Question

In trapezium ABCD, AB II DC, angle D and angle C are complementary angles, AE II BC and BFII AD.Find the measure of angle EGF, where G is the point of intersection of AE and BF.

Answer 1

see diagram...   it is really very simple....

AD ||  BF   and  CD is the transversal line.  So then
  ∠ADC = ∠BFC    = corresponding angles (on the same side of transversal...

Similarly,  AE || BC.   DC is the transversal line.  so then
   ∠AED =   ∠BCD    = corresponding angles  , right?
So     ∠BFC + ∠ AED =  ∠ADC + ∠BCD = 90°
 so     ∠GFE + ∠GEF = 90°

In the ΔGFE,   ∠EGF = 180° - (∠GFE + ∠ GEF) = 180° - 90° = 90°
right angle ... that is the answer.

Answer 2

ADC = 90°
Ab parallel to dc

ab = 15 cm

cd = 40cm

ac = 41 cm

Using Pythagoras Theorem 

ad² + dc² = ac²
ad² + 40² = 41²
ad² + 1600 = 1681
ad = √1681- 1600
ad = √81
as = 9 cm

area of trapezium =1/2 × (base1 + base2) × h
                              1/2× ( 15 + 41) × 9
                          = 247.5 cm²