Quantitative Aptitude Ques 2329
Question: The area of an isosceles trapezium is 176 and the height is $ \frac{2}{11},th $ of the sum of its parallel sides. If the ratio of the length of the parallel sides is 4 : 7, then the length of a diagonal (in cm) is
Options:
A) 24
B) $ \sqrt{137} $
C) 28
D) $ 2\sqrt{137} $
Show Answer
Answer:
Correct Answer: D
Solution:
- Let the length of sides be 4x and 7x.
Then, $ h=\frac{2}{11}(a+b) $
We have, $ \frac{1}{2}(a+b),h=176 $
Area of trapezium = 176
$ =\frac{1}{2}\times $ Height $ \times $ (sum of parallel sides) = 176
$ \frac{1}{2}(a+b)\times \frac{2}{11}(a+b)=176 $
$ {{(a+b)}^{2}}=176\times 11 $
$ a+b=44 $
$ 4x+7x=44 $
$ \Rightarrow $ $ x=4 $
Sides $ =4x=16cm $ and $ 7x=28cm $
$ h=\frac{2}{11}\text{(16+28)}=8cm $
Now, $ AE=6cm $
and $ DE=8cm $
$ AD=\sqrt{8^{2}+6^{2}}=10,cm $
Diagonal, $ AC=\sqrt{AF^{2}+CF^{2}} $
$ [\because CF\bot AB] $
$ =\sqrt{{{(a+6)}^{2}}+8^{2}} $
$ =\sqrt{22^{2}+8^{2}} $ $ [\because ,a=16] $
$ =\sqrt{484+64}=\sqrt{548} $
$ =2\sqrt{137},cm $
BETA
