Question

The angles A, B, C and D of a trapezium ABCD are in the ratio 3 : 4 : 5 : 6. Let. ∠A : ∠B : ∠C : ∠D = 3:4: 5 : 6.

Answer 1

let the common multiple be x
hence, angle A,B,C and D are 3x , 4x , 5x and 6x

now, sum of angles of trapezium are 360°
therefore,
3x+4x+5x+6x=360
7x+11x=360
18x=360
x=360/18
x=20

therefore
angle A =3x= 3×20 = 60°
angle B =4x= 4×20 = 80°
angle C =5x= 5×20 =100°
angle D =6x= 6×20 =120°


hope it helps :)

Answer 2

I think that we have to find the measures of the given angles.

We are given that the ratio of angles A, B, C, and D is 3 : 4 : 5 : 6.

Let us take the measures of the angles be 3x, 4x, 5x and 6x.

If ABCD is a trapezium, then AB must be parallel to CD and angle B and C must be supplementary angles.

[as opposite sides of a trapezium are parallel]

angle B + angle C = 180°
3x + 4x = 180°
9x = 180°
x = 180/9°
x = 20°
Now,
Angle A = 3x = 3×20° = 60°
Angle B = 4x = 4×20° = 80°
Angle C = 5x = 5×20° = 100°
Angle D = 6x = 6×20° = 120°

Hope it helps!!!

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