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CBSE Class 10 Answered

A military tent of height 8.25m is in the form of a right circular cylinder of base diameter 30m and height 5.5m surmounted by a right circular cone of same base radius.Find the lenght of the canvas use in making the tent,if the breadth of the canvas is 1.5m.
Asked by anusarika mohanty | 13 Mar, 2015, 11:24: AM
answered-by-expert Expert Answer
C o n s i d e r space t h e space f o l l o w i n g space f i g u r e.
G i v e n space t h a t space t h e space d i a m e t e r space o f space t h e space c y l i n r i c a l space p a r t space a n d space t h e space c o n i c a l space p a r t space o f space t h e space t e n t space i s space 30 space m D equals 30 m rightwards double arrow R equals 30 over 2 equals 15 space m W e space n e e d space t o space f i n d space t h e space c u r v e d space s u r v e d space a r e a space o f space t h e space t e n t. C u r v e d space S u r f a c e space A r e a space o f space t h e space t e n t space equals space C u r v e d space S u r f a c e space A r e a space o f space t h e space c y l i n d e r space plus space C u r v e d space S u r f a c e space A r e a space o f space t h e space c o n e space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 2 πRH plus πRL where space straight H space is space the space hieght space of space cylinder space and space straight L space is space the space slant space height space of space the space cone straight L equals square root of straight h squared plus straight R squared end root comma space where space straight h space is space the space height space of space the space cone rightwards double arrow straight L equals square root of 5.5 squared plus 15 squared end root rightwards double arrow straight L equals square root of 30.25 plus 225 end root rightwards double arrow straight L equals square root of 255.25 end root rightwards double arrow straight L equals 15.98 space straight m Thus comma space straight C straight u straight r straight v straight e straight d space straight S straight u straight r straight f straight a straight c straight e space straight A straight r straight e straight a space straight o straight f space straight t straight h straight e space straight t straight e straight n straight t space equals 2 πRH plus πRL space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals πR open parentheses 2 straight H plus straight L close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals πR open parentheses 2 cross times 8.25 plus 15.98 close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals πR open parentheses 32.48 close parentheses space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 22 over 7 cross times 15 cross times 32.48 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 1531.2 space straight m squared Given space that space the space breadth space of space the space canvas space is space 1.5 space straight m Area space of space the space canvas space required equals Length space cross times Breadth rightwards double arrow 1531.2 space space straight m squared equals Length space cross times 1.5 space straight m rightwards double arrow Length equals fraction numerator 1531.2 over denominator 1.5 end fraction equals 1020.8 space straight m
Answered by Vimala Ramamurthy | 14 Mar, 2015, 09:00: AM

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