Correct Answer: $2\sqrt{137}$

Solution :

Let the lengths of the parallel sides of the isosceles trapezium be as $a$ and $b$ $(a < b)$, and the height as $h$. Given that the area of the trapezium is 176 cm ² .
$⇒A = \frac{1}{2} (a + b) h = 176 $....................(1)
We have the height $h$ is $\frac{2}{11}$th of the sum of the parallel sides.
$⇒h = \frac{2}{11} (a + b)$...........................(2)
From (1) and (2)
$⇒176 = \frac{1}{2} (a + b) \times \frac{2}{11} (a + b)$
$⇒(a + b)^2 = 1936$
$⇒ a + b = 44$...................(3)
The ratio of the lengths of the parallel sides is 4 : 7,
Assume $a = 4x$ and $b = 7x$.
From equation (3),
$⇒4x + 7x = 44$
$⇒x = 4$
$⇒a = 16\;\mathrm{cm}$ and $b = 28\;\mathrm{cm}$
$⇒BE = CF = y$
$⇒28=2y+16$
$⇒y=6\;\mathrm{cm}$
$⇒BF = 28-6=22\;\mathrm{cm}$
From equation (2)
$⇒h = \frac{2}{11} (16+28)=8$
The diagonal (BD) of trapezium = $\sqrt{BF^2+h^2} = \sqrt{22^2+8^2}=\sqrt{548}=2\sqrt{137}$ cm
Hence, the correct answer is $2\sqrt{137}$.