Question 13 Marks
Lines /||m, p|| q, Find $a, b, c, d ?$
Questions
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Lines /||m, p|| q, Find $a, b, c, d ?$
Question 23 Marks
Try to identify a few transversals in your surroundings.
Answer
View full question & answer→Examples of transversals from our surroundings are as follow:
(i) A road crossing two or more roads.
(ii) A railway line crossing several other lines.
(iii) Grills of the window.
(iv) A towel stand.
(i) A road crossing two or more roads.
(ii) A railway line crossing several other lines.
(iii) Grills of the window.
(iv) A towel stand.
Question 33 Marks
Find the measures of the angles made by the intersecting lines at the vertices of an equilateral triangle.
Answer
View full question & answer→Let $\triangle A B C$ be an equilateral triangle.
Since, all the angles of an equilateral triangle are equal.
$
\therefore\angle A=\angle B=\angle C=x
$
We know that sum of all the angles of a triangle is $180^{\circ}$.
$
\begin{array}{rlrl}
\therefore\angle A+\angle B+\angle C=180^{\circ} \\
\Rightarrow x+x+x=180^{\circ} \\
\Rightarrow 3 x=180^{\circ} \\
\Rightarrow x=\frac{180^{\circ}}{3}=60^{\circ}
\end{array}
$
Hence, $\angle A=\angle B=\angle C=60^{\circ}$
Since, all the angles of an equilateral triangle are equal.
$
\therefore\angle A=\angle B=\angle C=x
$
We know that sum of all the angles of a triangle is $180^{\circ}$.
$
\begin{array}{rlrl}
\therefore\angle A+\angle B+\angle C=180^{\circ} \\
\Rightarrow x+x+x=180^{\circ} \\
\Rightarrow 3 x=180^{\circ} \\
\Rightarrow x=\frac{180^{\circ}}{3}=60^{\circ}
\end{array}
$
Hence, $\angle A=\angle B=\angle C=60^{\circ}$
Question 43 Marks
In the given figure, pll q. Find the unknown angles.
Answer
View full question & answer→Given, $p \| q$
$
\angle e+125^{\circ}=180^{\circ}
\qquad$[by linear pair]
$\begin{array}{ll}\Rightarrow & \angle e=180^{\circ}-125^{\circ} \\ \Rightarrow & \angle e=55^{\circ}\end{array}$
$\therefore \quad \angle f=\angle e=55^{\circ} \quad$ [vertically opposite angles]
Since, $p \| q$ and $t$ is a transversal.
$\therefore \quad \angle a=\angle f=55^{\circ} \quad$ [alternate interior angles]
$\angle d=125^{\circ}\qquad$[corresponding angles]
$\angle c=\angle a=55^{\circ} \quad$ [vertically opposite angles]
and $\angle b=\angle d=125^{\circ} \quad$ [vertically opposite angles]
Hence, $\angle a=55^{\circ}, \angle b=125^{\circ}, \angle c=55^{\circ}, \angle d=125^{\circ}$, $\angle e=55^{\circ}$ and $\angle f=55^{\circ}$.
$
\angle e+125^{\circ}=180^{\circ}
\qquad$[by linear pair]
$\begin{array}{ll}\Rightarrow & \angle e=180^{\circ}-125^{\circ} \\ \Rightarrow & \angle e=55^{\circ}\end{array}$
$\therefore \quad \angle f=\angle e=55^{\circ} \quad$ [vertically opposite angles]
Since, $p \| q$ and $t$ is a transversal.
$\therefore \quad \angle a=\angle f=55^{\circ} \quad$ [alternate interior angles]
$\angle d=125^{\circ}\qquad$[corresponding angles]
$\angle c=\angle a=55^{\circ} \quad$ [vertically opposite angles]
and $\angle b=\angle d=125^{\circ} \quad$ [vertically opposite angles]
Hence, $\angle a=55^{\circ}, \angle b=125^{\circ}, \angle c=55^{\circ}, \angle d=125^{\circ}$, $\angle e=55^{\circ}$ and $\angle f=55^{\circ}$.
Question 53 Marks
State the property that is used in each of the following statements.
(i) If $a \| b$, then $\angle 1=\angle 5$.
(ii) If $\angle 4=\angle 6$, then $a \| b$.
(iii) If $\angle 4+\angle 5=180^{\circ}$, then $a \| b$.
