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If V is the volume of a cuboid of dimensions a,b,c and S is its surface area then prove that = \frac{2}{S}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c} ★Also the Question is in the attachment.★Please don't spam :-/ |
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Answer» The dimensions of the CUBOID are a,b,c.We know that, Volume of the cuboid V=abc and surface area of the cuboid S=2(ab+bc+ac)To prove: V1 = S2 [ A1 + B1 + c1 ]Consider LHS, V1 = abc1 ...(1)Consider RHS.S2 [ a1 + b1 + c1 ]= 2(ab+bc+ac)2 [ a1 + b1 + c1 ] = ab+bc+ac1 [ a1 + b1 + c1 ] = ab+bc+ac1 [ abcab+bc+ac ] = abc1 S2 [ a1 + b1 + c1 ]= abc1 ...(2)Hence from (1) and (2) we get V1 = S2 [ a1 + b1 + c1 ] now MARK me as BRAINLIEST and follow me |
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