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Example 12 - The radii of ends of a frustum 45 cm high - Examples

Example 12 - Chapter 13 Class 10 Surface Areas and Volumes - Part 2
Example 12 - Chapter 13 Class 10 Surface Areas and Volumes - Part 3
Example 12 - Chapter 13 Class 10 Surface Areas and Volumes - Part 4
Example 12 - Chapter 13 Class 10 Surface Areas and Volumes - Part 5
Example 12 - Chapter 13 Class 10 Surface Areas and Volumes - Part 6
Example 12 - Chapter 13 Class 10 Surface Areas and Volumes - Part 7
Example 12 - Chapter 13 Class 10 Surface Areas and Volumes - Part 8
Example 12 - Chapter 13 Class 10 Surface Areas and Volumes - Part 9
Example 12 - Chapter 13 Class 10 Surface Areas and Volumes - Part 10
Example 12 - Chapter 13 Class 10 Surface Areas and Volumes - Part 11
Example 12 - Chapter 13 Class 10 Surface Areas and Volumes - Part 12

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Example 12 (Method 1 By deriving frustum formula) The radii of the ends of a frustum of a cone 45 cm high are 28 cm and 7 cm (see figure). Find its volume, the curved surface area and the total surface area(Take = 22/7) There are two cones OCD & OAB Volume of frustum ABDC = Volume of cone OAB Volume of cone OCD Curved Surface area of frustum ABDC = Curved Surface area of cone OAB Curved Surface area of cone OCD Total Surface area of frustum ABDC = Curved Surface area of frustum + Area of top circle + Area of bottom circle Now it is given Finding h, l of cone OCD In OQD & OPB DOQ = BOP OQD = OPB So, OQD OPB / = / / = 7/28 / = 1/4 4OQ = OP 4OQ = OQ + QP 4OQ OQ = QP 3OQ = QP OQ = /3 OQ = 45/3 = 15 cm So, height = h = 15 cm Now, we find slant height (l) We know that l2 = h2 + r2 l2 = 152 + 72 l2 = 225 + 49 l2 = 274 l = 274 l = 16.55 cm Finding H, L of cone OAB H = OP = OQ + QP = 45 + 15 = 60 cm Now, we find slant height (L) We know that L2 = h2 + r2 L2 = 602 + 282 L2 = 4384 L = 4384 L = (16 274) L = 16 274 L = (4^2 ) 274 L = 4 16.55 cm L = 66.20 cm Volume, CSA of cone OCD Volume of cone OCD = 1/3 2 = 1/3 (7)2 15 = 49 5 = 245 cm3 Curved Surface area of cone OCD = r = 7 16.55 = 115.85 cm2 Volume, CSA of cone OAB Volume of cone OAB = 1/3 R2H = 1/3 (28)2 60 = 784 20 = 15680 cm3 Curved Surface area of cone AOB = R = 28 66.20 = 1853.6 cm2 Volume of frustum = Volume of cone OAB Volume of cone OCD = 15680 245 = 15435 = 15435 22/7 = 48510 cm3 Curved Surface area of frustum = Curved Surface area of cone OAB Curved Surface area of cone OCD = 1853.6 115.85 = 1737.75 = 1737.75 " " 22/7 = 5461.5 cm2 Total Surface area of frustum = Curved Surface area of frustum + Area of top circle + Area of bottom circle = 5461.5 + (28)^2 + (7)^2 = 5461.5 + 784 + 49 = 5461.5 + 833 = 5461.5 + 833 " " 22/7 = 5461.5 + 2618 = 8079.5 cm2 Example 12 (Method 2 By frustum formula) The radii of the ends of a frustum of a cone 45 cm high are 28 cm and 7 cm (see figure). Find its volume, the curved surface area and the total surface area(Take = 22/7) Given data, Height = h = PQ = 45 cm Radius of larger end = r1 = PB = 28 cm Radius of smaller end = r2 = QD = 7 cm We find slant height l We know that l = ("h2 + (r1 r2)2" ) = ("(45)2 + (28 7)2" ) = ((15 3)2+(21)2) = ((15 3)2+(7 3)2) = 3 (152+72) = 3 (225+49) = 3 274 = 3 16.55 = 49.65 Now, Volume of the frustum = 1/3 ( 12+ 22+ 1 2) = 1/3 22/7 45((28)2+(7)2+28 7) = 22/7 15(784+49+196) = 22/7 15(1029) = 48510 cm3 Curved surface area of the frustum = ( 1+ 2) = 22/7 (28 + 7) (49.65) = 22/7 (35) (49.65) = 22 5 (49.65) = 5461.5 cm2 Total surface area of the frustum = ( 1+ 2) + 12+ 22 = Curved surface area of frustum + 12+ 22 = 5461.5 + 22/7 (28)2+22/7 (7)2 = 5461.5 + 22/7 28 28+22/7 7 7 = 5461.5 + 22 4 28+22 7 = 5461.5 + 2464 + 154 = 8079.5 cm3 Hence, Volume of frustum = 48510 cm3 Curved surface area of frustum = 5461.5 cm2 Total curved surface area of the frustum = 8079.5 cm2

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.