5 Marks Questions

Question 15 Marks

List ten figures around you and identify the acute, obtuse and right angles found in them.

Answer

1.		In this figure, right angles are formed.
2.		In this figure, acute angles are formed.
3.		In this figure, right angles are formed.
4.		In this figure, acute angles are formed.
5.		In this figure, obtuse angles are formed
6.		In this figure, acute and right angles are formed.
7.		In this figure, right angles are formed.
8.		In this figure, acute angle is formed.
9.		In this figure, obtuse angle is formed.
10.		In this figure, acute angles are formed.

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Question 25 Marks

Amisha makes a star with the help of line segments $a, b, c, d, e$ and $f$ in which $a\|d, b\| e$ and $c \| f$. Chhaya marks an angle as $120^{\circ}$, as shown in figure and asks Amisha to find $\angle x, \angle y$ and $\angle z$. Help Amisha in finding the angles.

Answer

Given, $a\|d, b\| e, c \| f, a, b, c, d, e$ and $f$ are line segments.
Give them points $A, B, C, D, E, F, G, H, I, J$ and $K$.

$
\begin{array}{ll}
\because \angle A K E=120^{\circ} \\
\therefore \angle J K I=\angle A K E=120^{\circ}
\end{array}
$
[vertically opposite anglol
Now,$\ a \| d$
$
\begin{array}{lr}
\therefore \angle J K L+\angle K C H=180^{\circ} \quad \text { [cointerior anda] } \\
\Rightarrow 120^{\circ}+\angle z=180^{\circ} \\
\Rightarrow \angle z=180^{\circ}-120^{\circ}=60^{\circ}
\end{array}
$
Also, $\angle F G H+\angle K C H=180^{\circ} \quad$ [cointerior anglo]
$
\Rightarrow \angle F G H=180^{\circ}-60^{\circ}=120^{\circ}
$
$
\begin{array}{l}
\angle F G H=\angle B G D \\
\text { [vertically opposite angles] } \\
\Rightarrow \angle B G D=120^{\circ} \\
\Rightarrow \angle y=120^{\circ} \\
\text { and } \angle x+\angle F G H=180^{\circ} \quad \text { [cointerior angles] } \\
\therefore \angle x=180^{\circ}-120^{\circ}=60^{\circ} \\
\text { Hence, } \angle x=60^{\circ}, \angle y=120^{\circ} \text { and } \angle z=60^{\circ} \text {. }
\end{array}
$

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Question 35 Marks

In the following figure, $I \| m$ and a line $t$, intersects these lines at $P$ and $Q$, respectively. Find the value of $2 a+b$.

Answer

Since, lines $l$ and $m$ are parallel to each other and line $t$ is a transversal.
$\therefore \angle132^{\circ}$ and $\angle b$ are vertically opposite angles.
So,$
\angle b=132^{\circ}
$
Also,$\angle132^{\circ}$ and $\angle a$ are corresponding angles.
So,$
\angle a=132^{\circ}
$
The value of $2 a+b=2 \times 132^{\circ}+132^{\circ}=264^{\circ}+132^{\circ}$
$
\therefore 2 a+b=396^{\circ}
$
Hence, the value of $2 a+b$ is $396$.

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Question 45 Marks

In the given figure, examine whether the following pairs of lines are parallel or not.

Q.1. EF and GH
Q.2. AB and CD

Answer

Given, $\angle P S R=115^{\circ} ; \angle R Q D=70^{\circ}$ and $\angle C P F=65^{\circ}$
(i)
$\begin{array}{l} \angle P Q R+\angle R Q D=180^{\circ} \\ \Rightarrow \quad \angle P Q R=180^{\circ}-70^{\circ} \text { [linear pair] } \\ \Rightarrow \quad \angle P Q R=110^{\circ}\end{array}$
$\angle C P F=\angle S P Q=65^{\circ}\quad$ [vertically opposite angley]
If $E F \| G H$
Then, according to the definition of cointerior
angles, the sum of $\angle S P Q+\angle R Q P$ should be $180^{\circ}$.
$
\therefore \quad 65^{\circ}+110^{\circ}=175^{\circ} \neq 180^{\circ}
$
So, $E F$ is not parallel to $G H$.
(ii) $\angle C P F+\angle C P S=180^{\circ}\quad$[linear pair]
$\Rightarrow 65^{\circ}+\angle C P S=180^{\circ}$
$\Rightarrow \quad \angle C P S=180^{\circ}-65^{\circ}=115^{\circ}$
If $A B \| C D$, then according to the definition of alternate angles,
$\angle R S P=\angle C P S$
It is satisfying the condition. $\left[\because \angle R S P=\angle C P S=115^{\circ}\right]$
So, $A B$ is parallel to $C D$.

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Question 55 Marks

A road crosses a railway line at an angle of $30^{\circ}$ as shown in figure. Find the values of $a, b$ and $c$.

Answer

Firstly, make a rough diagram of given figure.

