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# If angles A, B, C and D of the quadrilateral ABCD, taken in order, are in the ratio 3:7:6:4, then ABCD is a

a. rhombus

b. parallelogram

c. trapezium

d. kite

**Solution:**

It is given that

Ratio of angles is 3:7:6:4

Consider 3x, 7x, 6x and 4x as the angles

As the sum of all angles of a quadrilateral is 360º

We can write it as

3x + 7x + 6x + 4x = 360º

20x = 360º

Dividing both sides by 20

x = 18º

So the angles are

∠A = 3 x 18 = 54º

∠B = 7 X 18 = 126º

∠C = 6 x 18 = 108º

∠D = 4 x 18 = 72º

From the figure,

∠BCE = 180º - ∠BCD [Linear pair axiom]

∠BCE = 180º - 108º = 72º

As the corresponding angles are equal, BC || AD

So the sum of co interior angles is

∠A + ∠B = 54º + 126º = 180º

∠C + ∠D = 108º + 72º = 180º

So it is a trapezium

Therefore, ABCD is a trapezium.

**✦ Try This: **If angles P, Q, R and S of the quadrilateral PQRS, taken in order, are in the ratio 4:8:7:5, then PQRS is a a. rhombus, b. parallelogram, c. trapezium, d. kite

**☛ Also Check:** NCERT Solutions for Class 9 Maths Chapter 8

**NCERT Exemplar Class 9 Maths Exercise 8.1 Problem 6**

## If angles A, B, C and D of the quadrilateral ABCD, taken in order, are in the ratio 3:7:6:4, then ABCD is a , a. rhombus, b. parallelogram, c. trapezium, d. kite

**Summary:**

If angles A, B, C and D of the quadrilateral ABCD, taken in order, are in the ratio 3:7:6:4, then ABCD is a trapezium

**☛ Related Questions:**

- If bisectors of ∠A and ∠B of a quadrilateral ABCD intersect each other at P, of ∠B and ∠C at Q, of ∠ . . . .
- If APB and CQD are two parallel lines, then the bisectors of the angles APQ, BPQ, CQP and PQD form , . . . .
- The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is ,a. a rh . . . .

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