# In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.

Fig. 6.29

**Solution:**

Given: AB || CD, CD || EF and y : z = 3 : 7

To find: The value of x

When two parallel lines are cut by a transversal, co-interior angles formed are supplementary.

Also, we know that lines which are parallel to the same line are parallel to each other.

Thus, If AB || CD, CD || EF, we can say AB || EF.

Therefore, the angles x and z are alternate interior angles and hence are equal.

x = z.......(1)

AB and CD are parallel lines cut by a transversal. So the co-interior angles formed are supplementary.

x + y = 180°.

Since x = z,

We get y + z = 180°......... (2)

Let, y = 3a, z = 7a [Since, y : z = 3 : 7]

Substituting the values in equation (2),

3a + 7a = 180°

10a = 180°

a = 180°/10

a = 18°

∴ y = 3a = 3 × 18 = 54°

y = 54°

∴ x + y = 180°

x + 54° = 180°

x = 180° - 54°

x = 126°

Thus, x = 126°

**☛ Check: **NCERT Solutions for Class 9 Maths Chapter 6

**Video Solution:**

## In Fig. 6.29, if AB || CD, CD || EF and y : z = 3 : 7, find x.

NCERT Solutions Class 9 Maths Chapter 6 Exercise 6.2 Question 2

**Summary:**

If AB||CD, CD||EF and y : z = 3 : 7, then the value of x from the given figure is 126°.

**☛ Related Questions:**

- In Fig. 6.30, if AB||CD, EF ⊥ CD and ∠GED =126°, find ∠AGE, ∠GEF and ∠FGE.
- In Fig. 6.31, if PQ || ST, ∠PQR = 110° and ∠RST = 130°, find ∠QRS.[Hint: Draw a line parallel to ST through point R.]
- In Fig. 6.32, if AB || CD, ∠APQ = 50° and ∠PRD = 127°, find x and y.
- In Fig. 6.33, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes the mirror PQ at B, the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Prove that AB||CD.

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