322 Applied Math Mensuration of Solid Chapter 15 Mensuration of Solid 15.1 Solid: It is a body occupying a portion of three-dimensional space and therefore bounded by a closed surface which may be curved (e.g., sphere), curved and planer (e.g., cylinder) or planer (e.g., cube or prism). 15.2 Mensuration of Prisms: Prism: A solid bounded by congruent parallel bases or ends and the side faces (called the lateral faces) are the parallelograms, formed by joining the corresponding vertices of the bases. It is called a right prism if the lateral are rectangles. Otherwise an oblique prism. A common side of the two lateral faces is called a lateral edge. Prism are named according to the shape of ends. A prism with a square base, a rectangular base, a hexagonal base, and a parallelogram base, is called a square prism, a rectangular prism, a hexagonal prism and a parallelepiped respectively. Altitude: The altitude of a prism is the vertical distance from the centre of the top to the base of the prism. Fig. 15.1 Axis: The axis of a prism is the distance between the centre of the top to the centre of the base. In a right prism the altitude, the axis and the latera edge are the same lengths. In the Fig. 1 the lateral faces are OAEC, BDGF, OBCF dan ADEG. The bases are OABD and ECFG. LM = h is the axis of the prism, where L and M are the centres of the bases. 15.3 Surface Area of a Prims: Lateral surface area of a prism is the sum of areas of the lateral faces. From Fig. 1. Lateral surface area = Area of (OAEC + ADEG + BDGF + OBCF) = (OA)h + (AD)h + (DB)h + (OB)h 323 Applied Math Mensuration of Solid = (OA + AD + DB + BO)h = Perimter of the base x height of the prism (1) Total surface area = Lateral surface area + Area of the bases (2) Note: that, Total surface, surface area, total surface area and the surface of any figure represent the same meanings. 15.4 Volume of a Prism: A sold occupies an amount of space called its volume. Certain solids have an internal volume or cubic capacity. (The term capacity, when not associated with cubic, is usually reserved for the volume of liquids or materials which pour, and special sets of units e.g.; gallon, litre, are used). The volume of solid is measured as the total number of unit cubes that it contains. If the solid is a Prism the volume can be computed directly from the formula, V = l . b . h Where, l, b and h denoted the length, breadth and height of the prism respectively. Also l . b denotes the area A of the base of the prism. Then Volume of the Prims = Ah = Area of the base x height of the prism (4) Fig. 15.2 l = 6 units b = 3 units and h = 5 units The total number of unit cubes = l b h = 6 x 3 x 5 = 90 So volume of prism = 90 cubic unit Weight of solid = volume of solid x density of solid (density means weight of unit volume) 15.5 Types of Prism: 1. Rectangular Prism: If the base of prism is a rectangle, it is called a rectangular prism. Consider a rectangular prism with length a, breadth b and height c (Fig. 15.3) 324 Applied Math Mensuration of Solid Fig. 15.3 (i) Volume of rectangular prism = abc cu. Unit (ii) Lateral surface area = area of four lateral faces = 2 ac + 2bc sq. unit = (2a + 2b)c = Perimetr of base x height (iii) Total surface area = Area of six faces = 2ab + 2ac + 2ca = 2 (ab + bc + ca) sq. unit (iv) Length of the diagonal OG In the light triangle ODG, by Pythagorean theorem, OG2 = OD2 + DG2 ……………….. (1) = OD2 + C2 ………… (I) Also in the right triangle OAD, OD2 = OA2 + AD2 = a2 + b2 Put OD2 in equation (1) (the line joining the opposite corners of the rectangular prism is called its diagonal). OG2 = a2 + b2 + c2 Or |OG| = a2 b 2 c 2 2. Cube: A cube is a right prism will all sides equals. Let ‘a’ be the side of the cube Fig. 4 Fig. 15.4 325 Applied Math Mensuration of Solid (i) Volume of the cube = a . a. a = a3 cu . unit (ii) Lateral surface area = Area of four lateral faces = 2a . a + 2a . a = 4a2 sq. unit (iii) Total