The area of an isosceles trapezium is 176 cm² and the height is $${\frac{2}{{11}}^{{\text{th}}}}$$ 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

Solution (By Examveda Team)

Distance between two parallel line = $$\frac{2}{{11}}$$ (sum of both parallel line)
= $$\frac{2}{{11}}$$ × (7x + 4x)
= 2x
Area = $$\frac{1}{2}$$ (sum of parallel sides) × (distance between them)
⇒ $$\frac{1}{2}$$ (7x + 4x) × 2x = 176
⇒ 11x² = 176
⇒ x² = 16
⇒ x = 4
AB = 7 × 4 = 28 cm
CD = 4 × 4 = 16 cm
CM = 2 × 4 = 8 cm
AM = AN + NM
⇒ AM = AN + 16
⇒ AM = 6 + 16
⇒ AM = 22
$$\eqalign{ & \left( {{\text{AN}} = {\text{BM}} = \frac{{12}}{2} = 6} \right) \cr & {\text{A}}{{\text{C}}^2} = {\text{C}}{{\text{M}}^2} + {\text{A}}{{\text{M}}^2} \cr & \Rightarrow {\text{A}}{{\text{C}}^2} = {8^2} + {22^2} \cr & \Rightarrow {\text{AC}} = \sqrt {64 + 484} \cr & \Rightarrow {\text{AC}} = \sqrt {548} \cr & \Rightarrow {\text{AC}} = 2\sqrt {137} \cr} $$