(i) If $a \| b$, then $\angle 1=\angle 5$.
(ii) If $\angle 4=\angle 6$, then $a \| b$.
(iii) If $\angle 4+\angle 5=180^{\circ}$, then $a \| b$.
Answer
View full question & answer→(i) We have if $a \| b$, then $\angle1=\angle 5$.
If a transversal intersects two parallel lines, then the corresponding angles are equal.
$\therefore$ By corresponding angles property, it is true.
(ii) We have if $\angle 4=\angle 6$, then $a \| b$.
If a transversal intersects two parallel lines, then the alternate interior angles are equal.
$\therefore$ By alternate interior angles property, it is true.
(iii) We have if $\angle 4+\angle 5=180^{\circ}$, then $a \| b$.
If a transversal intersects two parallel lines, then the interior angles on the same side of the transversal are supplementary.
If a transversal intersects two parallel lines, then the corresponding angles are equal.
$\therefore$ By corresponding angles property, it is true.
(ii) We have if $\angle 4=\angle 6$, then $a \| b$.
If a transversal intersects two parallel lines, then the alternate interior angles are equal.
$\therefore$ By alternate interior angles property, it is true.
(iii) We have if $\angle 4+\angle 5=180^{\circ}$, then $a \| b$.
If a transversal intersects two parallel lines, then the interior angles on the same side of the transversal are supplementary.
Question 63 Marks
In the following figure, $P Q \| R T$. Find the value of $a+b$
Answer
View full question & answer→Since, $P Q \| R T$ and $R Q$ is a transversal. $\angle R P Q$ and $\angle a$ are corresponding angles.
$
\begin{aligned}
\therefore\angle R P Q=\angle a\\
\Rightarrow\angle a =45^{\circ}
\end{aligned}
$
Also, $\angle b$ and $\angle R Q P$ are alternate angles.
$
\therefore \angle b=55^{\circ}
$
Hence, $\angle a+\angle b=45^{\circ}+55^{\circ}=100^{\circ}$
$
\begin{aligned}
\therefore\angle R P Q=\angle a\\
\Rightarrow\angle a =45^{\circ}
\end{aligned}
$
Also, $\angle b$ and $\angle R Q P$ are alternate angles.
$
\therefore \angle b=55^{\circ}
$
Hence, $\angle a+\angle b=45^{\circ}+55^{\circ}=100^{\circ}$
Question 73 Marks
View full question & answer→
Question 83 Marks
In the given figure, $A B \| C D$ Find the reflex of $\angle E F G$.
Answer
View full question & answer→$\begin{array})\text{Given,}\angle 1 =34^{\circ}\\
\angle A E F=34^{\circ} & \text { [alternate angles] } \\
\because\angle F G D=135^{\circ} \\
\therefore \angle 2+135^{\circ}=180^{\circ}
\end{array}
$
[sum of pair of cointerior angles is $180^{\circ}$ ]
$
\begin{array}{ll}
\Rightarrow\angle 2=180^{\circ}-135^{\circ} \\
\Rightarrow \angle 2=45^{\circ}
\end{array}
$
$
\Rightarrow \angle1+\angle 2=34^{\circ}+45^{\circ}=79^{\circ}
$
Reflex of $\angle E F G=360^{\circ}-79^{\circ}=281^{\circ}$
[since, reflex angle is less than $360^{\circ}$ and more than $180^{\circ}$]
\angle A E F=34^{\circ} & \text { [alternate angles] } \\
\because\angle F G D=135^{\circ} \\
\therefore \angle 2+135^{\circ}=180^{\circ}
\end{array}
$
[sum of pair of cointerior angles is $180^{\circ}$ ]
$
\begin{array}{ll}
\Rightarrow\angle 2=180^{\circ}-135^{\circ} \\
\Rightarrow \angle 2=45^{\circ}
\end{array}
$
$
\Rightarrow \angle1+\angle 2=34^{\circ}+45^{\circ}=79^{\circ}
$
Reflex of $\angle E F G=360^{\circ}-79^{\circ}=281^{\circ}$
[since, reflex angle is less than $360^{\circ}$ and more than $180^{\circ}$]
Question 93 Marks
