top and the bottom faces. The lateral faces are parallelograms. The line joining the centres of the two parallel polygons is called the axis of the prism and the length of the axis is referred to as the height of the prism. If two parallel and congruent polygons are regular and, if the axis is perpendicular to the base, then the prism is called a right prism. The lateral surfaces of a right prism are rectangles. Consider two congruent and parallel triangular planes ABC and PQR. If we join the corresponding vertices of both the planes, i.e., A to P, B to Q and C to R, then the resultant solid formed is a triangular prism. A right prism, whose base is a rectangle, is called a cuboid and the one, the base of which is a pentagon, is called a pentagonal prism. If all the faces of a prism are congruent, it is a cube. In case of a cube or a cuboid, any face may be the base of the prism. A prism whose base and top faces are squares, but the lateral faces are rectangular is called a square prism (see Fig. 13.19).","Mensuration 13.11 C Hexagonal prism AB R PQ Triangular prism Square prism Cuboid Cube (Cuboid) Figure 13.19 \u2002Notes\u2002\u2002 The following points hold good for all prisms. 1.\t The number of lateral faces = The number of sides of the base. 2.\t The number of edges of a prism = The number of sides of the base \u00d7 3. 3.\t T\u0007 he sum of the lengths of the edges = 2\u0007 (The perimeter of base) + The number of sides \u00d7 Height. 4.\t \u0007For a prism, whose base is a polygon, Number of vertices + Number of faces = Number of edges + 2. This is known as Euler\u2019s formula, i.e., V + F = E + 2. Lateral Surface Area (LSA) of a Prism LSA = Perimeter of base \u00d7 Height = ph Total Surface area (TSA) of a Prism TSA = LSA + 2(Area of base) Volume of a Prism Volume = Area of base \u00d7 Height = Ah \u2002Note\u2002\u2002 The volume of water flowing in a canal in an hour = The cross-section of the canal \u00d7 The speed of water in an hour. Example 13.4 The base of a right prism is a right-angled triangle. The measure of the base of the right-angled triangle is 3 m and its height 4 m. If the height of the prism is 7 m, then find the following: 1.\t The number of edges of the prism. 2.\t The volume of the prism. 3.\t The total surface area of the prism.","13.12 Chapter 13 Solution 1.\t The number of the edges = The number of sides of the base \u00d7 3 = 3 \u00d7 3 = 9. 2.\t \u0007The volume of the prism = (Area of the base) \u00d7 (Height of the prism) = \uf8eb 1\uf8f6 (3 \u00d7 4) \u00d7 7 \uf8ec\uf8ed 2\uf8f8\uf8f7 = 42 m3. 3.\t TSA = LSA + 2(Area of base) \t = ph + 2(Area of base). \t Where, p = perimeter of the base = sum of lengths of the sides of the triangle. \t As hypotenuse of the triangle = 32 + 42 = 25 = 5 m. \t \u2234 Perimeter of the base = 3 + 4 + 5 = 12 m. \t \u21d2 LSA = ph = 12 \u00d7 7 = 84 m2. \t TSA = LSA + 2 (Area of base) \t = 84 + 2 \uf8eb 1 \u00d7 3 \u00d7 4\uf8f7\uf8f6\uf8f8 \uf8ec\uf8ed 2 \t = 84 + 12 = 96 m2. Cubes and Cuboids Cuboid In a right prism, if the base is a rectangle, then it is called a cuboid (see Fig. 13.20). A match box, a brick, a room, etc., are in the shape of a cuboid. The three dimensions of the cuboid, i.e., its length (l), breadth (b) and height (h) are, generally, denoted by l \u00d7 b \u00d7 h and read as l by b by h. 1.\t T\u0007 he lateral surface area of a cuboid = ph = 2(l + b)h sq. units, where p is the perimeter of the base. 2.\t \u0007The total surface area of a cuboid = LSA + 2(Base area) = 2(l + b)h Figure 13.20 A Cuboid + 2lb = 2(lb + bh + lh) sq. units. 3.\t T\u0007 he volume of a cuboid = Ah = (lb)h = lbh cubic units, where A is the area of the base. 4.\t Diagonal of a cuboid = l2 + b2 + h2 units. \u2002Note\u2002\u2002 If a closed wooden box of thickness t has inner dimensions of l, b and h, then the outer length = l + 2t, the outer breadth = b + 2t and the outer height = h + 2t. Cube In a cuboid, if all the dimensions, i.e., its length, breadth and height are equal, then it is called a cube. All the edges of a cube are equal in length (see Fig. 13.21). Thus, the size of a cube is completely determined by its edge. If the edge of a cube is a units, then 1.\t The lateral surface area of the cube = 4a2. 2.\t The total surface area of the cube = LSA + 2(Area of base) = 4a2 + 2a2 = 6a2.","Mensuration 13.13 3.\t The volume of the cube = a3. 4.\t The diagonal of the cube = a2 + a2 + a2 = 3 a. \u2002Note\u2002\u2002 If the inner edge of a cubical wooden box of thickness t is a units, then the outer edge of the cube is given by a + 2t units. Figure 13.21 A Cube Example 13.5 The dimensions l \u00d7 b \u00d7 h of a room are 12 m \u00d7 7 m \u00d7 5 m. Find: (a)\t The diagonal of the room. (b)\t The cost of flooring at the rate of `2 per m2. (c)\t T\u0007 he cost of white-washing of the inside of the room excluding the floor at the rate of `3 per m2. Solution (a)\t The diagonal of the room = l2 + b2 + h2 = 122 + 72 + 52 144 + 49 + 25 = 218 m. (b)\t To find the cost of flooring, we should know the area of the base. \t Base area = lb = 12 \u00d7 7 = 84 m2. \t \u2234 Cost of flooring = 84 \u00d7 2 = `168. (c)\t Total area to be white-washed \t = LSA + area of roof = 2(l + b)h + lb \t = 2(12 + 7)5 + 12 \u00d7 7 = 2(19)(5) + 84 = 190 + 84 = 274 m2. \t \u2234 Cost of white-washing = 274 \u00d7 3 = `822. Example 13.6 A box is in the form of a cube. Its outer edge is 5 m long. Find: (a)\t The total length of the edges. (b)\t Cost of painting the outside of the box, on all the surfaces, at the rate of `5 per m2. Solution (a)\t Length of edges = Number of edges \u00d7 3 \u00d7 Length of each edge = 4 \u00d7 3 \u00d7 5 = 60 m. (b)\t To find the cost of painting the box, we need to find the total surface area. \t TSA = 6a2 = 6 \u00d7 52 = 6 \u00d7 25 = 150 m2. \t \u2234 Cost of painting = 150 \u00d7 5 = `750. Right Circular Cylinder A cylinder consists of two congruent and parallel circular regions which are connected by a curved surface. Each of the circular regions is called a base of the cylinder. A road roller, water pipe, power cables, round pillars are some of the cylinder-shaped objects.","13.14 Chapter 13 In Fig. 13.22, a right circular cylinder is shown. Let A be the centre r \u2022A of the top face and A1 be the centre of the base. The line joining the B\u2022 centres (i.e., AA1) is called the axis of the cylinder. The length AA1 is called the height of the cylinder. The radius r of the base of the h cylinder and the height h, completely describe the cylinder. \u2022 Lateral (curved) surface area = Perimeter of base \u00d7 Height = 2\u03c0rh A1 \u2022 The total surface area = LSA + 2(Base area) B1 = 2\u03c0rh + 2(\u03c0r2) = 2\u03c0r(h + r) Figure 13.22 Volume = Area of base \u00d7 Height = \u03c0r2h. \u2002 Note\u2002 \u2002 If a plastic pipe of length l is such that its outer radius is R and the inner radius is r, then the volume of the plastic content of the pipe is l\u03c0 (R2 - r2) cubic units. Example 13.7 A closed cylindrical container, the radius of which is 7 cm and height 10 cm is to be made out of a metal sheet. Find: (a)\t The area of metal sheet required. (b)\t The volume of the cylinder made. (c)\t \u0007The cost of painting the lateral surface of the cylinder at the rate of `4 per cm2. (Assume that the thickness of the metal sheet is negligible). Solution (a)\t The area of the metal sheet required = The total surface area of the cylinder \t = 2\u03c0r(r + h) \t = 2 \u00d7 22 \u00d7 7(7 + 10) = 44(17) = 748 cm 2 . 7 (b)\t Volume = \u03c0r2h \t = 22 \u00d7 7 \u00d7 7 \u00d7 10 = 22 \u00d7 70 = 1540 cm 3 . 7 (c)\t \u0007To find the cost of painting the lateral surface, we need to find the curved (lateral) surface area. \t \u2234LSA = 2\u03c0 rh = 2 \u00d7 22 \u00d7 7 \u00d7 10 = 22 \u00d7 70 = 440 cm 2 . 7 \t Cost of painting = 440 \u00d7 4 = `1760. Pyramid A pyramid is a solid obtained by joining the vertices of a polygon to a point in the space by straight lines. The polygon is the base of the pyramid. Triangles are formed with each side of the base and the point in the space. The fixed point in space where all the triangles (i.e., lateral faces) meet is called the vertex of the pyramid.","Mensuration 13.15 In Fig. 13.23, base ABCD is a quadrilateral. All the vertices of the base O are joined to a fixed point O in space by straight lines. The resultant solid obtained is a pyramid with a quadrilateral as the base. The straight line joining the vertex of the pyramid and the centre of the base is called the axis of the pyramid. If the axis is not perpendicular to the base, it is an oblique D C pyramid. Right Pyramid AB Figure 13.23 If the line joining the vertex of the pyramid and the centre of the base is perpendicular to the base, then the pyramid is called a right pyramid (see Fig. 13.24). The length of the line segment joining the vertex of the pyramid and the centre of the base of a right pyramid is called the height of the pyramid. It is represented by h. The perpendicular distance between the vertex of the pyramid and the mid-point of any of the sides of the base (i.e., regular polygon) of a right pyramid is called its slant height. It is represented by l. For a right pyramid with perimeter of base = p, height = h and slant height = l: 1.