Question

In a trapezium ABCD, seg AB || seg DC, seg BD ⊥ seg AD, seg AC ⊥ seg BC. If AD = 15, BC = 15 and AB = 25, find A (ABCD).

Answer 1

Construction: Draw seg DE ⊥ seg AB, A – E – B 

and seg CF ⊥ seg AB, A – F- B.

In ∆ ACB, ∠ACB = 90° [Given] 

∴ AB² = AC² + BC² [Pythagoras theorem] 

∴ 25² = AC² + 15² 

∴ AC² = 625 – 225 = 40

∴ AC = √400 [Taking square root of both sides] 

= 20 units 

Now, A(∆ABC) = 1/2 × BC × AC 

Also, A(∆ABC) = 1/2 × AB × CF

∴ BC × AC = AB × CF 

∴ 15 × 20 = 25 × CF 

∴ CF = (15 x 20)/25 = 12 units 

In ∆CFB, ∠CFB 90° [Construction] 

∴ BC² = CF² + FB² [Pythagoras theorem] 

∴ 15² = 12² + FB² 

∴ FB² = 225 – 144 

∴ FB² = 81

∴ FB = √81 [Taking square root of both sides] 

= 9 units 

Similarly, we can show that, AE = 9 units Now, AB = AE + EF + FB [A – E – F, E – F – B] 

∴ 25 = 9 + EF + 9 

∴ EF = 25 – 18 = 7 units

In ⟂CDEF, 

seg EF || seg DC [Given, A – E – F, E – F – B] 

seg ED || seg FC [Perpendiculars to same line are parallel] 

∴ ⟂CDEF is a parallelogram. 

∴ DC = EF 7 units [Opposite sides of a parallelogram] 

A(⟂ABCD) = 1/2 × CF × (AB + CD) 

= 1/2 × 12 × (25 + 7) 

= 1/2 × 12 × 32 

∴ A(⟂ABCD) = 192 sq. units