Answer
View full question & answer→Given, $l\|m\| n$ and $p \| q$
$
\begin{aligned}
\therefore 6 a =120^{\circ} \qquad \text {[corresponding angles]} \end{aligned}
$
$
\begin{aligned}\Rightarrow a=\frac{120^{\circ}}{6}=20^{\circ}
\end{aligned}
$
Similarly, $4 c=120^{\circ}\qquad \text {[corresponding angles] } $
$
\Rightarrow c=\frac{120^{\circ}}{4}=30^{\circ} \text { and } 3 b=4 c\qquad \text {[corresponding angles] }
$
$
\Rightarrow b=\frac{4 \times 30^{\circ}}{3} \Rightarrow b=40^{\circ}
$
$
\begin{aligned}
\therefore 6 a =120^{\circ} \qquad \text {[corresponding angles]} \end{aligned}
$
$
\begin{aligned}\Rightarrow a=\frac{120^{\circ}}{6}=20^{\circ}
\end{aligned}
$
Similarly, $4 c=120^{\circ}\qquad \text {[corresponding angles] } $
$
\Rightarrow c=\frac{120^{\circ}}{4}=30^{\circ} \text { and } 3 b=4 c\qquad \text {[corresponding angles] }
$
$
\Rightarrow b=\frac{4 \times 30^{\circ}}{3} \Rightarrow b=40^{\circ}
$
Question 103 Marks
In the given figure, two parallel lines $l$ and $m$ are cut by two transversals $n$ and $p$. Find the value of $x$ and $y$.
Answer
View full question & answer→Given, $l \| m, n$ and $p$ are transversals, $\angle T B O=66^{\circ}$.
$
\begin{array}{lc}
\therefore\angle T B O+\angle x=180^{\circ} \\
{\left[\because l \| m \text { and cointerior angles has a sum of } 180^{\circ}\right]} \\
\Rightarrow\angle x=180^{\circ}-66^{\circ}=114^{\circ} \\
\because \angle O C A=\angle B D O
\end{array}
$
[alternate angles formed by transversal $m$ are equal]
$
\begin{array}{ll}
\because \angle O C A=48^{\circ} \\
\text { So, } \angle B D O=48^{\circ} \text { [alternate interior angles] }
\end{array}
$
$
\because \angle B D O+\angle Q D O=180^{\circ}\text { [linear pair] }
$
$
\Rightarrow 48^{\circ}+\angle y=180^{\circ}
$
$
\Rightarrow \angle y=180^{\circ}-48^{\circ}=132^{\circ}
$
$
\begin{array}{lc}
\therefore\angle T B O+\angle x=180^{\circ} \\
{\left[\because l \| m \text { and cointerior angles has a sum of } 180^{\circ}\right]} \\
\Rightarrow\angle x=180^{\circ}-66^{\circ}=114^{\circ} \\
\because \angle O C A=\angle B D O
\end{array}
$
[alternate angles formed by transversal $m$ are equal]
$
\begin{array}{ll}
\because \angle O C A=48^{\circ} \\
\text { So, } \angle B D O=48^{\circ} \text { [alternate interior angles] }
\end{array}
$
$
\because \angle B D O+\angle Q D O=180^{\circ}\text { [linear pair] }
$
$
\Rightarrow 48^{\circ}+\angle y=180^{\circ}
$
$
\Rightarrow \angle y=180^{\circ}-48^{\circ}=132^{\circ}
$
Question 113 Marks
In the following figure, $\angle 2=58^{\circ}$. Find $\angle 1$ and $\angle 3$.
View full question & answer→Question 123 Marks
In the following figure, if $/ \| m$, find the value of $\angle a$ and $\angle b$.
View full question & answer→Question 133 Marks
Answer
View full question & answer→Since, lines $l_1$ and $l_2$ are parallel to each other, where $l_3$ is transversal.
$\therefore \angle Q$ and $\angle 52^{\circ}$ are vertically opposite angles.
So, $\angle Q=52^{\circ}$
Also, angles $128^{\circ}$ and $\angle P$ form a linear pair angles.
So, $128^{\circ}+\angle P=180^{\circ}$
$
\Rightarrow \angle P=180^{\circ}-128^{\circ} \Rightarrow \angle P=52^{\circ}
$
$\therefore \angle Q$ and $\angle 52^{\circ}$ are vertically opposite angles.
So, $\angle Q=52^{\circ}$
Also, angles $128^{\circ}$ and $\angle P$ form a linear pair angles.