\t Lateral surface area = 1 ( perimeter of base) \u00d7 (slant height) = 1 pl 2 2 2.\t Total surface area = lateral surface area + area of base. 3.\t Volume of pyramid = 1 \u00d7 area of base \u00d7 height. 3 O O h C D FE A MD BC AB Hexagonal pyramid Square pyramid Figure 13.24 Example 13.8 A regular hexagonal pyramid is 20 m high. Side of the base is 5 m. Find the volume and the slant height of the pyramid. Solution Given, h = 20 m Side of base = a = 5 m","13.16 Chapter 13 \u2234Area of base = 3 a2 6 63 52 O 4 4 \u00d7 \u00d7 = \u00d7 = 75 3 m2. 2 r Volume = 1 Ah; where A = area of the base, and h = height. FE 3 A D 1 33 G = 3 \u00d7 2 (25) \u00d7 20 = 3 \u00d7 250 = 250 3 m3. H a B To find slant height, refer to the Fig. 13.25. In this figure, O is C the vertex of the pyramid and G is the centre of the hexagonal base. OG is the axis of the pyramid. OH is the slant height of the Figure 13.25 pyramid. \u2206OGH is a right-angled triangle. \u2234 OH\u20092 = GH\u20092 + OG\u20092. GH = Altitude of \u2206AGB = 3 a = 3 \u00d75 = 53 m. 2 2 2 \u2234OH 2 = \uf8eb 53 \uf8f62 + (20)2 = 25 \u00d7 3 + 400 \uf8ec\uf8ed\uf8ec 2 \uf8f8\uf8f7\uf8f7 4 = 75 + 1600 = 1675 4 4 \u21d2 OH = 1675 m. 2 \u2234Slant height = 5 67 m. 2 Circular Cone A O\u2022 A circular cone is a solid pointed figure with a circular base as shown in Fig. 13.26. A circular cone is a kind of pyramid whose base is a circle. A cone has one vertex, one plane surface (i.e., the base) and a curved surface. The line joining the vertex to the centre of base (i.e., AO) is called the axis of the cone. If the axis is not perpendicular to the base of a cone, then it is an oblique cone. An ice cream cone and a conical tent are some of the examples of conical objects. Right Circular Cone Figure 13.26 In a circular cone, if the line joining the vertex and the centre of the base of the cone is perpendicular to the base, then it is a right circular cone. In other words, if the axis of the cone is perpendicular to the base of the cone, then it is a right circular cone. The length of the line segment AO is called the height of the cone. In this chapter, we deal with problems on right circular cones.","Mensuration 13.17 A h l O r O \u2022 A B B Figure 13.27 Figure 13.28 We can also define a right circular cone as a solid obtained by the revolution of a right-angled triangle about one of its two perpendicular sides. If we consider any point B on the periphery of the base of the cone, then the line joining B and the vertex A is called the slant height of the cone and is denoted by l. Since AO is perpendicular to OB, DAOB is right-angled (see Fig. 13.27). \u2234l = r2 + h2 . Earlier, we have studied about sector. We may recall that sector is a figure bounded by an arc of a circle with its two (see Fig. 13.28). Now, consider the sector AOB. If we roll the sector up and bring (join) together the radii OA and OB such that they coincide, then the figure formed will be a cone. The radius of the circle becomes the slant height of the cone and the length of the arc of the sector becomes the perimeter of the base of the cone. For a cone of radius r, height h and slant height l: 1.\t Curved surface area of a cone = \u03c0rl sq. units. 2.\t Total surface area of a cone \t = Curved surface area + Area of base = \u03c0rl + \u03c0r2 = \u03c0r(r + l\u2009) sq. units. 3.\t Volume of a cone = 1 \u03c0r2h cubic units. 3 Cone Frustum (or a Conical Bucket) If a right circular cone is cut by a plane perpendicular to its axis (i.e., a plane parallel to the base), then the solid portion containing the base of the cone is called the frustum of the cone. \u2022R From Fig. 13.29, we observe that a frustum is in the shape of a bucket. Let radius of the upper base be R, radius of the lower base be h r, height of the frustum be h and slant height of the frustum be l. \u2022r H 1.\t Curved surface area of a frustum = \u03c0l(R + r) sq. units. 2.\t Total surface area of a frustum Figure 13.29 \t \u0007= Curved surface area + Area of upper base + Area of lower base = \u03c0l(R + r) + \u03c0r2 + \u03c0R2 sq. units.","13.18 Chapter 13 3.\t Volume of a frustum = 1 \u03c0rh(R2 + Rr + r2) cubic units. 3 4.\t Slant height (l\u2009) of a frustum = (R \u2212 r )2 + h2 units. Example 13.9 A Buffoon\u2019s cap is in the form of a cone of radius 7 cm and height 24 cm. Find the area of the paper required to make the cap. Solution Area of the paper required = Curved surface area of the cap (or cone) = \u03c0rl. Now, l = r 2 + h2 = 72 + 242 = 49 + 576 = 625 = 25 cm. \u21d2 Curved surface area = 22 \u00d7 7 \u00d7 25 = 550 cm 2 . 7 \u2234 Area of the paper required = 550 cm2. Example 13.10 The inner diameter of an ice cream cone is 7 cm and its height is 12 cm. Find the volume of ice cream that the cone can contain. Solution 1 3 Volume of ice cream (cone) = \u03c0r2h. = 1 \u00d7 22 \u00d7 \uf8eb 7\uf8f6 2 \u00d7 12 = 22 \u00d7 7 = 154 cm 3 . 3 7 \uf8ec\uf8ed 2\uf8f7\uf8f8 Example 13.11 The diameters of the top and bottom portions of a milk can are 56 cm and 14 cm, respectively. The height of the can is 72 cm. Find the: (a)\t Area of metal sheet required to make the can. (without lid) (b)\t Volume of milk which the can can hold. Solution The milk can is in the shape of a cone frustum with R = 28 cm, r = 7 cm and h = 72 cm. (a)\t Area of metal sheet required \t = Curved surface area + Area of bottom base = \u03c0l(R + r) + \u03c0r2. l = (R \u2212 r )2 + h2 = (28 \u2212 7)2 + 722 \t = 212 + 722 = 9(72 + 24)2 = 3 49 + 576 = 3 \u00d7 625 \t = 3 \u00d7 25 = 75 cm. 22 22 7 7 \t \u2234 Area of metal sheet = \u00d7 75(28 + 7) + \u00d7 72","Mensuration 13.19 \t = 22 \u00d7 75 \u00d7 5 + 22 \u00d7 7 = 22(375 + 7) \t = 22(382) = 8404 cm2. (b)\t Amount of milk which the container can hold = 1 \u03c0h(R2 + Rr + r2 ) 3 = 1 \u00d7 22 \u00d7 72 ( 282 + 7 \u00d7 28 + 72 ) 3 7 \t = 22 \u00d7 24 \u00d7(112 + 28 + 7) \t = 22 \u00d7 24 \u00d7 (147) = 77616 cm3. Example 13.12 From a circular canvas of diameter 56 m, a sector of 270\u00b0 was cut out and a conical tent was formed by joining the straight ends of this piece. Find the radius and the height of the tent. Solution As shown in Fig. 13.30, when the free ends of the canvas cut-out are joined to form a cone, the radius of the sector becomes the slant height (l\u2009) of the cone. \u2234l = 56 = 28 m. 2 The length of the arc of the sector becomes the circumference of the base of the cone. Let the radius of the base of the cone = r. \u21d2 2\u03c0 r = 22 \u00d7 56 \u00d7 \uf8eb 270 \uf8f6 7 2 \uf8ed\uf8ec 360 \uf8f8\uf8f7 \u21d2 2\u03c0 r = 22 \u00d7 28 \u00d7 3 \u21d2 r = 21 m. 270\u00b0 7 4 \u2234Height, h = l2 \u2212 r 2 28 m = 282 \u2212 212 = 72(42 \u2212 3)2 Figure 13.30 = 7 16 \u2212 9 = 7 7 m \u2234h = 7 7 m and r = 21 m. Torus (Solid Ring) A fully-inflated bicycle tube, tennikoit ring, and a life belt are some examples of a torus. A torus is a three-dimensional figure formed by the revolution of a circle about an axis lying in its plane, but not intersecting the circle. If r is the radius of a circle that rotates and a is the distance of the centre of rotating circle from the axis (see Figs. 13.31 and 13.32), then: Surface area of torus = 4\u03c02ra sq. units and Volume of torus = 2\u03c02r2a cubic units.","13.20 Chapter 13 a \u2022a 2r \u2022 2r Figure 13.31 Figure 13.32 Example 13.13 The radius of cross-section of a car tube is 7 cm. The outer radius of the tube is 16 cm. Find the surface area and the volume of the tube. Solution Here, r = 7 cm, hence a = 16 - r = 16 - 7 = 9 cm. Surface area = 4\u03c0 2ra = 4\u00d7 22 \u00d7 22 \u00d77\u00d79 = 17424 cm 2 . 7 7 7 Volume = 2\u03c02r2a = 2 \u00d7 22 \u00d7 22 \u00d7 7 \u00d7 7 \u00d7 9 = 8712 cm 3 . 7 7 Sphere Sphere is a set of points in the space which are equidistant from a fixed point. The fixed point is called the centre of the sphere, and the distance is called the radius of the sphere. A lemon, a football, the moon, globe, small lead balls used in cycle bearings are some example of spherical- shaped objects. A line joining any two points on the surface of sphere and passing through the centre of the sphere is called its diameter. The size of sphere can be completely determined by knowing its radius or diameter. Solid Sphere A solid sphere is the region in the space bounded by a sphere. The centre of a sphere is also a part of solid sphere, whereas the centre is not a part of hollow sphere. Marbles, lead shots, etc., are examples of solid spheres, while a tennis ball is a hollow sphere. O r \u2022 Hemisphere Figure 13.33 Hemisphere If a sphere is cut into two halves by a plane passing through the centre of sphere, then each of the halves is called a hemisphere (see Fig, 13.33).","Mensuration 13.21 Spherical Shell The region bounded in the space between two concentric solid spheres is called a spherical shell. The thickness of the shell is given by the difference in radii of the two spheres. Hemispherical Shell\u2002 A hemispherical shell is shown in Fig. 13.34. Formulas to Memorize Sphere 1.\t Surface area of a sphere = 4\u03c0r2 2.\t Volume of a sphere = 4 \u03c0 r 3 Figure 13.34 Hemispherical shell 3 Hemisphere 1.\t Curved surface area of a hemisphere = 2\u03c0r2 2.