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:
Answer & explanation
Correct answer: option 4
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