Additional Questions - Chapter 4 Geometry 9th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Exercise 4.1
Question $1 .$
Find the complement of each of the following angles
(i) $63^{\circ}$
(ii) $24^{\circ}$
(iii) $48^{\circ}$
Solution:
(i) The complement of $63^{\circ}=90^{\circ}-63^{\circ}=27^{\circ}$
(ii) The complement of $24^{\circ}=90^{\circ}-24^{\circ}=66^{\circ}$
(iii) The complement of $48^{\circ}=90^{\circ}-48^{\circ}=42^{\circ}$

Question $2 .$
Find the supplement of each of the following angles
(i) $58^{\circ}$
(ii) $148^{\circ}$
(iii) $120^{\circ}$
Solution:
(i) The supplement of $58^{\circ}=180^{\circ}-58^{\circ}=122^{\circ}$
(ii) The supplement of $148^{\circ}=180^{\circ}-148^{\circ}=32^{\circ}$
(iii) The supplement of $120^{\circ}=180^{\circ}-120^{\circ}=60^{\circ}$
 

Question $3 .$
Find the value of $x$

Solution:

(i) From the figure
$
\begin{aligned}
\angle \mathrm{AOC}+\angle \mathrm{COD}+\angle \mathrm{BOD} & =180 \\
x-20^{\circ}+x+40^{\circ} & =180^{\circ} \\
\Rightarrow 2 x+20^{\circ} & =180^{\circ} \\
\Rightarrow 2 x & =160^{\circ} \\
x & =80^{\circ}
\end{aligned}
$

(ii) From the figure
$
\begin{aligned}
\angle \mathrm{AOC}+\angle \mathrm{COD}+\angle \mathrm{BOD} & =180 \\
x+30^{\circ}+115^{\circ}-\not x+\not x & =180^{\circ} \\
x+145 & =180^{\circ} \\
x=180^{\circ}-145^{\circ} & =35^{\circ}
\end{aligned}
$
Question $4 .$
Find the values of $x, y$ in the following figures

Solution:
(i)
$\begin{aligned}
3 x+2 x &=180^{\circ} \quad \text { (linear pair) } \\
5 x &=180^{\circ} \\
x &=\frac{180^{\circ}}{5}=36^{\circ}
\end{aligned}$
(ii)
$\begin{aligned}
3 x+60^{\circ} &=180^{\circ} \text { (linear pair) } \\
\therefore 3 x &=180^{\circ}-60^{\circ}=120^{\circ} \\
\therefore x &=\frac{120^{\circ}}{3}=40^{\circ} \\
y+90^{\circ}+x &=180^{\circ} \\
\text { (i.e.) } y+90^{\circ}+40^{\circ} &=180^{\circ} \\
\therefore y &=180^{\circ}-40^{\circ}-90^{\circ}=50^{\circ}
\end{aligned}$

Question 5 .
In the given figure at right, side $\mathrm{BC}$ of $\triangle \mathrm{ABC}$ is produced to $\mathrm{D}$. Find $\angle \mathrm{A}$ and $\angle \mathrm{C}$.

Solution:
From the figure
$\begin{aligned}
&\text { Exterior angle }=120^{\circ} \\
&\Rightarrow \angle C=180^{\circ}-120^{\circ}=60^{\circ} \text { (linear pair) } \\
&\therefore \angle A=180^{\circ}-\left(40^{\circ}+60^{\circ}\right)=80^{\circ}
\end{aligned}$

Exercise 4.2
Question $1 .$
If the measures of three angles of a quadrilateral are $100^{\circ}, 84^{\circ}$ and $76^{\circ}$ then, find the measure of fourth angle.
Solution:
Let the measure of the fourth angle be $x^{\circ}$.
The sum of the angles of a quadrilateral is $360^{\circ}$
$\begin{aligned}
&\text { So, } 100^{\circ}+84^{\circ}+76^{\circ}+x^{\circ}=360^{\circ} \\
&260^{\circ}+x^{\circ}=360^{\circ} \\
&x=360^{\circ}-260^{\circ}=100^{\circ}
\end{aligned}$
Hence, the measure of the fourth angle is $100^{\circ}$.
 