\t Total surface area of a hemisphere = 3\u03c0r2 3.\t Volume of a hemisphere = 2 \u03c0 r 3 3 Spherical Shell 1.\t Thickness = R - r, where R = outer radius and r = inner radius. 2.\t Volume = 4 \u03c0 R3 \u2212 4 \u03c0r3 3 3 3.\t Total surface area of a hemispherical shell \t =\u0007 (\u0007curved surface area of outer hemisphere + curved surface area of inner hemisphere + area of ring) \t = 3\u03c0R2 + \u03c0r2. Example 13.14 The cost of painting a solid sphere at the rate of 50 paise per square metre is `1232. Find the volume of steel required to make the sphere. Solution Cost of painting = Surface area \u00d7 Rate of painting \u2234Surface area = Cost of painting Rate of painting \u21d2 1232 = 2464 m2 \u21d2 4\u03c0 r 2 = 2464 0.5 \u21d2 r2 = 2464 = 616 = 616 \u00d7 7 = 28 \u00d7 7 \u21d2 r = 7\u00d72 = 14 m. 4\u03c0 \uf8eb 22 \uf8f6 22 \uf8ed\uf8ec 7 \uf8f7\uf8f8 \u2234 Volume of steel required = 4 \u03c0 r 3 = 4 \u00d7 22 \u00d7 14 \u00d7 14 \u00d7 14 = 34496 m3. 3 3 7 3","13.22 Chapter 13 Example 13.15 A hollow hemispherical bowl, as shown in Fig. 13.35, of thickness 1 cm has an inner radius of 6 cm. Find the volume of metal required to make the bowl. Solution Inner radius, r = 6 cm Thickness, t = 1 cm \u2234 Outer radius, R = r + t = 6 + 1 = 7 cm. \u2234 Volume of steel required = 2 \u03c0 R3 \u2212 2 \u03c0 r 3 3 3 = 44 (73 \u2212 63 ) Figure 13.35 21 = 44 (343 \u2212 216) = 44 \u00d7 127 = 5588 cm 3 . 21 21 21 Example 13.16 A thin hollow hemispherical sailing vessel is made of metal surmounted by a conical canvas tent. The radius of the hemisphere is 14 m and the total height of the vessel (including the height of tent) is 28 m. Find the area of the metal sheet and the canvas required. Solution The vessel (with the conical tent) is shown in Fig. 13.36. Total height, H = 28 m Radius of hemisphere = r = 14 m. \u2234 Height of conical tent = h = H - r = 28 - 14 h = 14 m. We can observe that radius of the base of the \u2022 H = 28 m cone Figure 13.36 r = 14 m = Radius of the hemisphere = 14 m. \u2234 Area of canvas required = \u03c0rl. = 22 \u00d7 14 \u00d7 142 + 142 = 44 \u00d7 14 2 = 616 2 m2. 7 Area of metal sheet required = Surface area of hemisphere = 2\u03c0 r 2 = 2\u00d7 22 \u00d7 14 \u00d7 14 = 1232 m2. 7 Polyhedrons A solid bounded by plane polygons is a polyhedron. The bounding polygons are known as the faces and the intersection of the faces are edges. The points where three or more edges intersect are called the vertices. A polyhedron having four faces is a tetrahedron, one having six faces is a hexahedron and one having eight faces is an octahedron.","Mensuration 13.23 If all the faces of a polyhedron are regular polygons which are congruent, then it is a regular polyhedron. However, further discussion on regular polyhedrons can be made in higher classes. Regular Tetrahedron A regular tetrahedron is a tetrahedron having all of its faces as equilateral triangles (see Fig. 13.37). The line joining the vertex of the tetrahedron and the centre of the base is the vertical height of a regular tetrahedron. The line joining the mid- P point of the side of the base and the vertex of a regular tetrahedron is its slant height. The lateral surface area of a regular tetrahedron C = 1 \u00d7 Perimeter of base \u00d7 Slant height A 2 1 3 = 2 \u00d7 3a \u00d7 2 a = 33 a2 (where a is the edge). B 4 Figure 13.37 Regular Tetrahedron The total surface area of a regular tetrahedron = LSA + Area of base = 33 a2 + 3 a2 4 4 = 3 a2. Volume of a regular tetrahedron = 1 \u00d7 Area of base \u00d7 Vertical height 3 = 1 \u00d7 3 a2 \u00d7 2 a = a3 . 3 4 3 62 Regular Octahedron A regular octahedron is an octahedron whose faces are P all congruent equilateral triangles. AD Figure 13.38 is an example of regular octahedron, BC Q where ABCD is a square; ABP, BCP, CDP, DAP, ABQ, Figure 13.38 Regular Octahedron BCQ, CDQ and DAQ are eight equilateral triangles which are congruent. If each side of an equilateral triangle is a units, then surface area of the octahedron = 8\u00d7 3 a2 = 2 3 a2 sq. units. 4 Volume of a regular octahedron = 2 \uf8eb 1 \u00d7 a2 \u00d7 a\uf8f6 = 2 a3 cu units. \uf8ec\uf8ed 3 2 \uf8f7\uf8f8 3","13.24 Chapter 13 TEST YOUR CONCEPTS Very Short Answer Type Questions \t1.\t The diagonal of a square is 10 cm. What is its area? \t17.\t What is the volume of a pyramid whose base area is (2 + 3 ) cm2 and its height being (2 \u2212 3 ) cm? \t2.\t If the circumference of a circle is numerically equal to its area, then what is the radius of the circle? \t18.\t If the edge of a cube increases by 20%, then find the percentage increase in its volume. \t3.\t A solid object has 8 vertices and 12 edges. It has _____ faces. 1\t 9.\t If a sphere, cone, cuboid, pyramid and cube have the same surface area, then which of these solid \t4.\t The number of vertices of a pyramid, whose base figure have the maximum volume? is a pentagon, is _____. PRACTICE QUESTIONS \t5.\t A tetrahedron has _____ edges. 2\t 0.\t The volumes of two cones are in the ratio of their respective radii, if they have the same height. \t6.\t What is the number of edges of an octahedron? [True\/False] \t7.\t If p + q and p - q are the sides of a rectangle, then 2\t 1.\t The radii of two spheres are 2 cm and 3 cm respec- its diagonal is _____ units. tively. The ratio of their surface areas is _______. \t8.\t Area of an isosceles triangle, one of whose equal \t22.\t What is the volume of a hemispherical shell with sides is 5 units and base 6 units is _____. outer radius of x units and thickness of y units? \t9.\t A triangle has a perimeter of 9 cm. How many \t23.\t If the volume of a solid hemisphere and its total combinations of its sides exist, if their lengths in surface area are numerically equal, then find its cm are integers? radius. 1\t 0.\t Among an equilateral triangle, an isosceles triangle 2\t 4.\t The perimeter of the base of a pyramid is (2x + 2y) and a scalene triangle, _____ triangle has the max- cm and its slant height is (x \u2212 y) cm. What is the imum area if the perimeter of each triangle is the lateral surface area of the pyramid? (x > y) same. \t11.\t Ratio of the area of a triangle to the product 2\t 5.\t The sides of a triangle are 5 cm, 7 cm and 8 cm. of its sides is _____ times the reciprocal of its Find its area. circum-radius. 2\t 6.\t What is the volume of the prism whose base is a 1\t 2.\t Area of a quadrilateral, which has one of its diago- hexagon of side 6 cm and height 12 3 cm? nals as 10 cm and the lengths of the offsets drawn to its 3 cm and 5 cm, is _____. 2\t 7.\t If the dimensions of a cuboid are 5 cm \u00d7 4 cm \u00d7 3 cm, then find the maximum volume of the cube 1\t 3.\t By joining the mid-points of the adjacent sides of a that can be carved out of it. quadrilateral of area 26 cm2 a _____ is formed and its area, is _____. 2\t 8.\t Find the lateral surface area and the length of the diagonal of a cube of side 8 cm. \t14.\t The longest needle that can be placed in a cylinder of volume \u03c0r2h, is _____. \t29.\t Find the number of soaps of size 2 cm \u00d7 3 cm \u00d7 5 cm, that can be arranged in a cuboidal box 1\t 5.\t The base of a pyramid is an equilateral triangle of dimensions 6 cm \u00d7 3 cm \u00d7 15 cm. with side a and the height of the pyramid is h. Then the volume of the pyramid is _____. 3\t 0.\t What is the area of a ground that can be levelled by a cylindrical roller of radius 3.5 m and length 4 m 1\t 6.\t The heights of two cylinders are equal and their by making 10 rounds? radii are in the ratio of 3 : 2. The ratio of their curved surface areas is _____.","Mensuration 13.25 Short Answer Type Questions \t31.\t In the given figure, AB = 6.4 cm, CF = 2.6 cm, 3\t 7.\t A square plate of side 28 cm is made into a cylinder AD = 3.2 cm. Find the area of the triangle ABC by joining two edges of it. Find the base area of the and length of the side BC. cylinder formed. A 3\t 8.\t From a cylindrical wooden log of length 30 cm and base radius 7 2 cm, maximum cuboid of F square base is made. Find the volume of wood wasted. BD C 3\t 9.\t Find the number of cubes of side 2 m to be dropped \t32.\t From the given figure, find the length of the major in a cylindrical vessel of radius 14 m to increase the arc AB. water level by 5 m. 7 cm O \t40.\t If eight cubes are stacked to form a big cube, then A 90\u00b0 the percentage decrease in the total surface area is _______. B 4\t 1.\t Three cubes, each of side 3.2 cm, are joined end- \t33.\t Find the area of a circle, if the area of the isosceles to-end. Find the total surface area of the resulting PRACTICE QUESTIONS right-angle triangle inscribed in it is 18 sq. cm. cuboid. \t34.\t The perimeter of a rectangle is a two digits number 4\t 2.\t The base of a right pyramid is an equilateral whose units digit and tens digit represent the triangle, each side of which is 6 3 cm long and length and breadth of the rectangle respectively. height is 4 cm. Find the total surface area of the Find its area. pyramid in cm2. 3\t 5.\t Three equilateral triangles are cut from an 4\t 3.\t What is the volume of a square pyramid whose equilateral triangle of area 36 cm2, such that vertical height is 10 cm and the length of the side a regular hexagon is left. Find the area of the of the base is 6 cm? hexagon. 