Tiêu đề 04 Surface Area _ Volume(1).pdf ✅

SURFACE AREA & VOLUME 4 CHAPTER ➢ FORMULAE 1. If , b and h denote respectively the length, breadth and height of a cuboid, then - (i) total surface area of the cuboid = 2 (b+bh+h) square units. (ii) Volume of the cuboid = Area of the base × height = bh cubic units. (iii) Diagonal of the cuboid or longest rod = 2 2 2  + b + h units. (iv) Area of four walls of a room = 2 ( + b) h sq. units. 2. If the length of each edge of a cube is ‘a’ units, then- (i) Total surface area of the cube = 6a2 sq. units. (ii) Volume of the cube = a3 cubic units (iii) Diagonal of the cube = 3 a units. 3. If r and h denote respectively the radius of the base and height of a right circular cylinder, then - (i) Area of each end = r 2 (ii) Curved surface area = 2rh = (circumference) height (iii) Total surface area = 2r (h + r) sq. units. (iv) Volume = r 2h = Area of the base × height 4. If R and r (R > r) denote respectively the external and internal radii of a hollow right circular cylinder, then - (i) Area of each end = (R2 – r 2 ) (ii) Curved surface area of hollow cylinder = 2 (R + r) h (iii) Total surface area = 2 (R + r) (R + h – r) (iv) Volume of material = h (R2 – r 2 ) 5. If r, h and  denote respectively the radius of base, height and slant height of a right circular cone, then- (i)  2 = r2 + h2 (ii) Curved surface area = r (iii) Total surface area = r 2 + r (iv) Volume = 3 1 r 2h 6. For a sphere of radius r, we have (i) Surface area = 4r 2 (ii) Volume = 3 4 r 3 7. If h is the height,  the slant height and r1 and r2 the radii of the circular bases of a frustum of a cone then - (i) Volume of the frustum = 3  (r1 2 + r1 r 2 + r2 2 ) h (ii) Lateral surface area =  (r1 + r2 )  (iii) Total surface area = {(r1 + r2 )  + r1 2 + r2 2} (iv) Slant height of the frustum = 2 1 2 2 h + (r − r ) (v) Height of the cone of which the frustum is a part = 1 2 1 r r hr − (vi) Slant height of the cone of which the frustum is a part = 1 2 1 r r r −  (vii) Volume of the frustum
= 3 h A1 + A2 + A1A2  , where A1 and A2 denote the areas of circular bases of the frustum. ❖ EXAMPLES ❖ Ex.1 A circus tent is in the shape of a cylinder, upto a height of 8 m, surmounted by a cone of the same radius 28 m. If the total height of the tent is 13 m, find: (i) total inner curved surface area of the tent. (ii) cost of painting its inner surface at the rate of −j 3.50 per m2 . Sol. According to the given statement, the rough sketch of the circus tent will be as shown: (i) For the cylindrical portion : r = 28 and h = 8 m  Curved surface area = 2rh = 2 × 7 22 × 28 × 8 m2 = 1408 m2 8m 13m 28m For conical portion : r = 28 m and h = 13 m – 8 m = 5 m   2 = h2 + r2   2 = 52 + 282 = 809   = 809 m = 28.4 m  Curved surface area = r =  7 22 28 × 28.4 m2 = 2499.2 m2 Total inner curved surface area of the tent. = C.S.A. of cylindrical portion + C.S.A. of the conical portion 1408 m2 + 2499.2 m2 = 3907.2 m2 (ii) Cost of painting the inner surface = 3907.2 × −j 3.50 = −j 13675.20 Ex.2 A cylinder and a cone have same base area. But the volume of cylinder is twice the volume of cone. Find the ratio between their heights. Sol. Since, the base areas of the cylinder and the cone are the same.  their radius are equal (same). Let the radius of their base be r and their heights be h1 and h2 respectively. Clearly, volume of the cylinder = r 2h1 and, volume of the cone = 3 1 r 2h2 Given : Volume of cylinder = 2 × volume of cone  r 2h1 = 2 × 3 1 r 2h2  h1 = 2 h 3 2  3 2 h h 2 1 = i.e., h1 : h2 = 2 : 3 Ex.3 Find the formula for the total surface area of each figure given bellow : (i)  r (ii) h (iii) r  h (iv) r Sol.(i) Required surface area = C.S.A. of the hemisphere + C.S.A. of the cone
