question_answer

In a trapezium, the two non-parallel sides are equal in length, each being of 5 units. The parallel sides are at a distance of 3 units apart. If the smaller side of the parallel sides is of length 2 units, then the sum of the diagonals of the trapezium is

A) \[10\sqrt{5}\]units                      

B) \[6\sqrt{5}\] units

C) \[5\sqrt{5}\]units                        

D) \[3\sqrt{5}\]units

Correct Answer: B

Solution :

In\[\Delta BCF,\]by Pythagoras theorem.

\[{{(5)}^{2}}={{(3)}^{2}}+{{(BF)}^{2}}\]

\[\Rightarrow \]   \[BF=4\,\,cm\]

\[\therefore \]\[AB=2+4+4=10\,\,cm\]

Now, In \[\Delta ACF,\] \[A{{C}^{2}}=C{{F}^{2}}+F{{A}^{2}}\]

\[\Rightarrow \]\[A{{C}^{2}}={{3}^{2}}+{{6}^{2}}\]\[\Rightarrow \]\[AC=\sqrt{45}\,\,cm\]

Similarly,           \[BD=\sqrt{45}\,\,cm\]

\[\therefore \]Sum of diagonals \[=AC+BD=\sqrt{45}+\sqrt{45}\]

\[=2\sqrt{45}=6\sqrt{5}\,\,cm\]