Exercise 6.2 (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

Ex 6.2 Question 1.

$\text {In the given figure, if } A B|| C D, C D|| E F \text { and } y: z=3: 7 \text {, find } x \text {. }$

Answer.

We are given that $A B\|C D, C D\| E F$ and $y: z=3: 7$.

We need to find the value of $x$ in the figure given below.

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

We can conclude that $A B\|C D\| E F$.

Let $y=3 a$ and $z=7 a$.

We know that angles on same side of a transversal are supplementary.
$
\therefore x+y=180^{\circ} \text {. }
$
$x=z$ (Alternate interior angles)
$z+y=180^{\circ}$, or
$
\begin{aligned}
& 7 a+3 a=180^{\circ} \\
& \Rightarrow 10 a=180^{\circ} \\
& a=18^{\circ} .
\end{aligned}
$

$
\begin{aligned}
& z=7 a=126^{\circ} \\
& y=3 a=54^{\circ} .
\end{aligned}
$

Now $x+54^{\circ}=180^{\circ}$
$
x=126^{\circ} .
$

Therefore, we can conclude that $x=126^{\circ}$.

Ex 6.2 Question 2.

$\text {In the given figure, If } \mathbf{A B} \| \mathbf{C D}, E F \perp C D \text { and } \angle G E D=126^{\circ} \text {, find } \angle A G E, \angle G E F \text { and } \angle F G E \text {. }$

Answer.

We are given that $A B \| C D, E F \perp C D$ and $\angle G E D=126^{\circ}$.

We need to find the value of $\angle A G E, \angle G E F$ and $\angle F G E$ in the figure given below.
$
\begin{aligned}
& \angle G E D=126^{\circ} \\
& \angle G E D=\angle F E D+\angle G E F . \\
& \text { But, } \angle F E D=90^{\circ} . \\
& 126^{\circ}=90^{\circ}+\angle G E F \\
& \Rightarrow \angle G E F=36^{\circ} . \\
& \because \angle A G E=\angle G E D \text { (Alternate angles) } \\
& \therefore \angle A G E=126^{\circ} .
\end{aligned}
$

From the given figure, we can conclude that $\angle F E D$ and $\angle F E C$ form a linear pair.

We know that sum of the angles of a linear pair is $180^{\circ}$.
$
\angle F E D+\angle F E C=180^{\circ}
$

$
\begin{aligned}
& \Rightarrow 90^{\circ}+\angle F E C=180^{\circ} \\
& \Rightarrow \angle F E C=90^{\circ} \\
& \angle F E C=\angle G E F+\angle G E C \\
& \therefore 90^{\circ}=36^{\circ}+\angle G E C \\
& \Rightarrow \angle G E C=54^{\circ} .
\end{aligned}
$
$\angle G E C=\angle F G E=54^{\circ}$ (Alternate interior angles)

Therefore, we can conclude that $\angle A G E=126^{\circ}, \angle G E F=36^{\circ}$ and $\angle F G E=54^{\circ}$.

Ex 6.2 Question 4.

In the given figure, if $\mathbf{P Q} \| \mathbf{S T}, \angle P Q R=110^{\circ}$ and $\angle R S T=130^{\circ}$, find $\angle Q R S$.
[Hint: Draw a line parallel to ST through point R.]

Answer

We need to draw a line $R X$ that is parallel to the line $S T$, to get

Thus, we have $S T \| R X$.

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

We can conclude that $P Q\|S T\| R X$.
$\angle P Q R=\angle Q R X$, or(Alternate interior angles)
$
\angle Q R X=110^{\circ} \text {. }
$

We know that angles on same side of a transversal are supplementary.
$
\begin{aligned}
& \angle R S T+\angle S R X=180^{\circ} \Rightarrow 130^{\circ}+\angle S R X=180^{\circ} \\
& \Rightarrow \angle S R X=180^{\circ}-130^{\circ}=50^{\circ} .
\end{aligned}
$

From the figure, we can conclude that
$
\begin{aligned}
& \angle Q R X=\angle S R X+\angle Q R S \Rightarrow 110^{\circ}=50^{\circ}+\angle Q R S \\
& \Rightarrow \angle Q R S=60^{\circ} .
\end{aligned}
$

Therefore, we can conclude that $\angle Q R S=60^{\circ}$.

Ex 6.2 Question 4.

$\text {In the given figure, if } \mathbf{A B} \| \mathbf{C D}, \angle A P Q=50^{\circ} \text { and } \angle P R D=127^{\circ} \text {, find } \mathbf{x} \text { and } \mathbf{y} \text {. }$

Answer.

We are given that $A B \| C D, \angle A P Q=50^{\circ}$ and $\angle P R D=127^{\circ}$.

We need to find the value of $x$ and $y$ in the figure.
$\angle A P Q=x=50^{\circ}$. (Alternate interior angles)
$\angle P R D=\angle A P R=127^{\circ}$. (Alternate interior angles)
$\angle A P R=\angle Q P R+\angle A P Q$.
$127^{\circ}=y+50^{\circ} \Rightarrow y=77^{\circ}$.

Therefore, we can conclude that $x=50^{\circ}$ and $y=77^{\circ}$

Ex 6.2 Question 5.

In the given figure, PQ and RS are two mirrors placed parallel to each other. An incident ray $A B$ strikes the mirror $P Q$ at $B$, the reflected ray moves along the path $B C$ and strikes the mirror RS at $C$ and again reflects back along $C D$. Prove that $A B$ || CD.

Answer.

We are given that $P Q$ and $R S$ are two mirrors that are parallel to each other.

We need to prove that $A B \| C D$ in the figure.

Let us draw lines $B X$ and $C Y$ that are parallel to each other, to get

We know that according to the laws of reflection
$
\angle A B X=\angle C B X \text { and } \angle B C Y=\angle D C Y \text {. }
$
$\angle B C Y=\angle C B X$ (Alternate interior angles)

We can conclude that $\angle A B X=\angle C B X=\angle B C Y=\angle D C Y$.

From the figure, we can conclude that
$
\angle A B C=\angle A B X+\angle C B X \text {, and } \angle D C B=\angle B C Y+\angle D C Y \text {. }
$

Therefore, we can conclude that $\angle A B C=\angle D C B$.

From the figure, we can conclude that $\angle A B C$ and $\angle D C B$ form a pair of alternate interior angles corresponding to the lines $A B$ and $C D$, and transversal $B C$.

Therefore, we can conclude that $A B \| C D$