Discussion :: Applied Mechanics
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The centre of gravity of the trapezium as shown in below figure from the side is at a distance of
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A.
\(\frac { h } { 3 } \) x \([\frac { b+2a} { b+a } ]\) |
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B.
\(\frac { h } { 3 } \) x \([\frac { 2b+a} { b+a } ]\) |
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C.
\(\frac { h } { 2 } \) x \([\frac { b+2a} { b+2a } ] \) |
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D.
\(\frac { h } { 2 } \) x \([\frac { 2b+a} { b+a } ] \) |
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Answer : Option A
Explanation :
No answer description available for this question
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