Exercise 4.4 - Chapter 4 Geometry Term 1 6th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $4.4$
Miscellaneous Practice Problems
Question $1 .$
Find the type of lines marked in thick lines (Parallel, intersecting or perpendicular)

(i) Parallel lines
(ii) Parallel lines
(iii) Parallel lines and Perpendicular lines
(iv) Intersecting lines
 

Question $2 .$
Find the parallel and intersecting line segments in the picture given below.

Solution:
(a) Parallel line segments
- $\overline{\mathrm{YZ}}$ and $\overline{\mathrm{DE}}$
- $\overline{\mathrm{EA}}$ and $\overline{\mathrm{ZV}}$
- $\overline{\mathrm{VW}}$ and $\overline{\mathrm{AB}}$
- $\overline{\mathrm{WX}}$ and $\overline{\mathrm{BC}}$
- $\overline{\mathrm{YX}}$ and $\overline{\mathrm{DC}}$
- $\overline{\mathrm{YD}}$ and $\overline{\mathrm{XC}}$
- $\overline{\mathrm{XC}}$ and $\overline{\mathrm{WB}}$
- $\overline{\mathrm{WB}}$ and $\overline{\mathrm{VA}}$
- $\overline{\mathrm{VA}}$ and $\overline{\mathrm{ZE}}$
- $\overline{\mathrm{ZE}}$ and $\overline{\mathrm{YD}}$
(b) Intersecting line segments
- DE and ZV
- $\mathrm{WX}$ and $\mathrm{DC}$
 

Question $3 .$
Name the following angles as shown in the figure.

Solution:
(i) $\angle 1=\angle \mathrm{DBC}$ or $\angle \mathrm{CBD}$
(ii) $\angle 2=\angle \mathrm{DBE}$ or $\angle \mathrm{EBD}$
(iii) $\angle 3=\angle \mathrm{ABE}$ or $\angle \mathrm{EBA}$
(iv) $\angle 1+\angle 2=\angle \mathrm{EBC}$ or $\angle \mathrm{CBE}$
(v) $\angle 2+\angle 3=\angle \mathrm{ABD}$ or $\angle \mathrm{DHA}$
(vi) $\angle 1+\angle 2+\angle 3=\angle \mathrm{ABC}$ or $\angle \mathrm{B}$ or $\angle \mathrm{CBA}$
 

Question $4 .$
Measure the angles of the given figures using a protractor and identify the type of angle as acute, obtuse, right or straight.

Solution:
(i) $90^{\circ}$ - Right Angle
(ii) $45^{\circ}$ - Acute Angle
(iii) $180^{\circ}$ - Straight Angle
(iv) $105^{\circ}$ - Obtuse Angle

Question $5 .$
Draw the following angles using the protractor.
(i) $45^{\circ}$
(ii) $120^{\circ}$
(iii) $65^{\circ}$
(iv) $135^{\circ}$
(v) $0^{\circ}$
(vi) $180^{\circ}$
(vii) $38^{\circ}$
(viii) $90^{\circ}$

Construction:
1. Drawn the base ray PQ.
2. Placed the centre of the protractor at the vertex P. Lined up the ray $\overrightarrow{\mathrm{PQ}}$ with the $0^{\circ}$ line. Then drawn and labelled a pointed $(\mathrm{R})$ at the $45^{\circ}$ mark on the inner scale (a) anticlockwise and (b) outer scale (clockwise)
3. Removed the protractor and drawn at $\overrightarrow{\mathrm{PR}}$ to complete the angle
Now $\angle \mathrm{P}=\angle Q P R-\angle R P Q=45^{\circ}$.

Construction:
1. Placed the centre of the protractor at the vertex X. Lined up the ray $\overline{\mathrm{XY}}$ with the $0^{\circ}$ Line. Then draw and label a point $\mathrm{Z}$ at $120^{\circ}$ mark on the (a) inner scale (anti-clockwise) and (b) outer scale (clockwise).

2. Removed the protractor and draw $\overline{\mathrm{XZ}}$ to complete the angle. Now, $\angle \mathrm{X}=\angle \mathrm{ZXY}=\angle \mathrm{YXZ}=120^{\circ}$.

Construction:
1. Placed the centre of the protractor at the vertex A. Line up the ray $\overrightarrow{\mathrm{AB}}$ with the $0^{\circ}$ line. Then draw and label a point $\mathrm{C}$ at the $65^{\circ}$ mark on the (a) inner scale (anti-clockwise) (b) outer scale (clockwise).
2. Removed the protractor and draw $\overrightarrow{\mathrm{AB}}$ to complete the angle.
Now $\angle \mathrm{A}=\angle \mathrm{BAC}=\angle \mathrm{CAB}=65^{\circ}$.

