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If twice the width of a rectangle is equal to 5/6 times the length of
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16 Feb 2022, 10:54
Let W be width and L be length. So 1st eq is 2W=5/12 L -> W =5/12 L
Length of diagonal = \(\sqrt{L^2 + W^2 }\)
=\( L \sqrt{(1+ 25/144)}\)
= \(L \sqrt{(169/144)}\)
= \(\frac{13}{12} L\)
Perimeter = \(2(L+W)\)
= \(2 L ( \frac{5}{12} +1)\)
= \(2 L ( \frac{17}{12})\)
= \(\frac{34}{ 12} L\)
Solving Perimeter : Length of diagonal we get \(\frac{34}{13}\)
So ans is E