EXERCISE
Unless stated otherwise, take π =$22/7$1. 2 cubes each of volume 64 $cm^3$ are joined end to end. Find the surface area of the resulting cuboid.
If $l$ is the side of the cube, then volume of cube= $l^3$ Volume of cube is given as= 64 $cm^3$ Therefore $l^3$ = 64 i.e $l=4$ $cm$ If the 2 cubes are joined side by side, then the sides of the new cuboid are 4 $cm$, 8 $cm$ and 4 $cm$. If $l$ ,$b$ and $h$ are the length, breadth and height of the cuboid respectively, then, surface area of cuboid is given as = 2($l×b +b×h+ l×h$) = $2(4×8+8×4+4×4)$ = $2(32+32+16)$ =$160$ $cm^2$2. A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel. Curved Surface Area of Hollow Hemisphere=$2πr^2$ Curved Surface Area of circular Cylinder = $2πrh$ Total Surface Area of inner vessel= $2πr^2$ + $2πrh$ Given , Diameter of Hemisphere = 14 $cm$ Therefore radius $r$ of the hemisphere= 7 $cm$ Total height of the vessel = 13 $cm$ Therefore height of cylinder ($h$) = Total height- radius of height of hemisphere $h$ =(13-7) = 6 $cm$ Curved Surface Area of Hollow Hemisphere=$2πr^2$ = $2×22/7×7×7$ =$308$ $cm^2$ Curved Surface Area of circular Cylinder = $2πrh$ = $2×22/7×7×6$ =$264$ $cm^2$ Total Surface Area of inner vessel= $2πr^2$ + $2πrh$ = $308$ + $264$ = $572$ $cm^2$3. A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy. Total Surface area of the toy= curved surface area of hemisphere+ curved surface area of cone Given: radius of hemisphere = 3.5 $cm$ Total height of the toy= 15.5 $cm$ = height of cone+ height of hemisphere Therefore height of the cone = $15.5 - 3.5$ $cm$ $h$ = $12$ $cm$ Curved surface area of hemisphere = $2πr^2$ = $2×22/7×3.5×3.5 $ =$77$ $cm^2$ Curved surface area of cone= $πrl$ where $l=√(h^2+r^2)$ $l=√(12^2+3.5^2)$4. A cubical block of side 7 $cm$ is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid. Greatest diameter of the hemisphere which can be mounted on the cube = side of the cube. Hence diameter of hemisphere =$d$= 7 $cm$ Surface area of the solid= Surface area of cube+Curved Surface area of hemisphere - area of the base of hemisphere. = 6$a^2$ + $2πr^2$ - $πr^2$ = 6$a^2$ + $πr^2$ = $6 × 7×7 $+ $22/7×(7/2)×(7/2)$ = $294$ + $77/2$ = $665/2$ = 332.5 $cm^2$5. A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter $l$ of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid. Given, sides of the cube= diameter of the hemisphere = $l$ Surface area of remaining solid = S.A of cube + CSA of hemisphere - area of base of hemisphere. = $6l^2$ + $2πr^2$ - $πr^2$ = $6l^2$ + $πr^2$ = $6l^2$ + ${πl^2}/4$ = $6l^2$ + $22/7×{l^2}/4$ = $6l^2$ + $11/14×{l^2}$ = ${84+11}/14 ×{l^2}$ = ${95l^2}/14$6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see Fig.). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area. Let radius of hemisphere be $r$ = $5/2$ $cm$ Since the total length of the capsule = $14$ $cm$ Therefore the length of cylinder = length of capsule - radius of 1st hemisphere - radius of 2nd hemisphere. Lenght of the cylinder = $14-2.5 - 2.5$ = $9$ $cm$ Surface area of capsule = CSA of 1st hemisphere + CSA of cylinder + CSA of 2nd hemisphere = $2πr^2$ + $2πrh$ +$2πr^2$ = $4πr^2$ + $2πrh$ = $4π(5/2)(5/2)$ + $2π(5/2) (9)$ = $25π$ + $45π$ = $70π$ = $70× 22/7$ = $220$ $cm^2$
7. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of ₹500 per $m^2$. Curved surface area of the tent = CSA of the cylinder part of the tent + CSA of the connical part of the tent = $2πrh$ + $πrl$ = $πr(2h+l)$ = $22/7×2(2×2.1+4.8)$ = $22/7×2(4.2+4.8)$ = $22/7×2(7)$ = $22×2$ = $44$ $m^2$ Cost of canvas = ₹500 per $m^2$ Therefore total cost of canvas for the tent of $44$ $m^2$ = $500 × 44 $ = ₹ $22,000$8. From a solid cylinder whose height is 2.4 $cm$ and diameter 1.4 $cm$, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest $cm^2$. Total SA of the remaining solid = CSA of cylinder + area of the top of cylinder+ CSA of the cone. = $2πrh$ + $πr^2$ + $πrl$ Given: $h$= 2.4 $cm$ ; $r$ = $1.4/2$ $cm$ therefore slant height of the cone $l$ = $√(h^2+r^2)$ = $√({2.4}^2+0.7^2)$ = $√(5.74+0.49)$ = $√6.25$ = $2.5$ $cm$ Total SA of the remaining solid = $2πrh$ + $πr^2$ + $πrl$ = $2×22/7×0.7×2.4$ + $22/7×0.7×0.7$ + $22/7×0.7×2.5$ = $2×22×0.1×2.4$ + $22×0.1×0.7$ + $22×0.1×2.5$ = $22(2×0.1×2.4+ 0.1×0.7+0.1+2.5)$ = $22(0.48+ 0.07+0.25)$ = $22(0.8)$ = $17.6$ $cm^2$ 9. A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 13.11. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.