4\t 4.\t The area of the base of a pyramid, which is an \t36.\t If a + b, a - b and 2 ab are the sides of a cuboid, then find the longest stick that can be placed equilateral triangle, is 16 3 cm2, and its slant in it. height is 3 cm. What is its lateral surface area? 4\t 5.\t The base radius of a conical tent is 120 cm and its slant height is 750 cm. Find the area of the canvas required to make 10 such tents (in m2). (Take \u03c0 = 3.14). Essay Type Questions \t46.\t An open metallic conical tank is 6 m deep and its 4\t 8.\t The radius of a cross-section of a tube of a truck is circular top has a diameter of 16 m. Find the cost 28 cm. The outer radius of the tube itself is 63 cm. of tin plating in its inner surface at the rate of `0.8 What is the surface area of the tube? per 100 cm2. (Take \u03c0 = 3.14). \t49.\t If the thickness of a hemispherical bowl is 12 cm \t47.\t A drum in the shape of a frustum of a cone with and its outer diameter is 10.24 m, find the inner radii 20 ft and 14 ft and height 5 ft is full of water. surface area of the hemisphere. (Take \u03c0 = 3.14) The drum is emptied into a cuboidal tank of base 73 ft \u00d7 44 ft. Find the rise in the height of the \t50.\t The internal surface area of a hollow hemisphere water level in the tank. is 77 cm2 and its external surface area is 308 cm2. What is the thickness of the hollow hemisphere?","13.26 Chapter 13 CONCEPT APPLICATION Level 1 \t1.\t The perimeter of a right triangle is 72 cm and its \t9.\t The radius of the base and the slant height of a area is 216 cm2. Find the sum of the lengths of its cone is 5 cm and 13 cm respectively. Find the vol- ume of the cone. perpendicular sides. (in cm) \t\t(a) 36\t\t (b) 32 \t\t(a) 88\u03c0 cm3\t\t (b) 100\u03c0 cm3 \t\t(c) 42\t\t (d) 50 \t\t(c) 92\u03c0 cm3\t\t (d) 106\u03c0 cm3 \t2.\t Find the area of the regular pentagon of side 6 cm \t10.\t If the ratio of the volumes of the spheres is 8 : 27, and height 8 cm. (in cm2) then the ratio of their surface areas is ______. \t\t(a) 40\t\t (b) 60 \t\t(a) 4 : 9\t\t (b) 9 : 4 \t\t(c) 80 3 \t\t (d) 120 \t\t(c) 4 : 3\t\t (d) 2 : 9 \t3.\t Find the perimeter of a sector of a circle if the angle 1\t 1.\t The volume of a hemisphere is 18\u03c0 cm3. What is and radius of it are 30\u00b0 and 10.5 cm respectively. the total surface area of the hemisphere? \t\t(a) 26.5 cm\t\t (b) 21.5 cm \t\t(a) 18\u03c0 cm2\t\t (b) 27\u03c0 cm2 \t\t(c) 21\u03c0 cm2\t\t (d) 24\u03c0 cm2 \t\t(c) 23 cm\t\t (d) 8 cm \t4.\t The sides of a pentagonal prism are 10 cm, 12 cm, \t12.\t Find the volume of a regular octahedron of each 15 cm, 8 cm and 6 cm. Its height is 14 cm. Find edge 2 3 cm. the total length of its edges. \t\t(a) 4 3 cm3 \t(b) 8 3 cm3 \t\t(a) 172 cm\t\t (b) 162 cm \t\t(c) 4 6 cm3 \t(d) 8 6 cm3 \t\t(c) 182 cm\t\t (d) 152 cm PRACTICE QUESTIONS \t5.\t Three cubes of sides 3 cm, 4 cm and 5 cm, respec- 1\t 3.\t In a polyhedron, if the number of faces is 4 and the tively, are melted and formed into a larger cube. number of edges is 6, then the number of vertices What is the side of the cube formed? of that polyhedron is ______. \t\t(a) 7 cm\t\t (b) 6 cm \t\t(a) 1\t\t (b) 2 \t\t(c) 5 cm\t\t (d) 4 cm \t\t(c) 3\t\t (d) 4 \t6.\t The sum of the length, breadth and the height of 1\t 4.\t Fifteen identical spheres are made by melting a a cuboid is 20 cm and the length of its diagonal is solid cylinder of 10 cm radius and 5.4 cm height. 12 cm. Find the total surface area of the cuboid. Find the diameter of each sphere. \t\t(a) 156 cm2\t\t (b) 169 cm2 \t\t(a) 6 cm\t\t (b) 3 cm \t\t(c) 256 cm2\t\t (d) 269 cm2 \t\t(c) 2 cm\t\t (d) 4 cm \t7.\t The radius of the base of a cone is 7 cm and its \t15.\t Find the number of soaps of size 2.1 cm \u00d7 3.7 cm slant height is 25 cm. The volume of the cone \u00d7 2.5 cm that can be put in a cuboidal box of size is ______. 6.3 cm \u00d7 7.4 cm \u00d7 5 cm. \t\t(a) 3696 cm3\t (b) 1232 cm3 \t\t(a) 14\t\t (b) 12 \t\t(c) 2464 cm3\t (d) 1864 cm3 \t\t(c) 13\t\t (d) 11 \t8.\t The side of a square is equal to the diagonal of a \t16.\t If the radius of a sphere is increased by 25%, then cube. The square has an area of 1728 m2. Calculate the percentage increase in its volume is ______. (approximately) the side of the cube. \t\t(a) 12 m\t\t (b) 24 m \t\t(a) 90%\t\t (b) 95% \t\t(c) 27 m\t\t (d) 36 m \t\t(c) 83%\t\t (d) 78%","Mensuration 13.27 \t17.\t Find the volume of a hollow sphere of outer radius 2\t 4.\t ABCDEF is a regular hexagon. If AB is 6 cm long, 9 cm and inner radius 6 cm. then what is the area of triangle ABD? \t\t(a) 342\u03c0 cm3\t (b) 684\u03c0 cm3 \t\t(a) 8 3 cm2 \t\t (b) 12 3 cm2 \t\t(c) 36\u03c0 cm3\t\t (d) 128\u03c0 cm3 \t\t(c) 15 3 cm2 \t\t (d) 18 3 cm2 \t18.\t A hollow hemispherical bowl of thickness 1 cm 2\t 5.\t The area and the radius of a sector are 6.93 sq. cm has an inner radius of 4.5 cm. Find the curved sur- and 4.2 cm respectively. Find the length of the arc face area of the bowl. of the sector. \t\t(a) 106\u03c0 cm2\t (b) 108\u03c0 cm2 \t\t(a) 1.72 cm\t\t (b) 0.86 cm \t\t(c) 101\u03c0 cm2\t (d) 102\u03c0 cm2 \t\t(c) 3.3 cm\t\t (d) 6.6 cm \t19.\t If the length of each edge of a tetrahedron is 18 \t26.\t The perimeter of the base of a right prism, whose cm, then the volume of the tetrahedron is ______. base is an equilateral triangle, is 24 cm. If its total \t\t(a) 482 2 cm3 \t(b) 480 2 cm3 surface area is (288 + 32 3 ) cm2, then its height \t\t(c) 484 2 cm3 \t(d) 486 2 cm3 is ______. \t20.\t Find the total surface area of a regular octahedron, \t\t(a) 10 cm\t\t (b) 12 cm each edge of which is 10 cm. \t\t(c) 14 m\t\t (d) 14 cm \t\t(a) 100 3 cm2 \t(b) 200 3 cm2 2\t 7.\t A large sphere of radius 3.5 cm is carved from a \t\t(c) 300 3 cm2 \t(d) 400 3 cm2 cubical solid. Find the difference between their surface areas. \t21.\t The number of edges in a pyramid whose base has \t\t(a) 224 cm2\t\t (b) 140 cm2 20 edges is ______. \t\t(c) 176 cm2\t\t (d) 80.5 cm2 \t\t(a) 10\t\t (b) 20 \t\t(c) 30\t\t (d) 40 \t28.\t If one of the edges of a regular octahedron is 4 3 cm, then find its height. \t22.\t A copper cable, 32 cm long, having a diameter 6 cm, is melted to form a sphere. Find the radius of \t\t(a) 2 3 cm \t (b) 4 6 cm PRACTICE QUESTIONS the sphere. \t\t(c) 6 3 cm \t\t (d) 2 6 cm \t\t(a) 6 cm\t\t (b) 8 cm 2\t 9.\t The dimensions of a cuboidal container are 12 cm \t\t(c) 10 cm\t\t (d) 12 cm \u00d7 10 cm \u00d7 8 cm. How many bottles of syrup can \t23.\t The number of diagonals of a regular polygon in be poured into the container, if each bottle con- which each interior angle is 156\u00b0, is ______. tains 20 cm3 of syrup? \t\t(a) 24\t\t (b) 54 \t\t(a) 46\t\t (b) 54 \t\t(c) 90\t\t (d) 45 \t\t(c) 48\t\t (d) 58 Level 2 \t30.\t A circle of radius 2 cm is inscribed in an equilateral \t\t(a) 96\t\t (b) 128 triangle. Find the area of the triangle in cm2. \t\t(c) 48\t\t (d) 64 \t\t(a) 12 3 \t\t (b) 4 3 \t32.\t A parallelogram has two of its adjacent sides mea- \t\t(c) 6 3 \t\t (d) 24 3 suring 13 each. Find the sum of the squares of its diagonals. \t31.\t The circum-radius of a right triangle is 10 cm and one of the two perpendicular sides is 12 cm. Find \t\t(a) 169\t\t (b) 338 the area of the triangle in sq. cm. \t\t(c) 676\t\t (d) 507","13.28 Chapter 13 \t33.\t A classroom is 5 m long, 2.5 m broad and 3.6 m by 5 cm, its area increases by 20 cm2. Find the area high. If each student is given 0.5 m2 of the floor of the rectangle (in cm2). area, then how many cubic metres of air would \t\t(a) 20\t\t (b) 25 each student get? \t\t(c) 30\t\t (d) 35 \t\t(a) 1.4\t\t (b) 1.8 4\t 1.\t A cylinder-shaped tank is surmounted by a cone of equal radius. The height of the cone is 6 m and the \t\t(c) 1.2\t\t (d) 1.6 total height of the tank is 18 m. Find the volume of the tank if the base radius of the cylinder is 5 m. \t34.\t If the volume of a right equilateral triangular prism is 8500 3 dm3 whose height is 50 cm, then find \t\t(a) 1650 m3\t\t (b) 1244 m3 the side of its base. \t\t(c) 1100 m3\t\t (d) 2200 m3 \t\t(a) 10 17 cm \t(b) 10 17 dm \t\t(c) 20 17 dm \t(d) 20 17 cm \t42.\t The radius of the cross-section of an inflated cycle tyre is 7 cm. The distance of the centre of the 3\t 5.\t How many cubes, each of total surface area 54 cross-section from the axle is 20 cm. Find the vol- sq.dm, can be made from a cube of edge 1.2 metre. ume of air in the tyre. \t\t(a) 64\t\t (b) 81 \t\t(c) 125\t\t (d) 25 \t\t(a) 19360 cm3\t (b) 1760 cm3 \t\t(c) 6160 cm3\t (d) 880 cm3 \t36.\t Four times the sum of the areas of the two circular faces of a cylinder is equal to the twice its curved \t43.\t If h is the length of the perpendicular drawn from surface area. Find the diameter of the cylinder if its height is 8 cm. a vertex of a regular tetrahedron to the opposite face and each edge is of length s, then s2 is equal \t\t(a) 4 cm\t\t (b) 8 cm to ______. \t\t(c) 2 cm\t\t (d) 6 cm \t\t(a) 3h2 \t\t (b) 3h2 8 2 3\t 7.\t If the base of a right pyramid is a square of side 4 cm and its height is 18 cm, then the volume of the \t\t(c) h2\t\t (d) 8h2 pyramid is ______. 3 PRACTICE QUESTIONS \t\t(a) 90 cm3\t\t (b) 104 cm3 \t44.