Examples (Revised) - Chapter 6 - Lines & Angles - Ncert Solutions class 9 - Maths

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NCERT Solutions for Class 9 Maths Chapter 6: Lines & Angles | Free PDF Download

Example 1 :

In Fig. 6.9, lines PQ and RS intersect each other at point O. If $\angle \mathrm{POR}: \angle \mathrm{ROQ}=5: 7$, find all the angles.
Solution :

$\angle \mathrm{POR}+\angle \mathrm{ROQ}=180^{\circ}$
(Linear pair of angles)

But $\angle \mathrm{POR}: \angle \mathrm{ROQ}=5: 7$
(Given)

Therefore, $\quad \angle \mathrm{POR}=\frac{5}{12} \times 180^{\circ}=75^{\circ}$
Similarly, $\quad \angle \mathrm{ROQ}=\frac{7}{12} \times 180^{\circ}=105^{\circ}$
Now, $\quad \angle \mathrm{POS}=\angle \mathrm{ROQ}=105^{\circ}$
and $\quad \angle \mathrm{SOQ}=\angle \mathrm{POR}=75^{\circ}$

Example 2 :

In Fig. 6.10, ray OS stands on a line POQ. Ray OR and ray OT are angle bisectors of $\angle \mathrm{POS}$ and $\angle \mathrm{SOQ}$, respectively. If $\angle \mathrm{POS}=x$, find $\angle \mathrm{ROT}$.
Solution :

Ray OS stands on the line POQ.
Therefore, $\quad \angle \mathrm{POS}+\angle \mathrm{SOQ}=180^{\circ}$
But,
$\angle \mathrm{POS}=x$

Therefore,
So,
$
\begin{aligned}
\angle \mathrm{POS}+\angle \mathrm{SOQ} & =180^{\circ} \\
\angle \mathrm{POS} & =x \\
\angle+\angle \mathrm{SOQ} & =180^{\circ} \\
\angle \mathrm{SOQ} & =180^{\circ}-x
\end{aligned}
$
Now, ray OR bisects $\angle \mathrm{POS}$, therefore,
$
\begin{aligned}
\angle \mathrm{ROS} & =\frac{1}{2} \times \angle \mathrm{POS} \\
& =\frac{1}{2} \times x=\frac{x}{2} \\
\angle \mathrm{SOT} & =\frac{1}{2} \times \angle \mathrm{SOQ} \\
& =\frac{1}{2} \times\left(180^{\circ}-x\right) \\
& =90^{\circ}-\frac{x}{2}
\end{aligned}
$

Similarly,
$
\begin{aligned}
& =\frac{1}{2} \times x=\frac{x}{2} \\
\angle \mathrm{SOT} & =\frac{1}{2} \times \angle \mathrm{SOQ} \\
& =\frac{1}{2} \times\left(180^{\circ}-x\right) \\
& =90^{\circ}-\frac{x}{2}
\end{aligned}
$

Now,
$
\begin{aligned}
\angle \mathrm{ROT} & =\angle \mathrm{ROS}+\angle \mathrm{SOT} \\
& =\frac{x}{2}+90^{\circ}-\frac{x}{2} \\
& =90^{\circ}
\end{aligned}
$

Example 3 :

In Fig. 6.11, OP, OQ, OR and OS are four rays. Prove that $\angle \mathrm{POQ}+\angle \mathrm{QOR}+\angle \mathrm{SOR}+$ $\angle \mathrm{POS}=360^{\circ}$.

Solution :

In Fig. 6.11, you need to produce any of the rays OP, OQ, OR or OS backwards to a point. Let us produce ray OQ backwards to a point $\mathrm{T}$ so that TOQ is a line (see Fig. 6.12).
Now, ray OP stands on line TOQ.
Therefore,
$
\angle \mathrm{TOP}+\angle \mathrm{POQ}=180^{\circ}
$
(Linear pair axiom)
Similarly, ray OS stands on line TOQ.
Therefore,
$
\angle \mathrm{TOS}+\angle \mathrm{SOQ}=180^{\circ}
$

But
$
\angle \mathrm{SOQ}=\angle \mathrm{SOR}+\angle \mathrm{QOR}
$

So, (2) becomes
$
\angle \mathrm{TOS}+\angle \mathrm{SOR}+\angle \mathrm{QOR}=180^{\circ}
$

Now, adding (1) and (3), you get
$
\angle \mathrm{TOP}+\angle \mathrm{POQ}+\angle \mathrm{TOS}+\angle \mathrm{SOR}+\angle \mathrm{QOR}=360^{\circ}
$

But
$
\angle \mathrm{TOP}+\angle \mathrm{TOS}=\angle \mathrm{POS}
$

Therefore, (4) becomes
$
\angle \mathrm{POQ}+\angle \mathrm{QOR}+\angle \mathrm{SOR}+\angle \mathrm{POS}=360^{\circ}
$

Example 34:

$\text { In Fig. 6.19, if } P Q \| R S, \angle M X Q=135^{\circ} \text { and } \angle \mathrm{MYR}=40^{\circ} \text {, find } \angle \mathrm{XMY} \text {. }$

Solution :

Here, we need to draw a line $\mathrm{AB}$ parallel to line $\mathrm{PQ}$, through point $\mathrm{M}$ as shown in Fig. 6.20. Now, $A B \| P Q$ and $P Q \|$ RS.

