# Ex.6.2 Q2 Lines and Angles Solution - NCERT Maths Class 9

## Question

In the given figure, if \(AB\, ‖ \,CD\,,\; CD\, ‖ \,EF\) and \(y:z = 3:7\), find \(x\).

## Text Solution

**What is known?**

\(AB ‖ CD\)

\(CD ‖ EF\)

\(y:z = 3:7\) implies \(y = 3a\) and \(z = 7a\)

**What is unknown?**

Value of \(x.\)

**Reasoning:**

- Lines which are parallel to the same line are parallel to each other.
- When two parallel lines are cut by a transversal, co-interior angles formed are supplementary.

**Steps:**

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

If \(AB ‖ CD\), \(CD ‖ EF\), we can say that \(AB ‖ EF\).

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

\[x = z \qquad\ldots \ldots \ldots (1)\]

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

\[x + y =180^ {\circ}\]

\(\text{Since}\ x = z\), we get \(y + z =180^ {\circ}\)

Given \(y = 3a, \;z = 7a\)

\[\begin{align} 3 a + 7 a &= 180 ^ { \circ } \\ 10 a &= 180 ^ { \circ } \\ a &= \frac { 180^ { \circ } } { 10 } \\ a &= 18 ^ { \circ } \end{align}\]

\[\begin{align} \therefore \qquad y & = 3 a \\ y &= 3 \times 18 ^ { \circ } \\ y &= 54 ^ { \circ } \end{align}\]

\[\begin{align} \therefore \qquad x + y &= 180 ^ { \circ } \\ x + 54 ^ { \circ } &= 180 ^ { \circ } \\ x &= (180 ^ { \circ } - 54 ^ { \circ }) \\ x &= 126 ^ { \circ } \end{align}\]