Question $2 .$
In the parallelogram $\mathrm{ABCD}$ if $\angle \mathrm{A}=65^{\circ}$, find $\angle \mathrm{B}, \angle \mathrm{C}$ and $\angle \mathrm{D}$.
Solution:
Let $\mathrm{ABCD}$ be a parallelogram in which $\angle \mathrm{A}=65^{\circ}$
Since $\mathrm{AD}|| \mathrm{BC}$ we can treat $\mathrm{AB}$ as a transversal. So
$\angle \mathrm{A}+\angle \mathrm{B}=180^{\circ}$

$\begin{aligned}
&65^{\circ}+\angle B=180^{\circ} \\
&\angle B=180^{\circ}-65^{\circ} \\
&\angle B=115^{\circ} \\
&\text { Since the opposite angles of a parallelogram are equal, we have } \\
&\angle C=\angle A=65^{\circ} \text { and } \angle D=\angle B=115^{\circ} \\
&\text { Hence, } \angle B=115^{\circ}, \angle C=65^{\circ} \text { and } \angle D=115^{\circ} \\
&\text { Question } 3 . \\
&\text { If ABCD is a rhombus and if } \angle A=76^{\circ} \text {, find } \angle C D B \text {. } \\
&\text { Solution: } \\
&\angle A=\angle C=76^{\circ} \text { (Opposite angles of a rhombus) } \\
&\text { Let } \angle C D B=x^{\circ} \text {. In } \triangle C D B, C D=C B \\
&\angle C D B+\angle C B D+\angle D C B=180^{\circ} \text { (Angles of a triangle) }
\end{aligned}$

Question $4 .$
In a parallelogram, opposite sides are equal

Solution:
Given $\mathrm{ABCD}$ is a parallelogram
To Prove $\mathrm{ABCD}$ and $\mathrm{DA}=\mathrm{BC}$
Construction Join AC
Proof
Since $A B C D$ is a parallelogram

$\mathrm{AD} \| \mathrm{BC}$ and $\mathrm{AC}$ is the transversal
$\angle \mathrm{DAC}=\angle \mathrm{BCA} \rightarrow(1)$ (alternate angles are equal)
$\mathrm{AB} \| \mathrm{DC}$ and $\mathrm{AC}$ is the transversal
$\angle \mathrm{BAC}=\angle \mathrm{DCA} \rightarrow(2)$ (alternate angles are equal)
In $\triangle \mathrm{ADC}$ and $\triangle \mathrm{CBA}$
$\angle \mathrm{DAC}=\angle \mathrm{BCA}$ from (1)
$\mathrm{AC}$ is common
$\angle \mathrm{DCA}=\angle \mathrm{BAC}$ from $(2)$
$\triangle \mathrm{ADC} \cong \triangle \mathrm{CBA}$ (By ASA)
Hence $\mathrm{AD}=\mathrm{CB}$ and $\mathrm{DC}=\mathrm{BA}$ (Corresponding sides are equal)
 

Question $5 .$
The angles of a quadrilateral are $;$ the ratio $1: 2: 3: 4$. Find all the angles. Let each ratio be $x$.
Solution:
Then the angles are $\mathrm{x}^{\circ}, 2 \mathrm{x}^{\circ}, 3 \mathrm{x}^{\circ}, 4 \mathrm{x}^{\circ}$
$\begin{gathered}
x^{\circ}+2 x^{\circ}+3 x^{\circ}+4 x^{\circ}=360^{\circ} \\
10 x^{0}=360^{\circ} \\
x^{\circ}=\frac{360}{10}=36^{\circ}
\end{gathered}$

Exercise 4.3
Question 1.
The radius of a circle $15 \mathrm{~cm}$ and the length of one of its chord is $24 \mathrm{~cm}$. Find the distance of the chord from the centre.
Solution:
Distance of the chord from the centre.

