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A square sheet of paper ABCD is so folded that B falls on the mid-point M of CD. The crease will divide BC in the ratio

(a) 7 : 4 

(b) 5 : 3 

(c) 8 : 5 

(d) 4 : 1

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(b) 5 : 3.

By physically folding the square sheet of paper as per the conditions stated in the question, we can see that EF is the crease along which the fold is made. Also, 

MF = FB and MB ⊥ EF. 

We need to find BF : FC. 

Let each side of the square be a units and CF = x units 

Then, BF = (a – x) units and MF = (a – x) units. 

In ΔMFC, MF2 = MC2 + FC2 

⇒ FB2 = MC2 + FC2           (Since MF = FB) 

⇒ (a – x)2 = (a/2)2 + x2 

⇒ a2 + x2 – 2ax = \(\frac{a^2}{4}\) + x2

⇒ 2ax = a\(\frac{a^2}{4}\) = \(\frac{3a^2}{4}\)

⇒ x = \(\frac{3a}{8}\) ⇒ a - x = a - \(\frac{3a}{8}\) = \(\frac{5a}{8}\)

∴ BF : FC = (x - a) : x 

\(\frac{5a}{8}\) : \(\frac{3a}{8}\) = 5 : 3.

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