Therefore,
$
\begin{gathered}
\mathrm{AB} \| \mathrm{RS} \quad \text { (Why?) } \\
\angle \mathrm{QXM}+\angle \mathrm{XMB}=180^{\circ}
\end{gathered}
$

Now, $\quad \angle \mathrm{QXM}+\angle \mathrm{XMB}=180^{\circ}$
( $\mathrm{AB} \| \mathrm{PQ}$, Interior angles on the same side of the transversal $\mathrm{XM}$ )

But
$
\angle \mathrm{QXM}=135^{\circ}
$

So,
$
135^{\circ}+\angle \mathrm{XMB}=180^{\circ}
$
Now,
$
\angle \mathrm{XMB}=45^{\circ}
$

Therefore,
$
\angle \mathrm{BMY}=\angle \mathrm{MYR}
$
(AB \| RS, Alternate angles)
$
\angle \mathrm{BMY}=40^{\circ}
$

Adding (1) and (2), you get
$
\begin{aligned}
\angle \mathrm{XMB}+\angle \mathrm{BMY} & =45^{\circ}+40^{\circ} \\
\angle \mathrm{XMY} & =85^{\circ}
\end{aligned}
$

Example 5 :

If a transversal intersects two lines such that the bisectors of a pair of corresponding angles are parallel, then prove that the two lines are parallel.
Solution :

In Fig. 6.21, a transversal AD intersects two lines PQ and RS at points B and $\mathrm{C}$ respectively. Ray $\mathrm{BE}$ is the bisector of $\angle \mathrm{ABQ}$ and ray $\mathrm{CG}$ is the bisector of $\angle \mathrm{BCS}$; and $\mathrm{BE} \| \mathrm{CG}$.
We are to prove that $\mathrm{PQ} \| \mathrm{RS}$.
It is given that ray $\mathrm{BE}$ is the bisector of $\angle \mathrm{ABQ}$.
Therefore, $\quad \angle \mathrm{ABE}=\frac{1}{2} \angle \mathrm{ABQ}$
Similarly, ray $\mathrm{CG}$ is the bisector of $\angle \mathrm{BCS}$.
Therefore, $\quad \angle \mathrm{BCG}=\frac{1}{2} \angle \mathrm{BCS}$
But $\mathrm{BE} \| \mathrm{CG}$ and $\mathrm{AD}$ is the transversal.
Therefore, $\quad \angle \mathrm{ABE}=\angle \mathrm{BCG}$
(Corresponding angles axiom)

Substituting (1) and (2) in (3), you get
$
\frac{1}{2} \angle \mathrm{ABQ}=\frac{1}{2} \angle \mathrm{BCS}
$

That is
$
\angle \mathrm{ABQ}=\angle \mathrm{BCS}
$

But, they are the corresponding angles formed by transversal $\mathrm{AD}$ with $\mathrm{PQ}$ and $\mathrm{RS}$; and are equal.
Therefore,
$
\mathrm{PQ} \| \mathrm{RS}
$
(Converse of corresponding angles axiom)
Example 6 :

In Fig. 6.22, $\mathrm{AB} \| \mathrm{CD}$ and $\mathrm{CD} \| \mathrm{EF}$. Also $\mathrm{EA} \perp \mathrm{AB}$. If $\angle \mathrm{BEF}=55^{\circ}$, find the values of $x, y$ and $z$.

Solution :

$y+55^{\circ}=180^{\circ}$
(Interior angles on the same side of the transversal ED)

Therefore, $\quad y=180^{\circ}-55^{\circ}=125^{\circ}$
Again
$
x=y
$
(AB \| CD, Corresponding angles axiom)
Therefore
$
x=125^{\circ}
$

Now, since $\mathrm{AB} \| \mathrm{CD}$ and $\mathrm{CD} \| \mathrm{EF}$, therefore, $\mathrm{AB} \| \mathrm{EF}$.

So,
$
\angle \mathrm{EAB}+\angle \mathrm{FEA}=180^{\circ}
$
(Interior angles on the same side of the transversal EA)
Therefore,
$
\begin{aligned}
90^{\circ}+z+55^{\circ} & =180^{\circ} \\
z & =35^{\circ}
\end{aligned}
$