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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
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B
65cm
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C
45cm
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D
7 cm
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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.

1112478_1037647_ans_28ee969ddaf84dd3a8fa1cbc242e0013.PNG

Mathematics

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