Check sibling questions

A tent is in the shape of a cylinder surmounted by a conical top. If the height and radius of the cylindrical part are 3 m and 14 m respectively, and the total height of the tent is 13.5 m, find the area of the canvas required for making the tent, keeping a provision of 26 m^2 of canvas for stitching and wastage. Also, find the cost of the canvas to be purchased at the rate of β‚Ή 500 per m^2.

This question is similar to Ex 12.1, 7 – Chapter 12 Class 10

Please check https://www.teachoo.com/1892/551/Ex-13.1--7---A-tent-is-in-shape-of-a-cylinder-surmounted/category/Ex-13.1/

Slide40.JPG

Slide41.JPG
Slide42.JPG Slide43.JPG Slide44.JPG

Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class


Transcript

Area of the canvas = Curved surface area of cone + Curved surface area of the cylinder Curved surface area of cone Radius of cone = Radius of cylinder = r = 14 cm Height of cone = h = Total height – Height of cylinder = 13.5 – 3 = 10.5 m Finding Slant Height 𝑙 = √(β„Ž^2+ π‘Ÿ^2 ) 𝒍 = √(γ€–πŸπŸŽ.πŸ“γ€—^𝟐+ γ€–πŸπŸ’γ€—^𝟐 ) 𝑙 = √(110.25+196) 𝑙 = √306.25 𝒍 = 17.5 m Now, Curved surface area of the cone = πœ‹π‘Ÿπ‘™ = 22/7 Γ— 14 Γ— 17.5 = 22 Γ— 2 Γ— 17.5 = 770 m2 Curved surface area of the cylinder Radius of cylinder = r = 14 m Height of cylinder = h = 3 m Curved surface area of cylinder = 2πœ‹rh = 2 Γ— 𝟐𝟐/πŸ• Γ— πŸπŸ’ Γ— πŸ‘ = 2 Γ— 22 Γ— 2 Γ— 3 = 264γ€– π’Žγ€—^𝟐 Hence, Area of the canvas = Curved area of cone + Curved surface area of cylinder = 770 + 264 = 1034 m2 Now, from question… Keeping a provision of 26 π’Ž^𝟐 of canvas for stitching and wastage. Also, find the cost of the canvas to be purchased at the rate of β‚Ή 500 per π‘š^2. Total area of canvas to be purchased = 1034 + 26 = 1060 γ€– π’Žγ€—^𝟐 Finding Cost Cost of the canvas of the tent for 1γ€– π‘šγ€—^2 = Rs 500 Cost of the canvas of the tent for 1060γ€– π‘šγ€—^2 = 500 Γ— 1060 = Rs 5,30,000 Hence, Total cost = Rs 5,30,000

Ask a doubt
Davneet Singh's photo - Co-founder, Teachoo

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.