Given, the angles of a quadrilateral ABCD taken in an order are in the ratio 3:7:6:4

We have to find the type of the quadrilateral.

Let the angles of the quadrilateral ABCD be 3x, 7x, 6x and 4x.

We know that the sum of all angles of a quadrilateral is equal to 360 degrees.

Therefore

3x + 7x + 6x + 4x = 360°

20x = 360°

x = \dfrac{360°}{20}

x = 18°

Now four interior angles are calculated below

A = 3x = 3\cdot 18° = 54°

B = 7x = 7\cdot 18° = 126°

C = 6x = 6\cdot 18° = 108°

D = 4x = 4\cdot 18° = 72°

AD and BC are two lines cut by a transversal CD.

Now, sum of angles C and D on the same side of transversal,

C + D =108° + 72° =180°

This is the same for angle A and B

A + B =54° + 126° =180°

We know that the sum of interior angles on the same side of the transversal is equal to 180 \degree , then the two lines are parallel.

therefoer, AD|| BC

This implies ABCD is a quadrilateral in which one pair of opposite sides are parallel.

Therefore, ABCD is a trapezium.