So, $128^{\circ}+\angle P=180^{\circ}$
$
\Rightarrow \angle P=180^{\circ}-128^{\circ} \Rightarrow \angle P=52^{\circ}
$
Question 143 Marks
In the following figure, $I_1 \| I_2$. Find the value of $\angle M$.
Answer
View full question & answer→Since, lines $l_1$ and $l_2$ are parallel to each other, where $l_3$ is transversal.
$\therefore \angle P$ and $\angle 130^{\circ}$ form a linear pair angles.
So, $ \angle P+130^{\circ}=180^{\circ}$
$
\Rightarrow \angle P=180^{\circ}-130^{\circ} \Rightarrow \angle P=50^{\circ}
$
Also, $\angle P$ and $\angle M$ are alternate interior angles.
$
\angle M=\angle P=50^{\circ}
$
$\therefore \angle P$ and $\angle 130^{\circ}$ form a linear pair angles.
So, $ \angle P+130^{\circ}=180^{\circ}$
$
\Rightarrow \angle P=180^{\circ}-130^{\circ} \Rightarrow \angle P=50^{\circ}
$
Also, $\angle P$ and $\angle M$ are alternate interior angles.
$
\angle M=\angle P=50^{\circ}
$
Question 153 Marks
In the following figure, $I_1 \| I_2$. Find the value of $\angle P$ and $\angle Q$.
Answer
View full question & answer→Since, lines $l_1$ and $l_2$ are parallel to each other, where $l_3$ is transversal.
$\therefore \angle 130^{\circ}$ and $\angle P$ are corresponding angles.
So, $ \angle P=130^{\circ}$
Also, $\angle P$ and $\angle Q$ are vertically opposite angles .
So, $ \angle Q=\angle P=130^{\circ} \Rightarrow \angle Q=130^{\circ}$
$\therefore \angle 130^{\circ}$ and $\angle P$ are corresponding angles.
So, $ \angle P=130^{\circ}$
Also, $\angle P$ and $\angle Q$ are vertically opposite angles .
So, $ \angle Q=\angle P=130^{\circ} \Rightarrow \angle Q=130^{\circ}$
Question 163 Marks
In the following figure, $I_2 \| I_3$. Find the value of $\angle m$ and $\angle n$.
Answer
View full question & answer→Since, lines $I_2$ and $l_3$ are parallel to each other, where $l_1$ is transversal.
$\therefore \angle 60^{\circ}$ and $m$ are alternate interior angles.
So, $m=60^{\circ}$
Also, angles $m$ and $n$ form a linear pair angles.
So, $m+n=180^{\circ} \Rightarrow 60^{\circ}+n=180^{\circ}$
$\Rightarrow n=180^{\circ}-60^{\circ}$
$\Rightarrow n=120^{\circ}$
$\therefore \angle 60^{\circ}$ and $m$ are alternate interior angles.
So, $m=60^{\circ}$
Also, angles $m$ and $n$ form a linear pair angles.
So, $m+n=180^{\circ} \Rightarrow 60^{\circ}+n=180^{\circ}$
$\Rightarrow n=180^{\circ}-60^{\circ}$
$\Rightarrow n=120^{\circ}$
Question 173 Marks
In the following figure, $I_1 \| I_2$. Find the value of $\angle 1$ and $\angle 2$ :
Answer
View full question & answer→Since, line $l_1$ is parallel to the line $l_2$, where line $l_3$ is transversal and $\angle 30^{\circ}$ and $\angle 1$ are alternate interior angles.
So, $ \angle 1=30^{\circ}$
Also, $\angle 1$ and $\angle 2$ form a linear pair angles.
So, $\angle 1+\angle 2=180^{\circ} \Rightarrow 30^{\circ}+\angle 2=180^{\circ}$
$
\Rightarrow \angle 2=180^{\circ}-30^{\circ} \Rightarrow \angle 2=150^{\circ}
$
So, $ \angle 1=30^{\circ}$
Also, $\angle 1$ and $\angle 2$ form a linear pair angles.
So, $\angle 1+\angle 2=180^{\circ} \Rightarrow 30^{\circ}+\angle 2=180^{\circ}$
$
\Rightarrow \angle 2=180^{\circ}-30^{\circ} \Rightarrow \angle 2=150^{\circ}
$