Question

In a trapezium ABCD, AB is parallel to CD, BD is perpendicular to AD. AC is perpendicular to BC. If AD = BC = 15 cm and AB = 25 cm, then the area of the trapezium is 

(a) 192 cm² 

(b) 232 cm² 

(c) 162 cm² 

(d) 172 cm² 

Answer 1

(a) 192 cm².

By Pythagoras' Theorem,

AC = \(\sqrt{AB^2-BC^2}\) and BD = \(\sqrt{AB^2-BC^2}\)

⇒ AC = BD = \(\sqrt{625-225}\) = \(\sqrt{400}\) = 2 cm        (By prop.of isos. trap. AC = BD)

Now, Area of ΔDAB = \(\frac{1}{2}\) x AD x BD 

Also, Area of ΔDAB = \(\frac{1}{2}\) x DM x AB 

∴ AD x BD = DM x AB

⇒ DM = \(\frac{AD\times{BD}}{AB}\) = \(\frac{15\times20}{25}\) = 12 cm

Also, CN = DM = 12 cm

AM = \(\sqrt{AD^2-DM^2}\) = \(\sqrt{15^2-12^2}\) = \(\sqrt{225-144}\) 

= \(\sqrt{81}\) = 9 cm

Also, BN = AM = 9 cm MN = AB – (AM + BN) = 25 – (9 + 9) = 25 – 18 = 7 cm 

⇒ CD = MN = 7 cm 

∴ Area of trapezium ABCD = \(\frac{1}{2}\) × DM × (AB + CD) 

= \(\frac{1}{2}\) × 12 × (25 + 7) = \(\frac{1}{2}\) × 12 × 32 = 192 cm².