  Question

In a trapezium, the two non-parallel sides are equal in length, each being of $$5\ cm$$. The parallel sides are at a distance of $$3\ cm$$. If the smaller  side of the parallel side is of length $$2\ cm$$, then the length of the diagonal of the trapezium is:

A
35cm  B
65cm  C
45cm  D
7 cm  Solution

The correct option is A $$3\sqrt {5}cm$$Given - non-parallel side of trapezium are equal and measure $$5\,cm$$ and Perpendicular distance is  $$3\,cm$$ with smaller $$\parallel$$ side equals to $$2\,cm$$.Now let trapezium is $$ABCD$$In $$\Delta ACE$$ & $$\Delta BDF$$                                by Pythagoras theorem$$CE^{2} = AC^{2}-AE^{2}$$                                   $$FD^{2} = BD^{2}-BF^{2}$$$$= 25-9$$                                                                  $$= 25-9$$$$= 16$$                                                                         $$= 16$$$$CE = \sqrt{16}$$                                                     $$FD = \sqrt{16}$$$$CE = 4$$                                                           $$FD = 4$$Now $$AB = EF$$ because opposite sides of rectangle $$ABEF$$  So $$EF = 2\,cm$$Now see triangle $$BCF$$In $$\Delta BCF$$ by Pythagoras theorems$$CB^{2} = CF^{2}+BF^{2}$$$$CB^{2} = (6)^{2}+(3)^{2}$$$$CB^{2} = 36+9$$$$CB^{2} = 45$$$$CB = \sqrt{45} = \sqrt{9\times 5}$$$$\boxed{CB = 3\sqrt{5}}$$ CB is our diagonal. Mathematics

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