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Exercise - 13.1

Surface Area and Volume

**Question-1 :-** 2 cubes each of volume 64 cm^{3} are joined end to end. Find the surface area of the resulting cuboid. (Take π = 22/7)

Given that:- Volume of cubes = 64 cm^{3}(Edge)^{3}= 64 cm Edge = (64)^{1/3}cm Edge = 4cm If cubes are joined end to end, the dimensions of the resulting cuboid will be 4 cm, 4 cm, 8 cm. Therefore, Surface area of cuboids = 2(lb + bh + lh) = 2[(4 x 4) + (4 x 8) + (8 x 4)] = 2[16 + 32 + 32] = 2 x 80 = 160 cm^{2}

**Question-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. (Take π = 22/7)

Given that:- A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The Diameter of the hemisphere = 14 cm The Radius of the hemisphere = 14/2 = 7 cm Total height of the vessel = 13 cm Now, Height of hemispherical part = Radius = 7 cm (Because, height and radius both are equals in a hemispherical) Height of cylindrical part (h) = 13 − 7 = 6 cm Inner surface area of the vessel = CSA of cylindrical part + CSA of hemispherical part = 2πrh + 2πr^{2}= 2πr(h + r) = 2 x 22/7 x 7 x (6 + 7) = 44 x 13 = 572 cm^{2}

**Question-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. (Take π = 22/7)

Given that:- A toy is in the form of Cone. Radius(r) of the cone = 3.5 cm = Radius(r) of hemisphere Total height of the toy = 15.5 cm Now, Height of hemispherical part = Radius = 3.5 cm (Because, height and radius both are equals in a hemispherical) Height(h) of the cone = 15.5 - 3.5 = 12 cm Now, slant height(l) of the conical part = √r² + h² = √(3.5)^{2}+ (12)^{2}= √12.25 + 144 = √156.25 = 12.5 cm The total surface area of the toy = CSA of Cone part + CSA of hemispherical part = πrl + 2πr^{2}= πr(l + 2r) = 22/7 x 3.5 x [12.5 + (2 x 3.5)] = 22/7 x 35/10 x [12.5 + 7.0] = 11 x 19.5 = 214.5 cm^{2}

**Question-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. (Take π = 22/7)

Given that:- Side(Edge) of the Cube = 7cm It can be observed that the greatest diameter possible for such hemisphere is equal to the cube’s edge, i.e., 7cm. So, radius(r) of the hemispherical part = 7/2 = 3.5 cm Total surface area of solid = Surface area of cubical part + CSA of hemispherical part − Area of base of hemispherical part = 6 x (Edge)^{2}+ 2πr^{2}- πr^{2}= 6 x (Edge)^{2}+ πr^{2}= 6 x (7)^{2}+ 22/7 x 3.5 x 3.5 = 6 x 49 + 22 x 0.5 x 3.5 = 294 + 11 x 3.5 = 294 + 38.5 = 332.5 cm^{2}

**Question-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. (Take π = 22/7)

Given that:- Diameter of hemisphere = Edge of cube = l Radius(r) of the hemisphere = l/2 Therefore, Total surface area of solid = Surface area of cubical part + CSA of hemispherical part − Area of base of hemispherical part = 6 x (Edge)^{2}+ 2πr^{2}- πr^{2}= 6 x (Edge)^{2}+ πr^{2}= 6 x l^{2}+ π x (l/2)^{2}= 6 x l^{2}+ π x l^{2}/4 = 1/4(24 + π)l^{2}unit^{2}

**Question-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. (Take π = 22/7)

Given that:- In this figure, two hemisperical and one cylinderical part inculde. Diameter of capsule = 5 mm Radius (r) of cylindrical part = Radius (r) of hemispherical part = 5/2 mm Length of the entire capsule = 14 mm Now, length of cylinderical part(h) = Length of the entire capsule - 2r = 14 - (2 x 5/2) = 14 - 5 = 9 mm Therefore, Surface area of capsule = 2 × CSA of hemispherical part + CSA of cylindrical part = 2 x 2πr^{2}+ 2πrh = 2πr(2r + h) = 2 x 22/7 x 5/2 x [(2 x 5/2) + 9)] = 110/7 x 14 = 110 x 2 = 220 mm^{2}

**Question-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 Rs 500 per m^{2}. (Note that the base of the tent will not be covered with canvas.) (Take π = 22/7)

Given that:- A tent is in the shape of a cylinder surmounted by a conical top. Height(h) of the cylinderical part = 2.1 m Diameter of the cylinderical part = 4 m Radius(r) of the cylinderical part = 4/2 = 2 m = Radius(r) of the conical top surmounted on it Slant height(l) of the conical tent = 2.8 m Therefore, Area of canvas used (i.e., CSA Area of tent) = CSA of conical part + CSA of cylindrical part = πrl + 2πrh = πr(l + 2h) = 22/7 x 2 x [2.8 + (2 x 2.1)] = 22/7 x 2 x [2.8 + 4.2] = 44/7 x 7 = 44 m^{2}Now, Cost of 1 m^{2}canvas = Rs 500 Cost of 44 m^{2}canvas = 44 × 500 = 22000 Therefore, it will cost Rs 22000 for making such a tent.

**Question-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}. (Take π = 22/7)

Give that:- Height(h) of the solid cylinder = 2.4 cm Diameter of the solid cylinder = 1.4 cm Radius of the solid cylinder = 1.4/2 = 0.7 cm According to question, a conical cavity of the same height and same diameter is hollowed out. Slant height(l) of the conical part = √r² + h² = √(0.7)^{2}+ (2.4)^{2}= √0.49 + 5.76 = √6.25 = 2.5 cm Therefore, the total surface of the remaining solid will be = CSA of cylindrical part + CSA of conical part + Area of cylindrical base = 2πrh + πrl + πr^{2}= πr(2h + l + r) = 22/7 x 0.7 x [(2 x 2.4) + 2.5 + 0.7] = 2.2 x [4.8 + 3.2] = 2.2 x 8 = 17.6 cm^{2}The total surface area of the remaining solid to the nearest cm^{2}is 18 cm^{2}

**Question-9 :-** A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in Fig. 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. (Take π = 22/7)

Given that:- Height(h) of the cylinder = 10 cm Radius(r) of the cylinderical part = 3.5 cm = Radius (r) of hemispherical part Therefore, the total surface area of the article = CSA of cylindrical part + 2 × CSA of hemispherical part = 2πrh + 2 x 2πr^{2}= 2πr(h + 2r) = 2 x 22/7 x 3.5 x [10 + (2 x 3.5)] = 44 x 0.5 x [10 + 7.0] = 22 x 17 = 374 cm^{2}

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