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From Karnataka Open Educational Resources
966 bytes added ,  08:52, 31 October 2019
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*A bucket in the shape of a frustum with the top and bottom circles of rarii 15cm and 10cm. Its depth  is 12cm. Find its curved surface area and total surface area.
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(Express the answer in terms of π)
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=== Objectives ===
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[[Image:frustum of cone.png|link=http://karnatakaeducation.org.in/KOER/en/index.php/File:Frustum_of_cone.png]]
Content objectives  - what content areas
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*Statement of the problem
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A bucket in the shape  of a frustum with the top & bottom circles of radii 15cm and 10cm Its depth(height) is 12cm .CSA & TSA to be calculated length of frustum to be found.
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*Assumptions
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#Student should know CSA & TSA of frustum of cone
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#Student should know the value of л=22/7
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#Student should know the difference between radius , height(depth),lenght
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#Student should know the proper substitution simplification
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Concepts to be taught
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#Lenght of frustum l=
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#CSA & TSA of frustum of cone
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Skill objectives - what specific skills
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Gaps
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#difference between radius , height(depth) and lenght
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Classroom objectives - to demo peer learning, to make a classroom resource, etc -
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Skills
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#To imagine the frustum of a cone having unequal circles at the top & bottom.
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#To imagine that the frustum a cone is having height(depth) inside & length out side.
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Algorthem
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*Part-01
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#To calculate the  Lenght of frustum
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given:<math>r_{1}</math>=15cm,<math>r_{2}</math>=10cm, h=12cm
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All these kinds of objectives need not be there for every activity. And no need to list them as different headings.  This is only for our reference when we are developing activities.
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Lenght of frustum l=<math>\sqrt{h^2+(r_{1}-r_{2})^2}</math>
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===Estimated Time===
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l=<math>\sqrt{12^2+(15-10)^2}</math>
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=== Prerequisites/Instructions, prior preparations, if any ===
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l=<math>\sqrt{144+5^2}</math>
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===Materials/ Resources needed===
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l=<math>\sqrt{144+25}</math>
===Process (How to do the activity)===
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How to do the different steps of the activity?
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What kinds of questions you can ask for that activity
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l=<math>\sqrt{169}</math>
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What are the student follow-up activities/ questions you can give?
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l=13cm
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Categories:  (Subject) (Topic) (Sub-concept/topic) (Class 6) (Resource format)
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*Part-02
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#To calculate CSA
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CSA= <math> π(r_{1}+r_{2})l</math>  Where  l=<math>\sqrt{h^2+(r_{1}-r_{2})^2}</math>
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Example - (Mathematics) (Triangle) (Area) (Perimeter) (Class 6) (Class 8) (Geogebra) (Video)
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CSA=<math>π(15+10)13</math>
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CSA=<math>25 X 13 X π</math>
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CSA=<math>325πcm^2</math>
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#To calculate TSA
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TSA= <math>π({(r_{1}+r_{2})l+r_{1}^2+r_{2}^2})</math>
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TSA =<math>π({(15+ 10) 13 +15^2+10^2})</math>
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TSA = <math>π({25X13 + 225 + 100})</math>
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TSA = <math>({325 + 225 + 100})π</math>
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TSA =<math>650πcm^2</math>
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[[Category:Mensuration]]