$\begin{array}{l}
\text { Given, }\angle P Q R=30^{\circ} \\
\angle H S T=\angle P Q S \quad \text { [corresponding angles] } \\
\text { or } \angle P Q S=\angle C \\
\because\angle R Q S=180^{\circ} \qquad\text { [by linear pair] } \\
\therefore \angle P Q R+\angle P Q S=180^{\circ} \\
30^{\circ}+\angle c=180^{\circ} \\
\Rightarrow \angle c=180^{\circ}-30^{\circ} \\
\Rightarrow \angle c=150^{\circ} \\
\because \angle X Y Q=\angle P Q S \quad \text { [corresponding angles] } \\
\text { and } \angle X Y Q=\angle b \quad \text { [vertically opposite angles] } \\
\therefore \angle b=\angle c \text { or } \angle b=150^{\circ} \\
\text { Now, } \angle I X S=\angle X Y Q \quad \text { [corresponding angles] } \\
\therefore \angle I X S=150^{\circ} \\
\text { Here, } \angle I X S+\angle a=180^{\circ} \quad \text { [linear pair] } \\
\Rightarrow 150^{\circ}+\angle a=180^{\circ} \\
\Rightarrow \angle a=180^{\circ}-150^{\circ}=30^{\circ}\end{array}$

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Question 65 Marks

Iron rods $a, b, c, d, e$ and $f$ are making a design of a bridge as shown in figure, in which $a\|b, c\| d, e \| f$. Find the marked angles between

Q.1. b and c
Q.2. d and e
Q.3. d and f
Q.4. c and f

Answer

1. Angle between $b$ and $c=30^{\circ}$
[vertically opposite angles]
2. Angle between $d$ and $e+$ Angle between $e$ and $c=180^{\circ}\qquad$[by linear pair]
$\therefore$ Angle between $d$ and $e=180^{\circ}-75^{\circ}=105^{\circ}$$\qquad$$\left[\because\right.$ Angle between $c$ and $\left.e=75^{\circ}\right]$
3. Angle between $d$ and $f+$ Angle between $d$ and $e=180^{\circ}$$\qquad$[ $\because$ pair of cointerior angles]
$\therefore$ Angle between $d$ and $f=180^{\circ}-105^{\circ}=75^{\circ}$
4. Angle between $c$ and $f=$ Angle between $d$ and $f$ $\quad$[corresponding angles]
$\therefore$ Angle between $c$ and $f=75^{\circ}$

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Question 75 Marks

In the following figure, $E F \| GH , \angle E A B=70^{\circ}$ and $\angle A C H=120^{\circ}$. Then, find $\angle C A F$ and $\angle B A C$.

Answer

Since, lines $E F$ and $G H$ are parallel to each other.
Where, $\angle E A B=70^{\circ}$ and $\angle A C H=120^{\circ}$
$\angle E A B$ and $\angle C B A$ are alternate angles.
So, $\angle C B A=70^{\circ}$
$\because(\angle E A B+\angle B A C)$ and $\angle A C H$ are alternate angles.
so,$\angle E A B+\angle B A C=120^{\circ}$
$\Rightarrow 70^{\circ}+\angle B A C=120^{\circ}$
$\Rightarrow \angle B A C=120^{\circ}-70^{\circ}=50^{\circ}$
$\because \angle C A E=\angle E A B+\angle B A C$
$\therefore \angle C A E=70^{\circ} \div 50^{\circ}=120^{\circ}$
Also,$\angle F A C+\angle C A E=180^{\circ}\qquad$[linear pair]
$\Rightarrow \angle F A C=180^{\circ}-120^{\circ}$
$\therefore \angle F A C=60^{\circ}$

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Question 85 Marks

In the given figure, $A E\|G F\| B D, A B\|C G\| D F$, then find $\angle A B C$ and $\angle C D E$

Answer

Given, $A E\|G F\| B D, A B\|C G\| D F, \angle C H E=120^{\circ}$
$
\begin{array}{l}
\because A E \| B D \\
\Rightarrow \angle C H E+\angle H C D=180^{\circ} \\
\Rightarrow 120^{\circ}+\angle H C D=180^{\circ} \\
\Rightarrow \angle H C D=180^{\circ}-120^{\circ}=60^{\circ}.
\end{array}
$
$
\Rightarrow \angle H C D=\angle A B C=60^{\circ} \quad \text { [corresponding angles] }
$
$
\begin{array}{crl}
\text { Also, } \angle H C D+\angle C D E=180^{\circ} \\
\Rightarrow 60^{\circ}+\angle C D E=180^{\circ} \\
\Rightarrow \angle C D E =180^{\circ}-60^{\circ} \\
\therefore \angle C D E=120^{\circ}
\end{array}
$

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Question 95 Marks

In the following figure, $A B\|C D, A F\| E D, \angle A F C=70^{\circ}$ and $\angle F E D=40^{\circ}$, then find $\angle E F D$.

Answer

Since, $A F \| E D$ and $E F$ is a transversal.
So, $\angle A F E=\angle F E D=40^{\circ} \quad$ [alternate interior angles]
Also, $ A B \| C D$
$70^{\circ}$ and ( $\left.\angle A F E+\angle E F D\right)$ form a linear pair angles.
$\begin{aligned} 70^{\circ}+(\angle A F E+\angle E F D) & =180^{\circ} \\ \Rightarrow 70^{\circ}+40^{\circ}+\angle E F D & =180^{\circ} \\ \Rightarrow \quad \angle E F D & =180^{\circ}-110^{\circ} \\ \Rightarrow \quad \angle E F D & =70^{\circ}\end{aligned}$

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MATHS 5 Marks Questions

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