\t The radii of the ends of a frustum of a cone are 28 cm and 7 cm. The height of the cone is 40 cm. \t\t(c) 100 cm3\t\t (d) 96 cm3 Find its volume. \t38.\t A right circular conical tent is such that the angle \t\t(a) 32340 cm3\t (b) 43120 cm3 \t\t(c) 10780 cm3\t (d) None of these at its vertex is 60\u00b0 and its base radius is 14 m. Find 4\t 5.\t The diagonal of a cube is 6 3 cm. Find its volume. the cost of the canvas required to make the tent at the rate of `25 per m2. \t\t(a) `15,400\t\t (b) `30,800 \t\tThe following are the steps involved in solving the above problem. Arrange them in sequential order. \t\t(c) `16,400\t\t (d) `32,800 \t\t(A) \u0007\u2234 Volume of the cube = a3 cm3 = (6)3 cm3 = \t39.\t A goat is tied to a corner of a rectangular plot of 216 cm3. dimensions 14 m \u00d7 7 m with a 21 m long rope. It \t\t(B) Then, diagonal of the cube = a 3 cm. cannot graze inside the plot, but can graze outside it as far as it is permitted by the rope. Find the area it can graze (in m2). (Take \u03c0 = 22 ) \t\t(C) From the given data a 3 = 6 3 \u21d2 a = 6 cm. 7 \t\t(a) 240\t\t (b) 1560.5 \t\t(D) Let the side of the cube be a cm. \t\t(c) 1543.5\t\t (d) 1232 \t\t(a) DCBA\t\t (b) DBCA \t40.\t If the breadth of a rectangle is increased by 5 cm, \t\t(c) DACB\t\t (d) DBAC its area increases by 25 cm2. If its length is increased","Mensuration 13.29 \t46.\t Find the length of the arc of a sector of a circle \t\t(A) Then the diagonal of cube = a 3 cm. \t\t(B) \u2234 Total surface area = 6a2 = 6(8)2 = 384 cm2. whose angle at the centre is 120\u00b0 and area of the \t\t(C) Let the side of cube be a cm. sector is 462 cm2. \t\t(D) From the given data, a 3 = 8 3 \u21d2 a = 8 cm. \t\t(a) ACBD \t\tThe following are the steps involved in solving the \t\t(b) CADB above problem. Arrange them in sequential order. \t\t(c) ABCD \t\t(d) CBAD \t\t(A) \u0007Given, \u03b8 = 120\u00b0, area of the sector = 462 cm2. We know that A = \u03b8 \u00d7 \u03c0 r 2. 360\u00b0 \t\t(B) \u21d2 462 = 120\u00b0 \u00d7 22 (r )2 \u21d2r = 21 cm. 360\u00b0 7 \t\t(C) \u0007Length of the arc of the sector = \u03b8 \u00d7 2\u03c0 r \t49.\t The length of the diagonals of a rhombus are 10 360\u00b0 cm and 24 cm. Find its side. (in cm) = 120 \u00d72\u00d7 22 \u00d7 21. \t\t(a) 10\t\t (b) 13 360\u00b0 7 \t\t(D) \u2234 Length of the arc = 44 cm \t\t(d) 12\t\t (d) 11 \t\t(a) ABCD \t50.\t Area of a right-angled triangle is 6 cm2 and its \t\t(b) ACBD perimeter is 12 cm. Find its hypotenuse. (in cm) \t\t(c) BACD \t\t(a) 5\t\t (b) 6 \t\t(d) BCAD \t\t(c) 7\t\t (d) 8 4\t 7.\t The radii of the top and the bottom of a metal 5\t 1.\t A largest possible right-circular cylinder is cut can which is cone-shaped frustum, are 20 cm and out from a wooden cube of edge 7 cm. Find the 8 cm, respectively. The height of the can is 16 cm. volume of the wood left over after cutting the Find the area of the metal sheet required to make cylinder. (in cu cm) the can with a lid. \t\t(a) 73.5\t\t (b) 82.5 \t\tThe following are the steps involved in solving \t\t(c) 76\t\t (d) 92 PRACTICE QUESTIONS the above problem. Arrange them in sequential order. \t52.\t A solid sphere of radius 4 cm is melted and recast into \u2018n\u2019 solid hemispheres of radius 2 cm each. \t\t(A)\tArea of metal sheet = \u03c0l(R + r) + \u03c0R2 + \u03c0r2 = Find n. \u03c0 \u00d7 20(20 + 8) + \u03c0(20)2 + \u03c0(8)2. \t\t(a) 32\t\t (b) 16 \t\t(B) Given, R = 20 cm, r = 8 cm and h = 16 cm. \t\t(c) 8\t\t (d) 4 \t\t(C) l = (R \u2212 r )2 + h2 = (20 \u2212 8)2 + (16)2 = 20 cm. \t53.\t Three small metallic cubes whose edges are in the ratio 3 : 4 : 5 are melted to form a big cube. If \t\t(D) \u2234 Area of the sheet = 1024 \u03c0 sq. cm. the diagonal of the cube so formed is 18 cm, then \t\t(a) BCAD find the total surface area of the smallest cube. (in cm2) \t\t(b) ACBD \t\t(a) 154\t\t (b) 184 \t\t(c) BDAC \t\t(c) 216\t\t (d) 162 \t\t(d) BACD \t54.\t A solid hemisphere of radius 8 cm is melted and 4\t 8.\t The diagonal of a cube is 8 3 cm. Find its total recast into x spheres of radius 2 cm each. Find x. surface area. \t\tThe following are the steps involved in solving \t\t(a) 4\t\t (b) 8 the above problem. Arrange them in sequential order. \t\t(c) 16\t\t (d) 32","13.30 Chapter 13 Level 3 \t55.\t In the following figure, ABCD is a rectangle with the height of the other. The sum of the volumes AB = 9 cm and BC = 6 cm. O is the centre of the of both the cylinders is 250\u03c0 cm3. Find the sum of circle. Find the area of the shaded region. (in cm2) their curved surface areas. (in cm2) AB \t\t(a) 80\u03c0\t\t (b) 100\u03c0 \t\t(c) 60\u03c0\t\t (d) 120\u03c0 O\u2022 \t59.\t In a triangle, the average of any two sides is 6 cm more than half of the third side. Find area of the triangle. (in sq. cm) DC \t\t(a) 64 3 \t\t (b) 48 3 \t\t(a) 18\t\t (b) 24 \t\t(c) 72 3 \t\t (d) 36 3 \t\t(c) 27\t\t (d) 15 \t60.\t The lengths of the diagonals of a rhombus are 9 cm and 12 cm. Find the distance between any 5\t 6.\t A circus tent is cylindrical upto a height of 8 m and two parallel sides of the rhombus. conical above it. If its base diameter is 70 m and the slant height of the conical part is 50 m, the area \t\t(a) 7.2 cm\t\t (b) 8 cm of the canvas required to make the tent is _____ m2. \t\t(c) 7.5 cm\t\t (d) 6.9 cm \t\t(a) 6380\t\t (b) 7260 \t61.\t In a triangle, the sum of any two sides exceed the third side by 6 cm. Find its area (in sq. cm). \t\t(c) 6850\t\t (d) 7460 \t\t(a) 12 3 \t\t (b) 9 3 \t57.\t A water tank of dimensions 11 m \u00d7 6 m \u00d7 5 m is \t\t(c) 15 3 \t\t (d) 18 3 full of water. The tank is emptied through a pipe PRACTICE QUESTIONS of cross-section 33 cm2 in 20 hours. Find the rate 6\t 2.\t From each corner of a square sheet of side 8 cm, a square of side y cm is cut. The remaining sheet is of flow of water. (in kmph) folded into a cuboid. The minimum possible vol- ume of the cuboid formed is M cubic cm. If y is an \t\t(a) 2\t\t (b) 5 integer, then find M. \t\t(c) 6\t\t (d) 10 \t\t(a) 32\t\t (b) 18 \t58.\t P and Q are two cylinders having equal total sur- \t\t(c) 36\t\t (d) 12 face areas. The radius of each cylinder is equal to","Mensuration 13.31 TEST YOUR CONCEPTS Very Short Answer Type Questions \t1.\t 50 cm2 \t16.\t 3 : 2 \t2.\t 2 units \t3.\t 6 \t17.\t 1 c.c 3 1\t 8.\t 72.8% \t4.\t 6 1\t 9.\t Sphere \t5.\t 6 2\t 0.\t False \t6.\t 12 \t21.\t 4 : 9 \t7.\t 2( p2 + q2 ) \t22.\t 2 \u03c0 x3 \u2212 2 \u03c0(x \u2212 y )3 3 3 \t8.\t 12 sq. units \t23.\t 4.5 units \t9.\t 3 2\t 4.\t (x2 \u2212 y2) cm2 \t10.\t equilateral \t25.\t 10 3 cm2 1 1\t 1.\t 4 \t26.\t 1944 cm3 \t12.\t 40 cm2 \t27.\t 3 \u00d7 3 \u00d7 3 = 27 cm3 \t13. \tparallelogram, 13 cm2 \t28.\t (i)\t256 cm2 \t14.\t 4r 2 + h2 \t\t(ii)\t8 3 cm 1\t 5.\t 3 a2 \u00d7 h 2\t 9.\t 9 12 3\t 0.\t 880 m2 Short Answer Type Questions \t31.\t 5.2 cm 3\t 8.\t 3360 cm3 ANSWER KEYS 3\t 2.\t 33 cm \t39.\t 385 \t33.\t 18\u03c0 m2 \t40.\t 50% \t34.\t 8 cm2 4\t 1.\t 143.36 cm2 \t35.\t 24 m2 \t42.\t 72 3 cm2 4\t 3.\t 120 cm3 \t36.\t 2 (a + b) units \t44.\t 36 cm2 \t45.\t 282.6 m2 3\t 7.\t 686 cm 2 11 Essay Type Questions \t46.\t `20,096 \t49.\t 157 m2 \t50.\t 3.5 cm 4\t 7.\t 1 3 ft 7 4\t 8.\t 38720 cm2","13.32 Chapter 13 CONCEPT APPLICATION Level 1 \t1.\u2002(c) \t 2.\u2002 (d)\t 3.\u2002 (a)\t 4.\u2002 (a) \t 5.\u2002 (b)\t 6.\u2002 (c)\t 7.\u2002 (b)\t 8.\u2002 (b) \t 9.\u2002 (b)\t 10.\u2002 (a) \t11.\u2002 (b)\t 12.\u2002 (d) \t 13.\u2002 (d)\t 14.\u2002 (a)\t 15.\u2002 (b)\t 16.\u2002 (b) \t 17.\u2002 (b)\t 18.\u2002 (c) \t 19.\u2002 (d)\t 20.\u2002 (b) \t21.\u2002 (d)\t 22.\u2002 (a) \t 23.\u2002 (c)\t 24.\u2002 (d) \t 25.\u2002 (c)\t 26.\u2002 (b) \t 27.\u2002 (b)\t 28.\u2002 (b) \t 29.\u2002 (c) Level 2 \t30.\u2002 (a) \t 31.\u2002 (a)\t 32.\u2002 (c) \t 33.\u2002 (b)\t 34.\u2002 (c)\t 35.\u2002 (a)\t 36.\u2002 (b) \t 37.\u2002 (d)\t 38.\u2002 (b) \t 39.\u2002 (d) \t40.\u2002 (a) \t 41.\u2002 (c)\t 42.\u2002 (a) \t 43.\u2002 (b)\t 44.\u2002 (a) \t 45.\u2002 (b)\t 46.\u2002 (a)\t 47.\u2002 (a) \t 48.\u2002 (b)\t 49.\u2002 (b) \t50.\u2002 (a)\t 51.\u2002 (a) \t 52.\u2002 (b)\t 53.\u2002 (d) \t 54.\u2002 (d) Level 3 57.\u2002 (b) \t 58.\u2002 (b) \t 59.\u2002 (d) \t 60.\u2002 (a)\t 61.\u2002 (b) \t 62.\u2002 (d)\t \t55.\u2002 (a) \t 56.\u2002 (b)\t ANSWER KEYS","Mensuration 13.33 CONCEPT APPLICATION Level 1 \t1.\t Use Pythagoras theorem. \t20.\t TSA of the octahedron = 8 \u00d7 Area of each face. \t21.\t The number of the side of the base = The number \t2.\t Area of regular polygon = 1 \u00d7 Perimeter of 2 of lateral edges. polygon \u00d7 Height. 2\t 2.\t Volume of the sphere = Volume of the cylinder. \t3.\t Length of the arc = \u03b8 \u00d7 2\u03c0 r. \t23.\t (i)\tFind the number of sides n of the polygon. 360 n(n \u2212 3) \t4.\t Sum of the lengths of the edges = nh + 2s, where \t\t(ii)\tNumber of diagonals = 2 . h = height, s = perimeter of the base and n = num- ber of sides of the base. \t24.\t (i)\tABD is a right triangle and its area is 1 AB \u00b7 2 \t5.\t Volume of the larger cube = Sum of the volumes of smaller cubes. BD. \t6.\t Length of the diagonal of a cuboid = l2 + b2 + h2 . \t\t(ii)\tBCD is an isosceles triangle. \t7.\t Use the formula to find volume of the cone. \t\t(iii)\t\u2220ABD = 90\u00b0. Find BD by using the basics of isosceles triangle (i.e., \u2206BDC). \t8.