$=\sqrt{15^{2}-12^{2}}$
$=\sqrt{225-144}$
$=\sqrt{81}=9 \mathrm{~cm}$

Question 2.
The chord of length $32 \mathrm{~cm}$ is drawn at the distance of $12 \mathrm{~cm}$ from the centre of the circle. Find the radius of the circle.
Solution:

Radius of the circle
$\begin{aligned}
&=\sqrt{16^{2}+12^{2}} \\
&=\sqrt{256+144} \\
&=\sqrt{400}=\sqrt{20 \times 20}=20
\end{aligned}$
 

Question $3 .$
In a circle, $\mathrm{AB}$ and $\mathrm{CD}$ are two parallel chords with centre $\mathrm{O}$ and radius $5 \mathrm{~cm}$ such that $\mathrm{AB}=8 \mathrm{~cm}$ and $\mathrm{CD}=6$ $\mathrm{cm}$ determine the distance between the chords?
Solution:

The distance between the two chord
$\begin{aligned}
\mathrm{FE} &=\mathrm{OF}+\mathrm{OE} \\
\mathrm{OE} &=\sqrt{5^{2}-3^{2}}=\sqrt{25-9} \\
&=\sqrt{16}=4 \mathrm{~cm} \\
\mathrm{OF} &=\sqrt{5^{2}-4^{2}}=\sqrt{25-16}=\sqrt{9}=3 \mathrm{~cm} \\
\therefore \mathrm{DE} &=4 \mathrm{~cm}+3 \mathrm{~cm}=7 \mathrm{~cm}
\end{aligned}$
Distance between the chord is $7 \mathrm{~cm}$.

Question $4 .$
Find the value of $x^{\circ}$

Solution:
$\begin{aligned}
x^{\circ} &=\frac{1}{2} \angle \mathrm{BQC} \\
&=\frac{1}{2}\left(50^{\circ}+30^{\circ}\right) \\
&=\frac{1}{2} \times 80=40^{\circ}
\end{aligned}$

Question $5 .$
Find the value of $x^{\circ}$

$\angle \mathrm{QPR}=\frac{1}{2} \angle \mathrm{QOR}=\frac{1}{2} \times 70^{\circ}=35^{\circ}$
In $\triangle \mathrm{QPR}$
$\begin{aligned}
\angle \mathrm{R}+\angle \mathrm{P}+\angle \mathrm{Q} &=180^{\circ} \\
75^{\circ}+35^{\circ}+\angle \mathrm{Q} &=180^{\circ} \\
\angle \mathrm{Q} &=180^{\circ}-110^{\circ}=70^{\circ} \\
\angle \mathrm{Q} &=x^{\circ}+55^{\circ}=70^{\circ} \\
x^{\circ} &=70^{\circ}-55^{\circ}=15^{\circ}
\end{aligned}$

Exercise 4.4
Question $1 .$
Find the value of $\mathrm{x}$ in the figure.
Solution:

In the cyclic quadrilateral ABCD
$\begin{aligned}
&\angle \mathrm{ABC}-180^{\circ}-140^{\circ}=40^{\circ} \\
&\angle \mathrm{BCA}=90^{\circ} \\
&\therefore \mathrm{x}=\angle \mathrm{BAC}=180^{\circ}-\left(90^{\circ}+40^{\circ}\right)=50^{\circ}
\end{aligned}$

Question $2 .$
Find all the angles of the given cyclic quadrilateral $A B C D$ in the figure.
Solution:

$\begin{aligned}
&\text { In the cyclic quadrilateral } \angle \mathrm{A}+\angle \mathrm{C}=180^{\circ}\\
&y+4^{\circ}+3 y^{\circ}+8^{\circ}=180^{\circ}\\
&4 y^{\circ}+12^{\circ}=180^{\circ}\\
&4 y^{\circ}=180^{\circ}-12^{\circ}\\
&4 y^{\circ}=168\\
&y=\frac{168}{4}=42\\
&\angle \mathrm{B}+\angle \mathrm{D}=180^{\circ}\\
&8 x+12=180^{\circ}\\
&8 x=180^{\circ}-12^{\circ}\\
&8 x=168\\
&x=\frac{168}{8}=21^{\circ}\\
&\therefore \angle \mathrm{A}=y+4^{\circ}=42^{\circ}+4^{\circ}=46\\
&\angle \mathrm{C}=3 y+8=3 \times 42+8\\
&=126+8=134^{\circ}\\
&\angle \mathrm{B}=3 x+6=3 \times 21+6\\
&=63+6=69^{\circ}\\
&\angle \mathrm{D}=5 x+6=5 \times 21+6\\
&=105+6=111^{\circ}
\end{aligned}$