surface area = Area of six faces = 6a2 sq. unit (iv) The length of the diagonal |OG| = aaa222= a 3 Example 1: Find the volume, total surface, diagonal and weight of rectangular block of wood 7.5 cm long, 8.7 cm wide and 12 cm deep, 1 cu. cm = 0.7 gm. Solution: Let a = 7.5 cm, b = 8.7 cm, and c = 12cm Then (i) Volume = abc = 7.5 x 8.7 x 12 = 783.00 cu. cm. (ii) Total surface = 2 (ab + bc + ca) = 2 (65.25 + 104.4 + 90) = 519.3 sq. cm. (iii) diagonal = = 7.52 8.7 2 12 2 = 275.94 16.6 cm (iv) Weight = Volume x density = 783 x 0.7 = 548.1 gms Example 2: The side of a triangular prism are 25, 51 and 52 cm. and height is 60 cm. Find the side 2of a cube 2 of 2 equivalent volume. Solution: a b c Volume of a prism = Area of base x height = S(s a)(s b)(s c) x h When a = 25cm, b = 51cm, c = 52cm, h = 60cm a + b + c 25 51 52 S = 64 22 S – a = 64–25 = 39, S – b = 64–51 = 13, S – c = 64–52 = 12 Volume of prism = 64(39)(13)(12) x 60 37440 cu. cm Volume of a cube of side l cm = l3 cu. cm. l3 = 37440 l = (37440)13 = 33.45 cu. cm. 326 Applied Math Mensuration of Solid Example 3: The volume of the cube is 95 cu. cm. Find the surface area and the edge of the cube. Solution: Volume of cube = 95 cu. cm. Let ‘a’ be the side of cube, then Volume = a3 a3 = 95 a = (95)13 = 4.56 cm Surface area = 6a2 = 6(4.56)2 = 124.92 sq. cm Example 4: Find the number of bricks used in a wall 100 ft long, 10 ft high and 1 one and half brick in thickness. The size of each is 9x 4 x 3 2 Solution: a = Length of wall = 100 ft = 1200 inches 9 b = Breadth of wall = 9 = 13.5 inches 2 h = Height of wall = 10ft = 120 inches Volume of the wall = abh = 1200 x 13.5 x 120 = 1944000 cu. in Note: Length of brick = a = 9 inches 9 Breadth of brick = b = = 4.5 inches 2 Height of wall = h = 3 inches 243 Volume of the brick = abh = 9 x 9.5 x 3 = cu. in. 2 Volume of wall 1944000 Number of bricks = = = 16000 Volume of brick 243 2 Example 5: Find the edge of a cube whose volume is equal to that of a rectangular bar measuring 126 cm long, 4 cm wide and 2 cm thick. Solution: Given that a = length of bar = 126 cm b = breadth of bar = 4 cm h = 2 cm Volume of rectangular bar = abh = 126 x 4 x 2 = 1008 cu. cm Let ‘a’ be the edge of a cube 327 Applied Math Mensuration of Solid The volume of cube = a3 cu. cm. Since Volume of the cube = Volume of bar a3 = 1008 a = (1008)13= 10.002 cm 3. Polygonal Prism: If the base of prism is a polygon, the prism is called polygonal prism. a. Volume of the polygonal prism when base is a regular polygon of n sides and h is th height = Area of the base x height. n a2o 180 i. V = cot xh , when side a is given. 4n n R2o 360 ii. V = Sin x h , when radius R of circumscribed circle is 2n given. 180o iii. V = nr2 tan x h , when radius r of inscribed circle is given. n b. Lateral surface area = Perimeter of the base x height = n a x height, a is the side of the base c. Total surface area = Lateral surface area + Area of bases Example 6: A hexagonal prism has its base inscribed about a circle of radius 2cm and which has a height of 10cm is cast into a cube. Find the size of the cube. Solution: Hence n = 6, R = 2cm, h = 10cm Let ‘a’ be the side of the cube 360o Volume of the hexagonal prism = n R2 sin x h n = 6 x 4 sin 60o x 10 = 240 x 0.866 = 207.84 cu. Cm. Volume of the cube = Volume of the hexagonal prism a3 = 207..84 a = 5.923 cm Note : Weight of solid = Volume Density 328 Applied Math Mensuration of Solid 15.6 Frustum of a Prism: When a solid is cut by a plane parallel to its base (or perpendicular to its axis), the section of the solid is called cross-section. If however, the plane section is not parallel to the bases, the portion of the prism between the plane section and the base is called a frustum. In Fig. 5 ABCDEIGH represents a Frustum of a prism whose cutting plane EFGH is inclined an angle θ to the horizontal. In this case the Frustum can be taken as a prism whose base ABEF which is a Trapezium and height BC. i.