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Exercise: 13.1 (Page No: 244), , 1. 2 cubes each of volume 64 cm? are joined end to end. Find the surface area of the resulting, cuboid., , Answer:, The diagram is given as:, , , , , , , , , , , , - ae ee Py, Le J cm, , Given,, , The Volume (V) of each cube is = 64 cm?, This implies that a? = 64 cm?, , “a=4cm, , , , Now, the side of the cube =a=4cm, , Also, the length and breadth of the resulting cuboid will be 4 cm each. While its height will be 8, cm., , So, the surface area of the cuboid = 2(lb+bh+lh), , = 2(8x4+4x4+4x8) cm?, , = 2(32+16+32) cm?, , = (2x80) cm? = 160 cm?, , 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., , Answer:, The diagram is as follows:
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| ieee =~. |13 om, , , , , , , , , , Now, the given parameters are:, , The diameter of the hemisphere = D = 14 cm, , The radius of the hemisphere = r = 7 cm, , Also, the height of the cylinder = h = (13-7) =6 cm, , And, the radius of the hollow hemisphere = 7 cm, , Now, the inner surface area of the vessel = CSA of the cylindrical part + CSA of hemispherical, part, , => (2mrh+2mr?) cm? = 2nr(h+r) cm?, , => 2x(22/7)x7(6+7) cm? = 572 cm?, , 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., , Answer:, The diagram is as follows:, , A, fy’, , , , , , —35om Sele”, , Given that the radius of the cone and the hemisphere (r) = 3.5 cm or 7/2 cm, The total height of the toy is given as 15.5 cm., So, the height of the cone (h) = 15.5-3.5 =12 cm
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Slant height of the cone(I)= VA? + r?, => t= 12? + (3.5), , => 1=,/12? + (7/2), , , , => 1=,/144 + 49/4=./(576 + 49)/4 = (625/74, , => 1=25/2, “. The curved surface area of cone = mrl, => (22/7)x(7/2)x(25/2) = 275/2 cm?, , Also, the curved surface area of the hemisphere = 2mr?, => 2x(22/7)x(7/2)?, = 77 etn", , Now, the Total surface area of the toy = CSA of cone + CSA of hemisphere, = (275/2)+77 cm?, , = (275+154)/2 cm?, , = 429/2 cm? = 214.5cm?, , So, the total surface area (TSA) of the toy is 214.5cm?, , 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., , Answer:, It is given that each side of cube is 7 cm. So, the radius will be 7/2 cm.
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We know,, , The total surface area of solid (TSA) = surface area of cubical block + CSA of hemisphere - Area, of base of hemisphere, , . TSA of solid = 6x(side)?+2mr?-mr?, , = 6x(side)?+mr?, , = 6x(7)?+(22/7)x(7/2)x(7/2), , = (6x49)+(77/2), , = 294+38.5 = 332.5 cm?, , So, the surface area of the solid is 332.5 cm?, , 5. A hemispherical depression is cut out from one face of a cubical wooden block such that, the diameter | of the hemisphere is equal to the edge of the cube. Determine the surface area, , of the remaining solid., , Answer:, The diagram is as follows:, , , , , , Now, the diameter of hemisphere = Edge of the cube =|, , So, the radius of hemisphere = I/2, , .. The total surface area of solid = surface area of cube + CSA of hemisphere - Area of base of, hemisphere, , = TSA of remaining solid = 6 (edge)?+2mr?-mr?, , = 6? + mr?, , = 6[*+1(1/2)?, , = 6|7+nl2/4
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= |?/4(24+m) sq. units, , 6. A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its, ends. The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find, , its surface area., , 5mm, , 14mm, , Answer:, Two hemisphere and one cylinder are shown in the figure given below., , , , , t eo, ( a, eI!, =| €, “= ~, wy m, | pee ae, \, . *s, , ~, , elie, 5 mm, , , , Here, the diameter of the capsule = 5 mm, , ”. Radius = 5/2 =2.5 mm, , Now, the length of the capsule = 14 mm, , So, the length of the cylinder = 14-(2.5+2.5) =9 mm, , .. The surface area of a hemisphere = 2mr? = 2x(22/7)x2.5x2.5, = 275/7 mm?, , Now, the surface area of the cylinder = 2mrh, = 2x(22/7)x2.5x9, => (22/7)x45 = 990/7 mm?
