# Ex 13.5, 2 (Optional)

Last updated at March 27, 2018 by Teachoo

Last updated at March 27, 2018 by Teachoo

Transcript

Ex 13.5, 2 A right triangle, whose sides are 3 cm and 4 cm (other than hypotenuse) is made to revolve about its hypotenuse. Find the volume and surface area of the double cone so formed. (Choose value of π as found appropriate.) Let ABC be the right triangle where AB = 3cm and BC = 4 cm Since AC is the hypotenuse AC2 = AB2 + BC2 AC2 = 32 + 42 AC2 = 9 + 16 AC2 = 25 AC = √25 AC = 5 cm Now, revolving Δ ABC around hypotenuse We get a double cone as shown in the figure We need to find the volume and surface area of the double cone formed. It has two cones, ABD and CBD We need to find BO, AO and OC To find AO, BO and OC, we find Area Δ ABC Cone ABD Radius = BO Height = AO Cone CBD Radius = BO Height = OC Area of ∆ABC = 1/2 "×" Base × Height = 1/2 × AB × BC = 1/2 × 3 × 4 = 6 Area of ∆ABC = 1/2 "×" Base × Height = 1/2 × AC × BO = 1/2 × 5 × BO = 5/2 BO From (1) and (2) 5/2 BO = 6 BO = (6 × 2)/2 BO = 12/5 cm In ∆ABC By Pythagoras Theorem AB^2=AO^2+BO^2 AO^2 =AB^2−BO^2 AO^2=(3)^2 −(12/5)^2 AO^2= 9 − 144/25 AO^2= (225 − 144)/25 AO^2= 81/25 AO= √(81/25) = √(9^2/5^2 ) AO = 9/5 cm In ∆BOC By Pythagoras Theorem BC^2=BO^2+CO^2 CO^2 =BC^2−BO^2 CO^2 =(4)^2 −(12/5)^2 CO^2 = 16 − 144/25 CO^2 = (400 − 144)/25 CO^2 = 256/25 CO= √(256/25) = √(〖16〗^2/5^2 ) CO = 16/5 cm Now, BO = 12/5 cm AO = 9/5 cm CO = 16/5 cm Finding volume of the Double Cone Volume of double cone = Volume of cone ABD + Volume of cone CBD Volume of cone ABD = 1/3 𝜋 (𝑅𝑎𝑑𝑖𝑢𝑠)^2×𝐻𝑒𝑖𝑔ℎ𝑡 =1/3 𝜋 (𝐵𝑂)^2×𝐴𝑂 =1/3 𝜋 (12/5)^2×9/5 =(𝜋 × (12)^2 × 9)/(3 × (5)^2 × 5) =(𝜋 × 144 × 3)/(25 × 5) =432/125 𝜋 Volume of cone CBD = 1/3 𝜋 (𝑅𝑎𝑑𝑖𝑢𝑠)^2×𝐻𝑒𝑖𝑔ℎ𝑡 =1/3 𝜋 (𝐵𝑂)^2×𝐶𝑂 =1/3 𝜋 (12/5)^2×16/5 =(𝜋 × (12)^2 × 16)/(3 × (5)^2 × 5) =(𝜋 × 144 × 16)/(3 × 25 × 5) =768/125 𝜋 Volume of double cone = 432/125 𝜋+768/125 𝜋 = 1200/125 𝜋 = 48/5 × 3.14 = 30.14 〖𝒄𝒎〗^𝟐 Finding surface Area of the Double cone: Surface area of double cone = Curved surface area of cone ABD + Curved surface area of cone CBD Here BO = 12/5 cm AO = 9/5 cm CO = 16/5 cm Curved surface area of cone ABD = 𝜋×𝑟×𝑙 = 𝜋×𝐵𝑂×𝐴𝐵 = 𝜋×12/5 ×3 = 36/5 𝜋 𝑐𝑚^2 Curved surface area of cone CBD = 𝜋×𝑟×𝑙 = 𝜋×𝐵𝑂×𝐵𝐶 = 𝜋×12/5 ×4 = 48/5 𝜋 𝑐𝑚^2 Curved surface area of double cone = 35/5 𝜋+48/5 𝜋 = 84/5 𝜋 = 84/5 × 3.14 = 52.75 〖𝒄𝒎〗^𝟐 Therefore, Volume of double cone = 30.14 〖𝑐𝑚〗^3 Surface area of double cone = 52.75 〖𝑐𝑚〗^2

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