\t Find the side of the square. \t25.\t (i)\tl = 2A . r \t9.\t Find height of the cone. (lr ) \t10.\t Find r1 . \t\t(ii)\tUse the formula, A= 2 . r2 \t26.\t (i)\tLSA = 288 cm2. Hints and Explanation 1\t 1.\t Find r by using the formula of volume of a sphere. \t\t(ii)\tTake perimeter of the base as 3a units and find a side. \t12.\t Volume of the regular octahedron = 2 \u00d7 Volume of the pyramid. \t\t(iii)\tLSA of prism is half the product of base, perimeter and slant height. 1\t 3.\t Use Euler\u2019s formula. \t14.\t Volume of the cylinder = Volume of 15 spheres. \t\t(iv)\tTSA = LSA + 2(Area of the base). 1\t 5.\t Number of soaps = Volume of box . \t\t(v)\tEquate the given value to the above formula Volume of each soap and find h. 1\t 6.\t Percentage increase in volume \t27.\t (i)\tEdge of the cube = Diameter of the sphere. = Increase in volume \u00d7 100. \t\t(ii)\tRadius of sphere is 3.5 cm, edge of the cube is Original volume 2(3.5 cm), i.e., 7 cm. 1\t 7.\t Volume of hollow sphere = 4 \u03c0 ( R3 \u2212 r 3 ). \t\t(iii)\tDifference between their surface areas = 4\u03c0r2 3 - 6s2. \t18.\t CSA of hollow hemisphere = 2\u03c0(R2 + r2). \t28.\t (i)\tHeight of regular octahedron = 2(Height of regular tetrahedron). \t19.\t Volume of tetrahedron = 1 (Area of \u2009base) \u00d7 Height, 3 \t\t(ii)\tHeight of the octahedron (h) = 2 a. where height of the tetrahedron = 2 \u00d7 Length \t29.\t Number of bottles Volume of container . 3 Volume of each bottle of the edge. =","13.34 Chapter 13 Level 2 \t30.\t (i)\tRadius of the circle is one-third of the median 3\t 8.\t (i)\tBase radius, vertical height and slant height of the triangle. form a right triangle with angles 30\u00b0, 60\u00b0 and 90\u00b0. \t\t(ii)\tInradius is 2 cm. \t\t(iii)\tHeight of the equilateral triangle = 3(2 cm) \t\t(ii)\tBase radius (14 cm) is opposite to 30\u00b0. = 6 cm. h2 . \t\t(iii)\tCost of canvas = LSA \u00d7 Cost\/m2. 3 \t\t(iv)\tArea of an equilateral triangle = \t39.\t (i)\tDraw the figure and proceed. 3\t 1.\t (i)\tCircum-radius is half of the hypotenuse. Use \t\t(ii)\tThe required area is equal to the sum of the areas of a sector of an angle 270\u00b0 and radius 21 Pythagoras\u2019s theorem. m, a sector of an angle 90\u00b0 and radius 7 m and a sector of an angle 90\u00b0 and radius 14 m. \t\t(ii)\tCircum-radius is 10 cm, so hypotenuse is 2(10 cm). \u03b8 360\u00b0 \t\t(iii)\tUse Pythagoras\u2019s theorem and get the length of \t\t(iii)\tArea of sector = (\u03c0 r 2 ). the other perpendicular side and find the area. 4\t 0.\t l(b + 5) = lb + 25 and b(l + 5) = lb + 20. 3\t 2.\t (i)\tGiven parallelogram is a rhombus. 4\t 1.\t Required volume is the sum of the volumes of \t\t(ii)\tAs adjacent sides are equal, parallelogram conical part and cylindrical part. becomes rhombus. \t42.\t (i)\tVolume of torus = 27\u03c0r2a. \t\t(iii)\tSum of the squares of diagonals is equal to four times the square of the side. \t\t(ii)\tGiven, r = 7 cm and a = 20 cm. \t33.\t (i)\tFind the number of students. \t\t(iii)\tVolume of the tyre = 2\u03c0r2a. 2 \t43.\t (i)\tHeight of the regular tetrahedron, h = 3 \t\t(ii)\tNumber of students = Area of the floor . \t\t(ii)\tSquare the above equation and get s2. s. 0.5 Hints and Explanation \t\t(iii)\tThe amount of air each student would get 1 3 = Volume of the room . \t44.\t (i)\tVolume of frustum = \u03c0h(R2 + r2 + Rr). Number of students 3\t 4.\t (i)\tVolume of prism = (Area of the base)(Height). \t\t(ii)\tVolume of cone frustum = \u03c0h (R2 + Rr + r 2 ). 3 \t\t(ii)\tArea of equilateral triangle = 3a2 . \t45.\t DBCA is the required sequential order. 4 \t\t(iii)\t1 dm = 10 cm. 4\t 6.\t ABCD is the required sequential order. 3\t 5.\t (i)\tEdge (a) of small cube is to be found by using \t47.\t BCAD is the required sequential order. 6a2 = 54. \t48.\t CADB is the required sequential order. \t\t(ii)\tNumber of small cubes = Volume of the big cube . \t49.\t \t A EB Volume of each small cube 3\t 6.\t (i)\tUse relevant formula. \t\t(ii)\tGiven condition is 8\u03c0r2 = 4\u03c0rh. O \t\t(iii)\tSubstitute h = 8 cm in the above equation and DC solve for r. \t\tLet ABCD be a rhombus. \t37.\t (i)\tVolume = 1 (Area of base)(Height). \t\tLet CE be the height of the rhombus. 3 \t\tGiven AC = 10 cm and BD = 24 cm. \t\t(ii)\tArea of square = s2. \t\tBC = OC2 + OB2 = 52 + 122 = 13 cm. 1 \t\t(iii)\tVolume of the pyramid = 3 (Area of the base) (height).","Mensuration 13.35 \t50.\t Let the sides of the triangle (in cm) be a, b and c, \t\tVolume of the cylinder \u21d2 \u03c0 r 2h = 22 \u00d7 3.5 \u00d7 3.5 where a < b < c. 7 \t \t \u2234 \u0007c is the hypotenuse and a and b are the perpen- \t\t\u00d7 7 = 269.5. dicular sides. \t\t\u2234 T\u0007 he required volume = 343 \u2212 269.5 = 73.5 cm3. \t\tGiven, a + b + c = 12\b (1) \t52.\t Given that, n \u00d7 2 \u03c0 (2)3 = 4 \u03c0 (4)3 \u21d2 n = 16. \t\t21 (a) (b) = 6\b (2) 3 3 \t\tFrom Eq. (1) \u21d2 a + b = 12 \u2212 c 5\t 3.\t Let the edges of the cubes be 3x, 4x and 5x \t\tSquaring both sides, respectively. Volume of the cube formed = Sum of the volumes of the small cubes = (3x)3 + (4x)3 + (5x)3 = 216x3. \t\ta2 + b2 + 2ab = 122 \u2212 24c + c2 \t \t \u2234 Edges of the cube formed = 6x. \t\t= a2 + b2 + 4 \uf8eb 1 ab\uf8f6\uf8f7\uf8f8 = 122 \u2212 24c + c2 \t\tDiagonal of the cube formed = 3(6x) \u21d2 6 3x \uf8ec\uf8ed 2 = 18 \u21d2 x = 3. \t\t\u21d2 c2 + 4(6) = 122 \u2212 24c + c2 \t\t\u2234 TSA of smallest cub=e 6=a2 6(3 3 )2 = 162 cm2. \t\tc = 5. \t51.\t Diameter of the cylinder = 7 cm. 4 2 3 3 \t \t \u2234 Radius of the cylinder = 3.5 cm. \t54.\t Given that, x \u00d7 \u03c0 (2)3 = \u03c0 (8)3 \u21d2 x = 32. \t\tHeight of the cylinder = 7 cm. \t\tVolume of cube = 73 = 343 cm3. Level 3 Hints and Explanation \t59.\t Let the sides of the triangle be a, b and l. \t\tIn \u2206BOC, BO2 + OC2 = BC2 \t\tGiven, a+b = 6+ c , a +c = 6+ b and b+ c \t\t\u21d2 62 + (4.5)2 = BC2 \u21d2 BC = 7.5 cm. 2 2 2 2 2 \t\tLet CE be the required distance. a = 6 + 2 . \t\tArea of rhombus = 12 \u00d7 9 = 54 sq. cm 2 \t\t\u21d2 a + b = 12 + c\b(1) CE \t\tb + c = 12 + a\b(2) \t\t\u21d2 2 ( AB + CD ) = 54 (Area of parallelogram) \t\tc + a = 12 + b\b(3) \t\t\u21d2 CE = 108 = 7.2 cm. 15 \t\tBy solving, we get 6\t 1.\t Let the sides of the triangle be a cm, b cm and \t\ta = b = c = 12 c cm. \t\t\u21d2 Area of \u2206ABC = 3 \u00d7 122 \t\ta + b \u2212 c = a + c \u2212 b = b + c \u2212 a = 6 2 \t\t\u21d2 a + b \u2212 c + a + c \u2212 b + b + c \u2212 a = 18 = 36 3 sq. cm. \t\ta + b + c = 18. \t60.\t E D \t\tArea of the triangle = A O \uf8eb a + b + c \uf8f6\uf8eba + b + c \u2212 a\uf8f8\uf8f7\uf8f6 \uf8ed\uf8ec a + 2 + c \uf8f7\uf8f8 \uf8ec\uf8ed \uf8eb 2 + + c\u2212 \t\t \uf8eb \u2212 b\uf8f8\uf8f7\uf8f6 \uf8ec\uf8ed a b c\uf8f6\uf8f8\uf8f7 \uf8ec\uf8ed b 2 BC 2","13.36 Chapter 13 \uf8eba + b + c\uf8f6 \uf8ebb + c \u2212 a\uf8f6 \t\tLength = Breadth = (8 \u2212 2y) cm and height = \uf8ec\uf8ed 2 \uf8f8\uf8f7 \uf8ec\uf8ed 2 \uf8f8\uf8f7 y cm. \uf8eba + c \u2212b\uf8f6 \uf8eba + b \u2212 c\uf8f6 \t\t= \uf8ed\uf8ec 2 \uf8f8\uf8f7 \uf8ec\uf8ed 2 \uf8f8\uf8f7 \t\tIts volume = (8 \u2212 2y)(8 \u2212 2y)y \t \t\t = (8 \u2212 2y)2 y cubic cm. \t\t 41=(18)(6)(6)(6) 9 3 cm2. \t\t8 \u2212 2y > 0, i.e., y < 4 and y is an integer. 6\t 2.\t \t \t \u2234 y = 1 or 2 or 3. y cm (8 \u2212 2y) cm y cm \t\tAmong these values of y, volume is minimum when y = 3. When y = 3, volume = 12 cm3. y cm y cm \t \t \u2234 M = 12. (8 \u2212 2y) cm (8 \u2212 2y) cm y cm y cm y cm (8 \u2212 2y) cm y cm Hints and Explanation","1142CChhaapptteerr CKoinoermdiantaictse Geometry REmEmBER Before beginning this chapter, you should be able to: \u2022 Know planes, lines and angles \u2022 Remember different types of triangles and polygons KEY IDEAS After completing this chapter, you should be able to: \u2022 Find the coordinates of a point and conversion of signs \u2022 Study about points on a plane and the distance between points \u2022 Know the applications of distance formula, mid-point of a line segment and centroid of a triangle \u2022 Learn about equation of some standard lines Figure 1.1","14.2 Chapter 14 INTRODUCTION X\u2032 Y M x P (x, y) Let X\u2009\u2032OX and YOY\u2009\u2032 be two mutually perpendicular lines intersecting at point O in a plane. y O LX These two lines are called reference lines or coordinate axes. The horizontal reference line X\u2009\u2032OX is called X-axis and the vertical reference line YOY\u2009\u2032 is called Y-axis. The point of intersection of these two axes, i.e., O is called the origin. The plane containing the coordinate axes is called coordinate plane or XY-plane. COORDINATES OF A POINT Y\u2032 Figure 14.1 Let P be a point in the XY-plane. Draw perpendiculars PL and PM to X-axis and Y-axis respectively (see Fig. 14.1). Let PL = y and PM = x. The point P is taken as (x, y). Here, x and y are called the rectangular Cartesian coordinates or coordinates of the point P. x is called x-coordinate or abscissa and y is called y-coordinate or ordinate of the point P. Convention of Signs 1.\t T\u0007 owards the right side of the Y-axis, x-coordinate of any point on the graph paper is taken positive and towards the left side of the Y-axis, x-coordinate is taken negative. 2.