The area of an isosceles trapezium is 176 cm² and height is \(\frac{2}{11}\)th of sum of its parallel sides.  If ratio of length of parallel sides is 5 : 6; then length of diagonal is:

\(\sqrt {548}\) cm

Ratio of parallel side = 5 : 6

So, sum of parallel sides = 11x

Area of trapezium = \(\frac{1}{2}\) (sum of parallel sides) × height

⇒ 176 = \(\frac{1}{2}\) (11x) × \(\frac{2}{11}\) (11x)

⇒ x² = 16

⇒ x = 4

Hence,

(i) Parallel sides of trapezium =  5 × 4 & 6 × 4 = 20 & 24 cm

(ii) Height = \(\frac{2}{11}\) (44) = 8 cm

(iii) AD = EF = 20 cm

(iv) BE = FC = \(\frac{24\;-\;20}{2}\) = 2 cm

In triangle BDF;

BD²   = BF² + DF²

        = (2 + 20)² + (8)²

        = (22)² + (8)²

        = 484 + 64

        = 548

BD    = \(\sqrt {548}\)cm