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.
Answers
Answered by
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.
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.
Attachments:
kvnmurty:
clik on thanks .... select brainliest answe
Answered by
0
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
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
Similar questions