DIRECTIONS for questions: Select the correct alternative from the giv...
Consider the following figure of a rectangular paper ABCD.
The line EF is the fold, which is made such that the corner D meets the diagonally opposite comer B. As DF coincides with FB upon making the fold DF = FB. Similarly, DE = EB. Also, as DF is the part of the length of the rectangle that is being folded so that D coincides with the opposite vertex and BE is the part of the length, that is being folded so that B coincides with D, DF = BE (from symmetry), i.e., quadrilateral DFBE is a rhombus (I) Now assume, FC = x cm
In the right angled triangle △BFC, BF =√BC2 + FC2 = √(30)2 + x2 = DF(since, EDBFis a rhombus and BF = DF)
Hence√302 + x2 = 40 - x
=> 900+x2= 1600 + x2– 80x
=> x = 700 / 80 = 35 / 4
Now, consider G on DC, such that EG ⊥DC. In △EGF, GF = 40 - 2x (as AE = FC = x), EG = 30 and EF is the length of the fold.
Alternative Solution
Consider the conclusion (I), i.e., that EDBF is a rhombus. Let the diagonals of the rhombus meet at O. In a rhombus, the diagonals bisect each other at right angles. Hence △EOB is similar to △DAB (both are right angled, with a common angle at B).