Question $3 .$
$\mathrm{AB}$ and $\mathrm{CD}$ are two parallel sides of a cyclic quadrilateral $\mathrm{ABCD}$ in the figure. such that $\mathrm{AB}=12 \mathrm{~cm}, \mathrm{CD}=16$ $\mathrm{cm}$ and the radius of the circle is $10 \mathrm{~cm}$. Find the shortest distance between the two sides $\mathrm{AB}$ and $\mathrm{CD}$.
Solution:
In this figure,

$\begin{aligned}
\mathrm{OE} &=\sqrt{10^{2}-6^{2}} \\
&=\sqrt{100-36} \\
&=\sqrt{64}=8 \mathrm{~cm} \\
\mathrm{OF} &=\sqrt{10^{2}-8^{2}} \\
&=\sqrt{100-64}=\sqrt{36}=6 \mathrm{~cm}
\end{aligned}$
The shortest distance between the two sides $=8+6=14 \mathrm{~cm}$

Question $4 .$
In the given figure, $\mathrm{AB}$ and $\mathrm{CD}$ are the parallel chords of a circle with centre $\mathrm{O}$, such that $\mathrm{AB}=30 \mathrm{~cm}$ and $\mathrm{CD}$ $=40 \mathrm{~cm}$. If $\mathrm{OM} \perp \mathrm{AB}$ and $\mathrm{OL} \perp \mathrm{CD}$ distance between $\mathrm{LM}$ is $35 \mathrm{~cm}$. Find the radius of the circle?

Solution;

In the figure
$\begin{aligned}
\mathrm{LM} &=35 \mathrm{~cm} \\
\text { Let } \mathrm{OM} &=(35-x) \mathrm{cm} \\
\mathrm{MB} &=\frac{30}{2}=15 \mathrm{~cm} \\
\mathrm{OB} &=\sqrt{15^{2}+(35-x)^{2}} \\
\mathrm{OD} &=\sqrt{20^{2}+x^{2}} \\
\mathrm{OB} &=\mathrm{OD}(\because \text { radius }) \\
\sqrt{225+(35-x)^{2}} &=\sqrt{400+x^{2}}
\end{aligned}$
Squaring both sides
$\begin{aligned}
225+(35-x)^{2} &=400+x^{2} \\
225+1225+x^{2}-70 x &=400+x^{2} \\
1050 &=70 x \\
x &=\frac{1050}{70}=15 \mathrm{~cm} \\
\therefore \text { Radius OD } &=\sqrt{15^{2}+20^{2}}=\sqrt{225+400}=\sqrt{625}=25
\end{aligned}$

Exercise $4.5$
Question $1 .$
Construct an equilateral triangle of sides $6 \mathrm{~cm}$ and locate its orthocentre.
Solution:

Construction:
(1) Draw the $\triangle \mathrm{ABC}$ with the given measurements.
(2) Construct the altitudes from any two vertices (A and B) to their opposite sides $B C$ and $A C$ respectively.
(3) The point of intersection of the altitudes $H$ is the orthocentre of the given $\triangle \mathrm{ABC}$.

Question $2 .$
Draw and locate the orthocentre of a right triangle $\mathrm{PQR}$ right angled at $\mathrm{Q}$, with $\mathrm{PQ}=4.5 \mathrm{~cm}$ and $\mathrm{QR}=6 \mathrm{~cm}$.
Solution:

Construction:
(1) Duaw the $\triangle \mathrm{PQR}$ with the given aneasuacuents.
(2) Construct altitudes from any two vertices ( $Q$ and $R$ ) to their opposite sides $P R$ and $P Q$ respectively.
(3) The point of inlersection of the altitudes $H$ is the orthocentre of the given $\triangle \mathrm{PQR}$.

Question $3 .$
Construct the circumcentre of the $\triangle \mathrm{ABC}$ with $\mathrm{AB}=5 \mathrm{~cm}, \angle \mathrm{A}=60^{\circ}$ and $\angle \mathrm{B}=80^{\circ}$, also draw two circumcircle and find the circum radius of the $\triangle \mathrm{ABC}$.
Solution:

Solution:
Step 1: Draw the $\triangle \mathrm{ABC}$ with the given measurements.
Step 2 : Construct the perpendicular bisector of any two sides (AC and BC) and let them meet at $S$ which is the circumcentre.
Step $3: \mathrm{S}$ as centre and $\mathrm{SA}=\mathrm{SB}=\mathrm{SC}$ as radius, draw the Circumcircle to passes through $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$.
Circumradius $=3.9 \mathrm{~cm}$.

