question_answer

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 [SSC (CGL) Mains 2015]

A) 24                                

B) \[\sqrt{137}\]

C) 28                                

D) \[2\sqrt{137}\]

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 \[=4x=16\text{ }cm\]and \[7x=28\text{ }cm\]

\[h=\frac{2}{11}\text{(16+28)}=8cm\]

Now, \[AE=6\text{ }cm\]

and \[DE=8\text{ }cm\]

\[AD=\sqrt{{{8}^{2}}+{{6}^{2}}}=10\,cm\]

Diagonal, \[AC=\sqrt{A{{F}^{2}}+C{{F}^{2}}}\]

\[[\because CF\bot AB]\]

\[=\sqrt{{{(a+6)}^{2}}+{{8}^{2}}}\]

\[=\sqrt{{{22}^{2}}+{{8}^{2}}}\]                 \[[\because \,a=16]\]

\[=\sqrt{484+64}=\sqrt{548}\]

\[=\,\,2\sqrt{137}\,cm\]