Construction:
1. Placed the centre of the protractor at the Vertex A. Lined up the ray $\overrightarrow{E G}$ with the $0^{\circ}$ line. Then draw and label a point $\mathrm{F}$ at the $135^{\circ}$ mark on the (a) inner scale (anti-clockwise) and (b) outer scale (clockwise)
2. Removed the protractor and draw $\overrightarrow{\mathrm{EF}}$ to complete the angle.
Now $\angle \mathrm{E}=\angle \mathrm{FEG}=\angle \mathrm{GEF}=135^{\circ}$.

Construction:
1. Placed the centre of the protractor at the vertex G. Lined up the ray $\overrightarrow{\mathrm{GH}}$ with the $0^{\circ}$ line. Then draw and label a point $\mathrm{I}$ at the $0^{\circ}$ mark on the
(a) inner scale (anti-clockwise)
(b) outer scale (clockwise)
2. Removed the protractor and seen $\overrightarrow{G I}$ lies exactly on $\overrightarrow{\mathrm{GH}}$
Now $\angle \mathrm{G}=\angle \mathrm{HGI}=\angle \mathrm{IGH}=0^{\circ}$, which is a zero angle.

Construction:
1. Placed the centre of the protractor at the vertex I. Lined up the ray $\overrightarrow{I J}$ with the $0^{\circ}$ line. Then draw and labelled a point $\mathrm{K}$ at the $180^{\circ}$ mark on the (a) inner scale (anticlockwise) (b) outer scale (clockwise)
2. Removed the protractor and draw $\overrightarrow{I K}$ to complete the angle.
Now $\angle \mathrm{I}=\angle \mathrm{JHK}=\angle \mathrm{KIJ}=180^{\circ}$, which is a straight Angle.

Construction:
1. Placed the centre of the protractor at the vertex L. Lined up the ray $\overrightarrow{L M}$ with the $0^{\circ}$ line. Then draw and label a point $\mathrm{N}$ at $38^{\circ}$ mark on the (a) inner scale (anticlockwise) and (b)* huter scale (clockwise).
2. Removed the protractor and draw $\overrightarrow{\mathrm{LN}}$ to complete the angle. Now $\angle \mathrm{L}=\angle \mathrm{MLN}=\angle \mathrm{NLM}=38^{\circ}$.

Construction:
1. Placed the centre of the protractor at the vertex ' $\mathrm{O}$ '. Lined up the ray $\overrightarrow{\mathrm{OP}}$ with the $0^{\circ}$ line. Then draw and label a point $Q$ at $90^{\circ}$ mark on the (a) inner scale (anticlockwise) and (b) outer scale (clockwise)
2. Removed the protractor and draw $\overrightarrow{\mathrm{OQ}}$ to complete the angle.
Now $\angle \mathrm{O}=\angle \mathrm{POQ}=\angle \mathrm{QOP}=90^{\circ}$.
 

Question 6 .
From the figures given below, classify the following pairs of angles into complementary and noncomplementary.

Solution:
We know that the two angles are complementary if they add up to $90^{\circ}$.
Therefore (a) (i) is complementary.
In (v) $\angle \mathrm{ABC}$ and $\angle \mathrm{CBD}$ are complementary
(b) (ii), (iii), (iv) and (v) are non-complementary
 

Question $7 .$
From the figures given below, classify the following pairs of angles into supplementary and nonsupplementary.

Solution:
If two angles add up to $180^{\circ}$, then they are supplementary angles.

(a) In (ii) $\angle \mathrm{AOB}$ and $\angle \mathrm{BOD}$ are supplementary. In (iv) the pair is supplementary
(b) (i) and (iii) are not supplementary.
 

Question $8 .$
From the figure.

(i) name a pair of complementary angles
(ii) name a pair of supplementary angles
Solution:
(i) $\angle \mathrm{FAE}$ and $\angle \mathrm{DAE}$ are complementary
(ii) $\angle F A D$ and $\angle D A C$ are supplementary
 

Question $9 .$
Find the complementary angle of
(i) $30^{\circ}$
(ii) $26^{\circ}$
(iii) $85^{\circ}$
(iv) $0^{\circ}$

Solution:
When we have an angle, how far we need to go to reach the right angle is called the complementary angle.
(i) Complementary angle of $30^{\circ}$ is $90^{\circ}-30^{\circ}=60^{\circ}$
(ii) Complementary angle of $26^{\circ}$ is $90^{\circ}-26^{\circ}=64^{\circ}$
(iii) Complementary angle of $85^{\circ}$ is $90^{\circ}-85^{\circ}=5^{\circ}$
(iv) Complementary angle of $0^{\circ}$ is $90^{\circ}-0^{\circ}=90^{\circ}$
(v) Complementary angle of $90^{\circ}$ is $90^{\circ}-90^{\circ}=0^{\circ}$
 