Exercise $4.6$
Question $1 .$
Draw the circumcircle for an equilateral triangle of side $6 \mathrm{~cm}$.
Solution:

Construction:
(1) Draw the $\triangle \mathrm{ABC}$ with the given measurements.
(2) Construct the perpendicular bisectors of $\mathrm{AC}$ and $\mathrm{BC}$ and let them meet at $\mathrm{S}$ which is the circumcentre.
(3) With $\mathrm{S}$ as centre and $\mathrm{SA}=\mathrm{SB}=\mathrm{SC}$ as radius, draw the circumcircle to pass through $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$.

Question $2 .$
Construct the centroid of $\triangle \mathrm{PQR}$ such that $\mathrm{PQ}=9 \mathrm{~cm}, \mathrm{PQ}=7 \mathrm{~cm}, \mathrm{RP}=8 \mathrm{~cm}$.
Solution:
In $\triangle \mathrm{PQR}$,
$\mathrm{PQ}=5 \mathrm{~cm}$
$\mathrm{PR}=6 \mathrm{~cm}$
$\angle \mathrm{QPR}=60^{\circ}$

Construction :
Step 1 : Draw $\triangle \mathrm{PQR}$ using the given measurements $\mathrm{PQ}=9 \mathrm{~cm}, \mathrm{QR}=7 \mathrm{~cm}$ and $R P=8 \mathrm{~cm}$ and construct the perpendicular bisector of any two sides $(P Q$ and $Q R)$ to find the mid-points $M$ and $N$ of $P Q$ and $Q R$ respectively.
Step 2 : Draw the medians $P N$ and RM and let them meet at $G$. The point $G$ is the centroid of the given $\triangle P Q R$.

Question $3 .$
Draw and locate the centroid of the triangle $A B C$ where right angle at $A, A B=8 \mathrm{~cm}$ and $A C=6 \mathrm{~cm}$.
Solution:

Step 1 : Draw $\triangle \mathrm{ABC}$ with the given measurements $\mathrm{AB}=8 \mathrm{~cm}, \angle \mathrm{A}=90^{\circ}$ and $\mathrm{AC}=6 \mathrm{~cm}$ and construct the perpendicular bisector of any two sides ( $\mathrm{AB}$ and $\mathrm{AC}$ ) to find the mid points $\mathrm{M}$ and $\mathrm{N}$ of $\mathrm{AB}$ and $\mathrm{BC}$ respectively.
Step 2 : Draw the medians ( $\mathrm{C}$ and $\mathrm{BN}$ and let them meet at $\mathrm{G}$. The point $\mathrm{G}$ is the centroid of the given $\triangle \mathrm{ABC}$

.
Question $4 .$
Construct the centroid of $\square \mathrm{PQR}$ whose sides are $\mathrm{PQ}=8 \mathrm{~cm}, \mathrm{QR}=6 \mathrm{~cm}, \mathrm{RP}=7 \mathrm{~cm}$.
Solution:
Side $=6.5 \mathrm{~cm}$

Construction:
Step 1 : Draw $\triangle \mathrm{ABC}$ with $\mathrm{AB}=\mathrm{BC}=\mathrm{CA}=6.5 \mathrm{~cm}$
Step 2 : Construct angle bisectors of any two angles (A and B) and let them meet at $1.1$ is the incentre of $\triangle \mathrm{ABC}$.
Step 3 : Draw perpendicular from I to any one of the side (AB) to meet $\mathrm{AB}$ at D.
Step 4 : With I as centre, ID as radius draw the circle. This circle touches all the sides of triangle internally.
Step 5 : Measure in radius. In radius $=1.9 \mathrm{~cm} .$
 

Exercise 4.7
Multiple Choice Questions:
Question $1 .$
If an angle is equal to one third of its supplement, its measure is equal to
(1) $40^{\circ}$
(2) $50^{\circ}$
(3) $45^{\circ}$
(4) $55^{\circ}$
Hint:
$x^{0}+\frac{1}{3} x^{0}=60^{\circ} \Rightarrow \frac{4 x}{3}=60^{\circ} x=45^{\circ}$
Solution:
(3) $45^{\circ}$

