Last updated at Feb. 25, 2017 by Teachoo

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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 π = 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 Δ OQD & Δ OPB ∠ DOQ = ∠ BOP ∠ OQD = ∠ OPB So, Δ OQD ∼ Δ OPB ∴ 𝑂𝑄/𝑂𝑃 = 𝑄𝐷/𝑃𝐵 𝑂𝑄/𝑂𝑃 = 7/28 𝑂𝑄/𝑂𝑃 = 1/4 4OQ = OP 4OQ = OQ + QP 4OQ – OQ = QP 3OQ = QP OQ = 𝑄𝑃/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 𝜋𝑟2ℎ = 1/3 π×(7)2×15 = π×49×5 = 245π cm3 Curved Surface area of cone OCD = πr𝑙 = π×7×16.55 = 115.85π cm2 Volume, CSA of cone OAB Volume of cone OAB = 1/3 πR2H = 1/3 π×(28)2×60 = π×784×20 = 15680π cm3 Curved Surface area of cone AOB = πR𝐿 = π×28×66.20 = 1853.6π cm2 Volume of frustum = Volume of cone OAB – Volume of cone OCD = 15680π – 245π = 15435π = 15435 × 22/7 = 48510 cm3 Curved Surface area of frustum = Curved Surface area of cone OAB – Curved Surface area of cone OCD = 1853.6π – 115.85π = 1737.75π = 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 + π(28)^2 + π(7)^2 = 5461.5 + 784π + 49π = 5461.5 + 833π = 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 π = 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 𝜋ℎ(𝑟12+𝑟22+𝑟1𝑟2) = 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 = 𝜋(𝑟1+𝑟2)𝑙 = 22/7 (28 + 7)×(49.65) = 22/7×(35)×(49.65) = 22×5×(49.65) = 5461.5 cm2 Total surface area of the frustum = 𝜋(𝑟1+𝑟2)𝑙+𝜋𝑟12+𝜋𝑟22 = Curved surface area of frustum + 𝜋𝑟12+𝜋𝑟22 = 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

Example 2
Important

Example 3 Important

Ex 13.1, 4 Important

Ex 13.1, 9 Important

Example 6 Important

Example 7 Important

Example 9 Important

Example 11 Important

Ex 13.3, 8 Important

Ex 13.3, 9 Important

Example 12 Important You are here

Ex 13.4, 4 Important

Ex 13.4, 5 Important

Surface Area and Volume Formulas Important

Class 10

Important Questions for Exam - Class 10

- Chapter 1 Class 10 Real Numbers
- Chapter 2 Class 10 Polynomials
- Chapter 3 Class 10 Pair of Linear Equations in Two Variables
- Chapter 4 Class 10 Quadratic Equations
- Chapter 5 Class 10 Arithmetic Progressions
- Chapter 6 Class 10 Triangles
- Chapter 7 Class 10 Coordinate Geometry
- Chapter 8 Class 10 Introduction to Trignometry
- Chapter 9 Class 10 Some Applications of Trignometry
- Chapter 10 Class 10 Circles
- Chapter 11 Class 10 Constructions
- Chapter 12 Class 10 Areas related to Circles
- Chapter 13 Class 10 Surface Areas and Volumes
- Chapter 14 Class 10 Statistics
- Chapter 15 Class 10 Probability

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.