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Exercise - 13.4

Surface Area and Volume

**Question-1 :-** A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass. (Take π = 22/7)

Given that:- Height(h) of a frustum of a cone = 14 cm The diameters of its two circular ends are 4 cm and 2 cm. Radius(r_{1}) of upper base of the glass = 4/2 = 2 cm Radius(r_{2}) of lower base of the glass = 2/2 = 1 cm Therefore, Capacity of glass (Volume of frustum of cone) = 1/3 x πh x (r_{1}^{2}+ r_{2}^{2}+ r_{1}r_{2}) = 1/3 x 22/7 x 14 x [(2 x 2) + (1 x 1) + (2 x 1)] = 1/3 x 22/7 x 14 x [4 + 1 + 2] = 1/3 x 22/7 x 14 x 7 = 308/3 = 102.66 cm^{3}Therefore, the capacity of the glass is 102.66 cm^{3}.

**Question-2 :-** The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum. (Take π = 22/7)

Given that:- Slant height of a frustum of a cone = 4 cm Perimeter of upper circular end of frustum = 18 cm 2πr_{1}= 18 r_{1}= 18/2π r_{1}= 9/π cm Perimeter of lower end of frustum = 6 cm 2πr_{2}= 6 r_{2}= 6/2π r_{2}= 3/π cm Therefore, CSA of frustum = π(r_{1}+ r_{2})l = π x (9/π + 3/π) x 4 = π x 12/π x 4 = 48 cm^{2}Therefore, the curved surface area of the frustum is 48 cm^{2}.

**Question-3 :-** A fez, the cap used by the Turks, is shaped like the frustum of a cone (see Fig.). If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it. (Take π = 22/7)

Given that:- Radius(r_{2}) at upper circular end = 4 cm Radius(r_{1}) at lower circular end = 10 cm Slant height (l) of frustum = 15 cm Therefore, Area of material used for making the fez = CSA of frustum + Area of upper circular end = π[r_{1}+ r_{2}]l + πr_{2}^{2}= 22/7 x [10 + 4] x 15 + 22/7 x 4 x 4 = 22/7 x 14 x 15 + 22/7 x 4 x 4 = 22/7 x (210 + 16) = 22/7 x 226 = 4972/7 = 710.28 cm^{2}Therefore, the area of material used for making it is 710.28 cm^{2}.

**Question-4 :-** A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of Rs. 20 per litre. Also find the cost of metal sheet used to make the container, if it costs Rs. 8 per 100 cm^{2}. (Take π = 3.14)

Given that:- Radius(r_{1}) of upper end of container = 20 cm Radius(r_{2}) of lower end of container = 8 cm Height(h) of container = 16 cm Slant height(l) of frustum = √(r_{1}- r_{2})² + h_{2}² = √(20 - 8)² + 16² = √144 + 256 = √400 = 20 cm Therefore, Capacity of container (Volume of frustum) = 1/3 x πh x (r_{1}^{2}+ r_{2}^{2}+ r_{1}r_{2}) = 1/3 x 3.14 x 16 x [400 + 64 + (20 x 8)] = 1/3 x 3.14 x 16 x [464 + 160] = 1/3 x 3.14 x 16 x 624 = 10449.92 cm^{3}= 10.45 Litres Cost of 1 litre milk = Rs 20 Cost of 10.45 litre milk = 10.45 × 20 = Rs 209 Area of metal sheet used to make the container = π[r_{1}+ r_{2}]l + πr_{2}^{2}= π(20 + 8) x 20 + π(8)^{2}= π[(28 x 20) + 64] = π[560 + 64] = 624π cm^{2}Cost of 100 cm^{2}metal sheet = Rs 8 Cost of 624π cm^{2}metal sheet = (624 x 3.14 x 8)/100 = Rs 156.75 Therefore, the cost of the milk which can completely fill the container is Rs 209 The cost of metal sheet used to make the container is Rs 156.75.

**Question-5 :-** A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum so obtained be drawn into a wire of diameter 1/16 cm, find the length of the wire. (Take π = 22/7)

Given that:- Height of right circular cone = 20 cm In ΔAEG, EG/AG = tan 30^{o}EG/AG = 1/√3 EG = AG/√3 EG = 10/√3 = 10√3/3 cm In ΔABD, BD/AD = tan 30^{o}BD/AD = 1/√3 BD = AD/√3 BD = 20/√3 = 20√3/3 cm Radius(r_{1}) of upper end of frustum = 10√3/3 cm Radius(r_{2}) of lower end of container = 20√3/3 cm Height(h) of container = 10 cm Volume of frustum = 1/3 x πh x (r_{1}^{2}+ r_{2}^{2}+ r_{1}r_{2}) = 1/3 x 22/7 x 10 x [(10√3/3)^{2}+ (20√3/3)^{2}+ (10√3/3 x 20√3/3)] = 1/3 x 22/7 x 10 x [100/3 + 400/3 + 200/3] = 220/21 x 700/3 = 22000/9 cm^{3}Now, Radius(r) of wire = 1/16 x 1/2 = 1/32 cm Volume of wire = Area of cross-section × Length = πr^{2}x l = π(1/32)^{2}x l Volume of frustum = Volume of wire 22000/9 = 22/7 x (1/32)^{2}x l 7000/9 x 1024 = l l = 796444.44 cm = 7964.44 m Therefore, The length of the wire = 7964.44 m

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