\t A\u0007 bove the X-axis, the y-coordinate of any point on the graph paper is taken positive and below the X-axis, y-coordinate is taken negative. If (x, y) is a point in the plane and Q1, Q2, Q3 and Q4 are the four quadrants of rectangular coordinate system, then: 1.\t If x > 0 and y > 0, then (x, y) \u2208 Q1. 2.\t If x < 0 and y > 0, then (x, y) \u2208Q2. 3.\t If x < 0 and y < 0, then (x, y) \u2208Q3. 4.\t If x > 0 and y < 0, then (x, y) \u2208Q4. Example 14.1 If x > 0 and y < 0, then (x, -y) lies in which quadrant? Solution y<0 \u21d2 -y > 0 \u2234 The point (x, -y) lies in first quadrant, i.e., Q1. Example 14.2 If (x, -y) \u2208 Q2, then (x, y) belongs to which quadrant? Solution Given, (x, -y) \u2208 Q2 \u21d2 x < 0, y < 0. \u2234 (x, y) belongs to third quadrant, i.e., Q3.","Coordinate Geometry 14.3 Example 14.3 Plot the points A(2, 3), B(\u22121, 2), C(\u22123, \u22122) and D(4, \u22122) in the XY-plane. Solution Y 3 \u2022 A(2, 3) B( \u2212 1, \u20222) 2 1 X\u2032 \u2212 3 \u2212 2 \u22121 01 2 3 4X \u2022 \u22121 \u2022 C( \u2212 3, \u2212 2) \u22122 D( 4, \u2212 2) \u22123 Y\u2032 Figure 14.2 POINTS ON THE PLANE Y X Point on X-axis and Y-axis P\u2032 (0, y) P (x, 0) Let P be a point on X-axis, so that its distance from X-axis is zero. Hence, point P can be taken as (x, 0). O Let P\u2009\u2032 be a point on Y-axis, so that its distance from Y-axis is zero. Hence, point P\u2009\u2032 can be taken as (0, y) (see Fig. 14.3). Distance Between Two Points Figure 14.3 Consider two points A(x1, y1) and B(x2, y2). Draw perpendiculars AL and BM from A and B to X-axis. AN is the perpendicular drawn from A on to BM (see Fig. 14.4). From right triangle ABN, AB = AN 2 + BN 2, (we have AB2 = AN2 + BN2). Here, AN = x2 - x1 and BN = y2 - y1. AB = (x2 \u2212 x1 )2 + (y2 \u2212 y1 )2 . Hence, the distance between two points A(x1, y1) and B(x2, y2) is AB = (x2 \u2212 x1 )2 + (y2 \u2212 y1 )2 units. \u2002Note\u2002\u2002 The distance of a point A(x1, y1) from origin O(0, 0) is OA = x12 + y12 .","14.4 Chapter 14 Y A(x1, y1) \u2022 x1\u2212 x2 B(x2\u2022, y2) y1\u2212 y2 N O L(x1, O) M(x2, O) X Figure 14.4 Example 14.4 Find the distance between points (-4, 5) and (2, -3). Solution Let the given points be A(-4, 5) and B(2, -3) AB = (2 \u2212 (\u22124))2 + (\u22123 \u2212 5)2 = 36 + 64 = 10 units. Example 14.5 Find a, if the distance between points A(8, -7) and B(-4, a) is 13 units. Solution Given, AB = 13 \u21d2 (\u22124 \u2212 8)2 + (a + 7)2 = 13 Taking squares on both sides, we get (a + 7)2 = 169 - 144 = 25 a + 7 = 25 a + 7 = \u00b15 \u2234 a = -2 or -12. Example 14.6 Find the coordinates of a point on Y-axis which is equidistant from points (13, 2) and (12, -3). Solution Let P(0, y) be the required point and the given points be A(12, -3) and B(13, 2). Then, PA = PB (given)","Coordinate Geometry 14.5 Y B(13, 2) P(0, y) X X\u2032 O A(12,\u2212 3) Y\u2032 Figure 14.5 (12 \u2212 0)2 + (\u22123 \u2212 y)2 = (13 \u2212 0)2 + (2 \u2212 y)2 \u21d2 144 + (y + 3)2 = 169 + (2 \u2212 y)2 Taking squares on both sides, we get 169 + 4 + y2 - 4y = 144 + 9 + y2 + 6y \u21d2 10y = 20 \u21d2 y = 2 \u2234 The required point on Y-axis is (0, 2). Collinearity of Three Points Let A, B and C be three given points. The distances AB, BC and CA can be calculated using distance formula. If the sum of any two of these distances is found to be equal to the third distance, then points A, B and C are said to be collinear. \u2002Notes\u2002 1.\t If AB + BC = AC, then points A, B and C are collinear. AB C 2.\t If AC + CB = AB, then points A, C and B are collinear. AC B 3.\t BA + AC = BC, then points B, A and C are collinear. BA C By Notes (1), (2) and (3), we can find the position of the points in collinearity. Applications of Distance Formula Example 14.7 Show that points P(5, 6), Q(4, 5) and R(3, 4) are collinear. Solution Given, P = (5, 6), Q = (4, 5) and R = (3, 4). PQ = (4 \u2212 5)2 + (5 \u2212 6)2 = 2 units.","14.6 Chapter 14 QR = (3 \u2212 4)2 + (4 \u2212 5)2 = 2 units. PR = (3 \u2212 5)2 + (4 \u2212 6)2 = 8 = 2 2units. Now, PQ + QR = 2 + 2 = 2 2 = PR. That is, PQ + QR = PR. Hence, points P, Q and R are collinear. Example 14.8 Show that points A(3, -1), B(-1, 2) and C(6, 3) form an isosceles right-angled triangle when joined. Solution Given, A = (3, -1), B = (-1, 2) and C = (6, 3). AB = (\u22121 \u2212 3)2 + (2 + 1)2 = 5 units BC = (6 \u2212 (\u22121))2 + (3 \u2212 2)2 = 50 units AC = (6 \u2212 3)2 + (3 \u2212 (\u22121))2 = 5 units Clearly, BC2 = AB2 + AC2. Also, AB = AC. Hence, the given points form the vertices of a right-angled isosceles triangle. Example 14.9 Show that points (2 - 3 , 3 + 1), (1, 0) and (3, 2) form an equilateral triangle. Solution Let A(2 - 3, 3 + 1), B(1, 0) and C(3, 2) be the given points. ( )AB = (1 \u2212 2 + 3 )2 + 0 \u2212 ( 3 + 1) 2 AB = 8 units. = ( 3 \u2212 1)2 + ( 3 + 1)2 . BC = (3 \u2212 1)2 + (2 \u2212 0)2 = 8 units ( ) ( )AC = 3 \u2212 (2 \u2212 3) 2 + 2 \u2212 ( 3 + 1) 2 = (1 + 3 )2 + (1 \u2212 3 )2 = 8 units. \u2234 AB = BC = AC = 8 units. Hence, the given points form an equilateral triangle.","Coordinate Geometry 14.7 Example 14.10 Show that points A(-1, 0), B(-2, 1), C(1, 3) and D(2, 2) form a parallelogram. Solution Given, A(-1, 0), B(-2, 1), C(1, 3) and D(2, 2). AB = (\u22122 + 1)2 + (1 \u2212 0)2 = 2 units BC = (1 \u2212 (\u22122))2 + (3 \u2212 1)2 = 13 units CD = (2 \u2212 1)2 + (2 \u2212 3)2 = 2 units DA = (2 \u2212 (\u22121))2 + (2 \u2212 0)2 = 13 units AC = (1 \u2212 (\u22121))2 + (3 \u2212 0)2 = 13 units BD = (2 \u2212 (\u22122))2 + (2 \u2212 1)2 = 17 units Clearly, AB = CD, BC = DA and AC \u2260 BD. That is, the opposite sides of the quadrilateral are equal and diagonals are not equal. Hence, the given points form a parallelogram. Example 14.11 Find the circum-centre and the circum-radius of a triangle ABC formed by the vertices A(2, -2), B(-1, 1) and C(3, 1). Solution (1) Let S(x, y) be the circum-centre of \u0394ABC. \u2234 SA2 = SB2 = SC2 Consider, SA2 = SB2 \t \u21d2 (x - 2)2 + (y + 2)2 = (x + 1)2 + (y - 1)2 x2 - 4x + 4 + y2 + 4y + 4 = x2 + 2x + 1 + y2 - 2y + 1 \t -4x + 4y + 8 = 2x - 2y + 2 \t6x - 6y - 6 = 0 \tx - y - 1 = 0\b SB2 = SC2 \t \u21d2 (x + 1)2 + (y - 1)2 = (x - 3)2 + (y - 1)2 x2 + 2x + 1 + y2 - 2y + 1 = x2 - 6x + 9 + y2 - 2y + 1 \t2x - 2y + 2 = -6x - 2y + 10 \t8x - 8 = 0 \t \u21d2 x = 1. Substituting x = 1 in Eq. (1), we get y = 0. \u2234 The required circum-centre of \u0394ABC is (1, 0). Circum-radius, SA = (2 \u2212 1)2 + (\u22122 \u2212 0)2 = 5 units.","14.8 Chapter 14 Example 14.12 Find the area of the circle whose centre is (-1, -2), and (3, 4) is a point on the circle. Solution Let the centre of the circle be A(-1, -2), and the point on the circumference be B(3, 4). Radius of circle = AB = (3 \u2212 (\u22121))2 + (4 \u2212 (\u22122))2 = 52 units. \u2234 The area of the circle = pr2 = p( 52 )2 = 52p sq. units. Example 14.13 Find the area of the square whose one pair of opposite vertices are (2, -3) and (4, 5). Solution D C(4, 5) Let the given vertices be A(2, -3) and C(4, 5). Length of AC = (4 \u2212 2)2 + (5 + 3)2 = 68 units. Area of the square AC 2 = ( 68 )2 34 sq. units. A(2,\u2013 3) B 2 2 \u2234 = = Figure 14.6 STRAIGHT LINES Y Inclination of a Line \u03b8 X The angle made by a straight line with positive direction of O X-axis in the anti-clockwise direction is called its inclination. Figure 14.7 Slope or Gradient of a Line If q is the inclination of a line L, then its slope is denoted by m Yl and is given by m = tan\u2009q (see Fig. 14.7). Example: The inclination of the line l in adjacent Fig. 14.8 is 45\u00b0. 45\u00b0 X O \u2234 The slope of the line is m = tan45\u00b0 = 1. Example: The line L in Fig. 14.9 makes an angle of 45\u00b0 in clockwise Figure 14.8 direction with X-axis. So, the inclination of the line L is 180\u00b0 \u2212 45\u00b0 = 135\u00b0. \u2234 The slope of the line L is m = tan 135\u00b0 = \u22121. Some Results on the Slope of a Line 1.\t The slope of a horizontal line is zero. Hence, \t(i)\tSlope of X-axis is zero. \t(ii)\tSlope of any line parallel to X-axis is also zero.","Coordinate Geometry 14.9 2.