Question 10 .
Find the supplementary angle of
(i) $70^{\circ}$
(ii) $35^{\circ}$
(iii) $165^{\circ}$
(iv) $90^{\circ}$

(v) $0^{\circ}$
(vi) $180^{\circ}$
(vii) $95^{\circ}$
Solution:
How far we should go in the same direction to reach the straight angle $\left(180^{\circ}\right)$ is called supplementary angle.
(i) Supplementary angle of $70^{\circ}=180^{\circ}-70^{\circ}=110^{\circ}$
(ii) Supplementary angle of $35^{\circ}$ is $180^{\circ}-35^{\circ}=145^{\circ}$
(iii) Supplementary angle of $165^{\circ}$ is $180^{\circ}-165^{\circ}=15^{\circ}$
(iv) Supplementary angle of $90^{\circ}$ is $180^{\circ}-90^{\circ}=90^{\circ}$
(v) Supplementary angle of $0^{\circ}$ is $180^{\circ}-0^{\circ}=180^{\circ}$
(vi) Supplementary angle of $180^{\circ}$ is $180^{\circ}-180^{\circ}=0^{\circ}$
(vii) Supplementary angle of $95^{\circ}$ is $180^{\circ}-95^{\circ}=85^{\circ}$

Challenging Problems
Question $11 .$
Think and write and object having.
(i) Parallel Lines
1 .
$2 .$
$3 .$
(ii) Perpendicular lines
1 .
$2 .$
$3 .$
(iii) Intersecting lines
$1 .$
$2 .$
$3 .$
Solution:
(i) 1. Opposite edges of a Table.
2. Path traced by the wheels of a car on a straight road
3. Opposite edges of a black board
(ii) 1 . Adjacent edges of a Table.
2. Hands of the block when it shows $3.30$
3. Strokes of the letter 'L'
(iii) 1 . Sides of a triangle
2. Strokes of letter ' $\mathrm{V}$ '
3. Hands of a scissors

Question $12 .$
Which angle is equal to twice of its complement?

Solution:
We know that the sum of complementary angles $90^{\circ}$ Given Angle $=2 \times$ Complementary angle

By trial and error, we find that Angle $=2 \times$ Complement for $60^{\circ}$
The required angle $=60^{\circ}$
Another method:
Let the angle be $\mathrm{x}$ given
$\mathrm{x}=2(90-\mathrm{x})$
$\Rightarrow \mathrm{x}=180-2 \mathrm{x}$
$\Rightarrow \mathrm{x}+2 \mathrm{x}=180$
$\Rightarrow 3 \mathrm{x}=180$
$\Rightarrow \mathrm{x}=60^{\circ}$
 

Question 13 .
Which angle is equal to two-thirds of its supplement.
Solution:
Supplementary angles sum upto $180^{\circ}$
Given Angle $=\frac{2}{3} \times$ Supplement.
Forming the Table.

By trial and error, we find that angle $=\frac{2}{3} \times$ supplement for $72^{\circ}$.
The required angle $72^{\circ}$.

Question $14 .$
Given two angles are supplementary and one angle is $20^{\circ}$ more than other. Find the two angles.
Solution:
Given two angles are supplementary i.e. their sum $=180^{\circ}$.
Let the angle be $x$
Then another angle $=x+20$ (given)

The two angles are $80^{\circ}$ and $100^{\circ}$.
 

Question 15 .
Two complementary angles are in the ratio $7: 2$. Find the angles.
Solution:
Let the angles be $7 \mathrm{x}, 2 \mathrm{x}$
According to the problem,
$\begin{aligned}
&7 x+2 x=90 \\
&9 x=90 \\
&x=\frac{90}{9} \\
&x=10
\end{aligned}$

$7 \mathrm{x}=7 \times 10$ $=70$ $2 \mathrm{x}=2 \times 10$ $=20$ $\therefore$ Two angles are $70^{\circ}$ and $20^{\circ}$ Question 16 . Two supplementary angles are in ratio $5: 4$. Find the angles. Solution: Total of two supplementary angles $=180^{\circ}$ Given they are in the ratio $5: 4$ Dividing total angles to $5+4=9$ equal parts. One angle $=\frac{5}{9} \times 180=100^{\circ}$ Another angle $=\frac{4}{9} \times 180=80^{\circ}$ Two angles are $100^{\circ}$ and $80^{\circ} .$
 

Question $16 .$
Two supplementary angles are in ratio $5: 4$. Find the angles.
Solution:
Total of two supplementary angles $=180^{\circ}$
Given they are in the ratio $5: 4$
Dividing total angles to $5+4=9$ equal parts.
One angle $=\frac{5}{9} \times 180=100^{\circ}$
Another angle $=\frac{4}{9} \times 180=80^{\circ}$
Two angles are $100^{\circ}$ and $80^{\circ}$.