Quantitative Aptitude Ques 1869
Question: 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
Options:
A) $ 10\sqrt{5} $ units
B) $ 6\sqrt{5} $ units
C) $ 5\sqrt{5} $ units
D) $ 3\sqrt{5} $ units
Show Answer
Answer:
Correct Answer: B
Solution:
- In $ \Delta BCF, $ by Pythagoras theorem. $ {{(5)}^{2}}={{(3)}^{2}}+{{(BF)}^{2}} $
$ \Rightarrow $ $ BF=4cm $
$ \therefore $ $ AB=2+4+4=10cm $ Now, In $ \Delta ACF, $ $ AC^{2}=CF^{2}+FA^{2} $
$ \Rightarrow $ $ AC^{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 $
BETA