\t The slope of a vertical line is not defined. Hence, Y \t(i)\tSlope of Y-axis is undefined. L \t(ii)\tSlope of any line parallel to Y-axis is also undefined. Theorem 1\u2002 Two non-vertical lines are parallel, if and only if, 45\u00b0 X their slopes are equal. O X X Proof: Let L1 and L2 be two non-vertical lines with slopes m1 Figure 14.9 and m2 respectively. If q1 and q2 are the inclinations of the lines, Y L1 L1 L1 and L2 respectively, then m1 = tan\u2009q1 and m2 = tan\u2009q2, since L1 || L2 .Then, q1 = q2 (see Fig. 14.10). ( They form a pair of corresponding angles) \u21d2 tan\u2009q1 = tan\u2009q2 \u21d2 m1 = m2 Conversely: Let m1 = m2 \u21d2 tan\u2009q1 = tan\u2009q2 \u03b8 1 \u03b82 \u21d2 q1 = q2 O \u21d2 L1 || L2 Figure 14.10 ( q1 and q2 form a pair of corresponding angles.) Hence, two non-vertical lines are parallel, if and only if, Y L2 their slopes are equal. L1 Theorem 2\u2002 Two non-vertical lines are perpendicular to each other, if and only if, the product of their slopes is \u22121. Proof: Let L1 and L2 be two non-vertical lines with slopes m1 and m2. If q1 and q2 are the inclinations of the lines L1 and L2 respectively, then m1 = tan\u2009q1 and m2 = tan\u2009q2 (see Fig. 14.11). If L1 \u22a5 L2, then q2 = 90\u00b0 + q1 90\u00b0 \u03b81 \u03b82 ( The exterior angle of a triangle is equal to the sum of two O opposite interior angles.) Figure 14.11 \u21d2 tan\u2009q2 = tan(90\u00b0 + q1) \u21d2 tan\u2009q2 = \u2212cotq1 \u21d2 tan\u2009q2 = \u22121 [ q1 \u2260 0] tan \u03b81 \u21d2 tanq1 \u00d7 tan\u2009q2 = \u22121 \u2234 m1m2 = \u22121. Conversely: Let m1m2 = \u22121 \u21d2 tanq1tanq2 = \u22121 \u21d2 tanq2 = \u22121 [ q1 \u2260 0] tan \u03b81 \u21d2 tanq2 = \u2212cotq1 \u21d2 tanq2 = tan(90\u00b0 + q1) \u21d2 q2 = 90\u00b0 + q1 \u21d2 L1 \u22a5 L2","14.10 Chapter 14 Hence, two non-vertical lines are perpendicular to each other, if and only if, the product of their slopes is \u22121. The Slope of a Line Passing through Points (x1, y1) and (x2, y2) Let A(x1, y1) and B(x2, y2) be two given points. Y Let AB be the straight line passing through points A and B(x2, y2) B. Let q be the inclination of line AB. A(x1, y1) \u03b8 N Draw perpendiculars AL and BM on to X-axis from A \u03b8 M X and B respectively. Also, draw AN \u22a5 BM (see Fig. 14.12). OL Then, let \u2220NAB = q. Figure 14.12 Here, BN = BM \u2212 MN = BM \u2212 AL = y2 \u2212 y1 AN = LM = OM \u2212 OL = x2 \u2212 x1 \u2234 The slope of the line L is m = tan\u2009q = BN = y2 \u2212 y1 . AN x2 \u2212 x1 Hence, the slope of a line passing through points (x1, y1) and (x2, y2) is m = y2 \u2212 y1 . x2 \u2212 x1 The following table gives inclination (q) of the line and its corresponding slope (m) for some particular values of q. q 0\u00b0 30\u00b0 45\u00b0 60\u00b0 90\u00b0 120\u00b0 135\u00b0 150\u00b0 m = tan\u2009q \u22123 \u22121 \u2212 1 01 1 3\u221e 3 3 \u2002Note\u2002\u2002 If points A, B and C are collinear, then the slope (m1) of AB = the slope (m2) of BC. m1 m2 AB C Example 14.14 Find the slope of the line joining points (3, 8) and (-9, 6). Solution Let A(3, 8) and B(-9, 6) be the given points. Then, the slope of AB = y2 \u2212 y1 x2 \u2212 x1 = 6\u22128 = 1 . \u22129 \u2212 3 6","Coordinate Geometry 14.11 Example 14.15 \u22121 3 Find the value of p if the slope of the line joining points (5, -p) and (2, -3) is\u2009 . Solution Let the given points be A(5, -p) and B(2, -3). Given, the slope of AB = \u22121 . 3 That is, \u22123 \u2212 (\u2212 p) = \u22121 2\u22125 3 p\u22123 \u22121 \u21d2 \u22123 = 3 \u21d2 p\u22123=1 \u21d2 p = 4. Example 14.16 Find the value of k, if lines AB and CD are perpendicular, where A = (4, 5), B = (k + 2, -3), C = (-3, 2) and D = (2, 4). Solution y2 \u2212 y1 x2 \u2212 x1 The slope of AB (m1) = = \u22123 \u2212 5 4 = \u22128 . (k + 2) \u2212 k\u22122 Slope of CD (m2) = 4\u22122 = 2 . 2 \u2212 (\u22123) 5 Since, AB \u22a5 PQ \u21d2 m1m2 = \u22121 That is, \u22128 \u00d7 \uf8eb 2 \uf8f6 = \u22121 k\u22122 \uf8ed\uf8ec 5 \uf8f8\uf8f7 \u21d2 16 =1 5k \u2212 10 \u21d2 16 = \u221210 + 5k \u21d2 k = 26 . 5 Example 14.17 Find the value of k, if points (-2, -4), (k, -2) and (3, 4) are collinear. Solution Let the given points be A(-2, -4), B(k, -2) and C(3, 4). The slope of AB = \u22122 + 4 = k 2 2 . k+2 +","14.12 Chapter 14 The slope of BC = 4+ 2 = 3 6 k . 3\u2212 k \u2212 Since the points A, B and C are collinear, The slope of AB = the slope of BC \u21d2 k 2 2 = 3 6 + \u2212k \u21d2 2(3 \u2212 k) = 6(k + 2) \u21d2 3 \u2212 k = 3k + 6 \u21d2 4k = \u22123 \u21d2 k = \u22123 . 4 Example 14.18 Find the ortho-centre of the DABC formed by vertices A(1, 6), B(5, 2) and C(12, 9). Solution The given vertices of DABC are A(1, 6), B(5, 2) and C(12, 9). Slope of AB = 2\u22126 = \u22124 = \u22121 5\u22121 4 Slope of BC = 9\u22122 = 7 = 1 12 \u2212 5 7 Slope of AC = 9\u22126 = 3 12 \u2212 1 11 Slope of AB \u00d7 Slope of BC = -1 \u2234 AB \u22a5 BC Hence, ABC is a right triangle, right angle at B. Hence, ortho-centre is the vertex containing right angle, i.e., B(5, 2). Intercepts of a Straight Line Y Say a straight line L meets X-axis in A and Y-axis in B. \u2022B X Then, OA is called the x-intercept and OB is called the \u2022 y-intercept (see Fig. 14.13). OA \u2002Note\u2002\u2002 OA and OB are taken as positive or negative, based on whether the line meets positive or negative axes. Figure 14.13 Example 14.19 The line l in Fig. 14.14 meets X-axis at A(-5, 0) and Y-axis at B(0, -3). Solution Hence, x-intercept = -5 and y-intercept = -3.","Coordinate Geometry 14.13 Y \u00d7A(\u20135, 0) O X l \u00d7 B(0,\u20133) Figure 14.14 Equation of a Line in General Form An equation of the form, ax + by + c = 0 (where |a| + |b| \u2260 0, i.e., a and b are not simultaneously equal to zero), which is satisfied by every point on a line is called the equation of a line. Equations of Some Standard Lines Equation of X-axis We know that the y-coordinate of every point on X-axis is zero. So, if P(x, y) is any point on X-axis, then y = 0. Hence, the equation of X-axis is y = 0. Equation of Y-axis We know that the x-coordinate of every point on Y-axis is zero. So, if P(x, y) is any point on Y-axis, then x = 0. Hence, the equation of Y-axis is x = 0. Equation of a Line Parallel to X-axis Y Let L be a line parallel to X-axis and at a distance of k units away L from X-axis. k OX Then the y-coordinate of every point on the line L is k. So, if P(x, y) is any point on the line L, then y = k. Figure 14.15 Hence, the equation of a line parallel to X-axis at a distance of k units from it, is y = k (see Fig. 14.15). \u2002Note\u2002\u2002 For the lines lying below X-axis, k is taken as negative. Equation of a Line Parallel to Y-axis Y L\u2032 k Let L\u2032 be a line parallel to Y-axis and at a distance of k units away from it. Then the x-coordinate of every point on the line L\u2032 is k. O X So, if P(x, y) is any point on the line L\u2032, then x = k. Figure 14.16 Hence, the equation of a line parallel to Y-axis and at a distance of k units from it, is x = k (see Fig. 14.16). \u2002Note\u2002\u2002 For the lines lying on the left side of Y-axis, k is taken as negative.","14.14 Chapter 14 Oblique Line A straight line which is neither parallel to X-axis nor parallel to Y-axis is called an oblique line or an inclined line. Different Forms of Equations of Oblique Lines Gradient Form (or) Slope Form\u2002 The equation of a straight line with slope m and passing through the origin is given by y = mx. Point\u2013Slope Form\u2002 The equation of a straight line passing through point (x1, y1) and with slope m is given by y \u2212 y1 = m(x - x1). Slope\u2013intercept Form\u2002 The equation of a straight line with slope m and having y-intercept as c is given by y = mx + c. \u2002Note\u2002\u2002 Area of triangle formed by the line y = mx + c is 1 c2 \u2009sq. units. 2 m Two-point Form\u2002 The equation of a straight line passing through points (x1, y1) and (x2, y2) is given by y \u2212 y1 = y2 \u2212 y1 (x \u2212 x1 ) or y \u2212 y1 = x \u2212 x1 . x2 \u2212 x1 y2 \u2212 y1 x2 \u2212 x1 Intercept Form\u2002 The equation of a straight line with x-intercept as a and y-intercept as b is given x y by a + b =1. \u2002Note\u2002\u2002 Area of triangle formed by line x y = 1 with the coordinate axes is 1 ab \u2009sq. units. a +b 2 Example 14.20 Find the equation of the line parallel to Y-axis and passing though point (5, -7). Solution The equation of a line parallel to Y-axis is x = k. Given, the line passes through point (5, -7) \u21d2 k = 5. Hence, the equation of the required line is x = 5. That is, x - 5 = 0. Example 14.21 Find the equation of the line passing through (3, 4) and having a slope\u200945 . Solution The equation of the line passing through (x1, y1) and having slope m is given by y - y1 = m(x - x1). Hence, the equation of the required line is y - 4 = 4 (x - 3) 5 5y - 20 = 4x - 12 4x - 5y + 8 = 0.","Coordinate Geometry 14.15 Example 14.22 Find the equation of a line making intercepts 4 and 5 on the coordinate axes. Solution Given, x-intercept (a) = 4 and y-intercept (b) = 5. \u2234 The equation of the required line is\u2009 x + y = 1. a b That is, x + y =1 4 5 \u21d2 5x + 4y - 20 = 0. Equation of a Line Parallel or Perpendicular to the Given Line Let ax + by + c = 0 be the equation of a straight line, then: 1.\t The equation of a line passing through point (x1, y1) and parallel to the given line: \t The slope of the required line (m) = The slope of ax + by + c = 0 \u2212a \t = b \u2003 (Since the lines are parallel) \t \u2234 The required line is y \u2212 y1 = m(x \u2212 x1 ) \u21d2 y \u2212 y1 = \u2212a (x \u2212 x1 ) b \u21d2 b(y \u2212 y1 ) = \u2212a(x \u2212 x1 ) \u21d2 a(x \u2212 x1 ) + b(y \u2212 y1 ) = 0. 2.\t \u0007The equation of a line passing through point (x1, y1) and perpendicular to the given line: \t The slope of ax + by + c = 0 is \u2212a . b \t \u2234 The slope of the required line is b . (Since the line are perpendicular.) a \t \u2234 The required line is (y \u2212 y1 ) = m(x \u2212 x1 ) \t \u21d2 (y \u2212 y1 ) = b (x \u2212 x1 ) a \u21d2 b(x \u2212 x1 ) \u2212 a(y \u2212 y1 ) = 0. Example 14.23 Find the equation of a line passing through the point P(-3, 2) and parallel to line 4x - 3y - 7 = 0. Solution Here, (x1, y1) = (-3, 2), a = 4 and b = 2. \u2234 Equation of the line passing through P(-3, 2) and parallel to 4x - 3y - 7 = 0."]