= 2r 2 + r = r (2r + ) (ii) Required surface area = 2 × C.S.A. of a hemisphere + C.S.A. of the cylinder = 2 × 2r 2 + 2rh = 2r (2r + h) (iii) Required surface area = C.S.A. of the hemisphere + C.S.A. of the cylinder + C.S.A. of the cone = 2r 2 + 2rh + r = r (2r + 2h + ) (iv) If slant height of the given cone be  =  2 = h2 + r2   = 2 2 h + r And, required surface area = 2r 2 + r = r (2r + ) = r       + + 2 2 2r h r Ex.4 The radius of a sphere increases by 25%. Find the percentage increase in its surface area. Sol. Let the original radius be r.  Original surface area of the sphere = 4r 2 Increase radius = r + 25% of r = r + 4 5r r 100 25 =  Increased surface area =4 4 25 r 4 5r 2 2   =      Increased in surface area = 2 2 – 4 r 4 25 r   = 4 9 r 4 25 r 16 r 2 2 2  =  −  and, percentage increase in surface area 100% Original aera Increase in area =  16 9 100% 4 r 4 9 r 2 2  =   = × 100% = 56.25% Alternative Method : Let original radius = 100  Original C.S.A. = (100)2 = 10000 Increased radius = 100 + 25% of 100 = 125  Increased C.S.A. = (125)2 = 15625 Increase in C.S.A. = 15625 – 10000 = 5625  Percentage increase in C.S.A. = 100% Original C.S.A Increase in C.S.A.  = 100% 1000 5625    = 56.25% If the radius increases by 25%, the diameter also increases by 25%. Conversely, if diameter decreases by 20%, the radius also decreases by 20%. Ex.5 Three solid spheres of radii 1 cm, 6 cm and 8 cm are melted and recasted into a single sphere. Find the radius of the sphere obtained. Sol. Let radius of the sphere obtained = R cm.  3 3 3 3 (8) 3 4 (6) 3 4 (1) 3 4 R 3 4  =  +  +  . R 3 = 1 + 216 + 512 = 729  R = (93 ) 1/3 = 9 cm Ans. Ex.6 In given figure the top is shaped like a cone surmounted by a hemisphere. The entire top is 5 cm in height and the diameter of the top is 3.5 cm. Find the area it has to colour. (Take  = 7 22 ) 3.5 cm 5 cm Sol. TSA of the top = CSA of hemisphere + CSA of cone Now, the curved surface of the hemisphere = 2 1 (4r 2 ) = 2r 2 =          2 3.5 2 3.5 7 22 2 cm2 Also, the height of the cone
= height of the top – height (radius) of the hemispherical part =       − 2 3.5 5 cm = 3.25 cm So, the slant height of the cone (l) = 2 2 r + h = 2 2 (3.25) 2 3.5  +      cm = 3.7 cm (approx.) Therefore, CSA of cone = rl =        3.7 2 3.5 7 22 cm2 This gives the surface area of the top as          2 3.5 2 3.5 7 22 2 cm2+        3.7 2 3.5 7 22 cm2 = 7 22 × 2 3.5 (3.5 + 3.7)cm2 = 2 11 ×(3.5 + 3.7) cm2 = 39.6 cm2 (approx) Ex.7 The decorative block shown in figure is made of two solids — a cube and a hemisphere. The base of the block is a cube with edge 5 cm, and the hemisphere fixed on the top has a diameter of 4.2 cm. Find the total surface area of the block. (Take  = 7 22 ) 5 cm 5 cm 5 cm 4.2cm Sol. The total surface area of the cube = 6 × (edge)2 = 6 × 5 × 5 cm2 = 150 cm2 . Note that the part of the cube where the hemisphere is attached is not included in the surface area. So, the surface area of the block = TSA of cube – base area of hemisphere + CSA of hemisphere = 150 – r 2 + 2r 2 = (150 + r 2 ) cm2 = 150 cm2 +         2 4.2 2 4.2 7 22 cm2 = (150 + 13.86) cm2 = 163.86 cm2 Ex.8 A wooden toy rocket is in the shape of a cone mounted on a cylinder, as shown in figure. The height of the entire rocket is 26 cm, while the height of the conical part is 6 cm. The base of the conical portion has a diameter of 5 cm, while the best diameter of the cylindrical portion is 3 cm. If the conical portion is to be painted orange and the cylindrical portion yellow, find the area of the rocket painted with each of these colours. (Take  = 3.14) 26 cm 6 cm 3 cm base of cone base of cylinder 5 cm Sol. Denote radius of cone by r, slant height of cone by , height of cone by h, radius of cylinder by r' and height of cylinder by h'. Then r = 2.5 cm, h = 6 cm, r' = 1.5 cm, h' = 26 – 6 = 20 cm and l = 2 2 r + h = 2 2 2.5 + 6 cm = 6.5 cm Here, the conical portion has its circular base resting on the base of the cylinder, but the base of the cone is larger than the base of the cylinder. So a part of the base of the cone (a ring) is to be painted. So, the area to be painted orange = CSA of the cone + base area of the cone – base area of the cylinder = rl + r 2 – (r')2 = [(2.5 × 6.5) + (2.5)2 – (1.5)2 ] cm2 = [20.25] cm2 = 3.14 × 20.25 cm2 = 63.585 cm2 Now, the area to be painted yellow