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, \[AC=\sqrt{A{{F}^{2}}+C{{F}^{2}}}\] |
\[[\because CF\bot AB]\] |
\[=\sqrt{{{(a+6)}^{2}}+{{8}^{2}}}\] |
\[=\sqrt{{{22}^{2}}+{{8}^{2}}}\] \[[\because \,a=16]\] |
\[=\sqrt{484+64}=\sqrt{548}\] |
\[=\,\,2\sqrt{137}\,cm\] |
You need to login to perform this action.
You will be redirected in
3 sec