Question $2 .$
In the given figure, $\mathrm{OP}$ bisect $\angle \mathrm{BOC}$ and $\mathrm{OQ}$ bisect $\angle \mathrm{AOC}$. Then $\angle \mathrm{POQ}$ is equal to
(1) $90^{\circ}$
(2) $120^{\circ}$
(3) $60^{\circ}$
(4) $100^{\circ}$

Solution:
(1) $90^{\circ}$
 

Question $3 .$
The complement of an angle exceeds the angle by $60^{\circ}$. Then the angle is equal to
(1) $25^{\circ}$
(2) $30^{\circ}$
(3) $15^{\circ}$
(4) $35^{\circ}$
Solution:
(3) $15^{\circ}$
 

Question $4 .$
$\mathrm{ABCD}$ is a parallelognaun, $\mathrm{E}$ is the mid-point of $\mathrm{AB}$ aud $\mathrm{CE}$ biscuts $\angle \mathrm{BCD}$. Thec $\angle \mathrm{DEC}$ is
(1) $60^{\circ}$
(2) $90^{\circ}$
(3) $100^{\circ}$
(4) $120^{\circ}$
Solution:
(2) $90^{\circ}$
 

Question $5 .$
If the length of a chord decreases, then its distance from the centre.
(1) increases
(2) decreases
(3) same
(4) cannot say
Solution:
(1) increases

Question $6 .$
In the figure, $\mathrm{O}$ is the centre of the circle and $\angle \mathrm{ACB}=60^{\circ}$ then $\angle \mathrm{AOB}=$
(1) $60^{\circ}$
(2) $90^{\circ}$
(3) $120^{\circ}$
(4) $180^{\circ}$

Solution:
(3) $120^{\circ}$

Question $7 .$
The angle subtend by a semicircle at the centre is.
(1) $60^{\circ}$
(2) $90^{\circ}$
(3) $120^{\circ}$
(4) $180^{\circ}$
Solution:
(4) $180^{\circ}$

Text Book Activities
Activity $-3$
Angle sum for a polygon.
Draw any quadrilateral $\mathrm{ABCD}$.
Mark a point $P$ in its interior. Join the segments PA, PB, PC and PD.
You have 4 triangles now.
How much is the sum of all the angles of the 4 triangles?
How much is the sum of the angles at their vertex, now $\mathrm{P}$ ?
Can you now find the 'angle sum' of the quadrilateral $\mathrm{ABCD}$ ?
Can you extend this idea to any polygon?
Solution:

Sum of the angles of the 4 triangle $=180^{\circ} \times 4=720^{\circ}$
Sum of the angles at their vertex, now $p=360^{\circ}$
Angle sum of the quadrilateral $\mathrm{ABCD}=720^{\circ}-360^{\circ}=360^{\circ}$
Yes we can extend this idea to any polygon.

Activity - 4
Procedure.

1. Draw a circle with centre $\mathrm{O}$ and with suitable radius.

2. Make it a semi-circle through folding. Consider the point $A, B$ on it.
3. Make crease along $\mathbf{A B}$ in the semi circles and open it.
4. We get one more crease line on the another part of semi circle, name it as $\mathrm{CD}$ (observe $\mathrm{AB}=\mathrm{CD}$ )

5. Join the radius to get the $\triangle \mathrm{OAB}$ and $\triangle \mathrm{OCD}$.

6. Using trace paper, take the replicas of triangle $\triangle O A B$ and $\triangle O C D$.
7. Place these triangles $\triangle \mathrm{OAB}$ and $\triangle \mathrm{OCD}$ one on the other.

Activity - 6
Procedure:
1. Draw a circle of any radius with centre $O$.
2. Mark any four points $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and $\mathrm{D}$ on the boundary. Make a cyclic quadrilateral $\mathrm{ABCD}$ and name the angles as in figure.

5. Paste the angle cutout $\angle 1, \angle 2, \angle 3$ and $\angle 4$ adjacent to the angles opposite to $\mathrm{A}, \mathrm{B}, \mathrm{C}$ and $\mathrm{D}$ as in Figure.
6. Measure the angles $\angle 1+\angle 3$, and $\angle 2+\angle 4$.
Solution:
$\begin{aligned}
&\angle 1+\angle 3=180^{\circ} \\
&\angle 2+\angle 4=180^{\circ}
\end{aligned}$