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 \[=AC+BD=\sqrt{45}+\sqrt{45}\] |
\[=2\sqrt{45}=6\sqrt{5}\,\,cm\] |
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