Exercise 4.1 - Chapter 4 Geometry Term 2 7th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Question $1 .$
Can $30^{\circ}, 60^{\circ}$ and $90^{\circ}$ be the angles of a triangle?
Solution:
Given angles $30^{\circ}, 60^{\circ}$ and $90^{\circ}$
Sum of the angles $=30^{\circ}+60^{\circ}+90^{\circ}=180^{\circ}$
$\therefore$ The given angles form a triangle.

Question $2 .$
Can you draw a triangle with $25^{\circ}, 65^{\circ}$ and $80^{\circ}$ as angles?
Solution:
Given angle $25^{\circ}, 65^{\circ}$ and $80^{\circ}$.
Sum of the angles $=25^{\circ}+65^{\circ}+80^{\circ}=170^{\circ} \neq 180$
$\therefore$ We cannot draw a triangle with these measures.

Question $3 .$
In each of the following triangles, find the value of $x$.

Solution:
(i) Let $\angle \mathrm{G}=\mathrm{x}$
By angle sum property we know that,
$\angle \mathrm{E}+\angle \mathrm{F}+\angle \mathrm{G}=180^{\circ}$
$80^{\circ}+55^{\circ}+x=180^{\circ}$
$135^{\circ}+x=180^{\circ}$
$\mathrm{x}=45^{\circ}$
(ii) Let $\angle \mathrm{M}=\mathrm{x}$
By angle sum property of triangles we have
$\angle \mathrm{M}+\angle \mathrm{M}+\angle \mathrm{O}=180^{\circ}$
$x+96^{\circ}+22^{\circ}=180^{\circ}$
$x+118^{\circ}=180^{\circ}$
$\mathrm{X}=180^{\circ}-118^{\circ}=620$
(iii) Let $\angle Z=(2 x+1)^{\circ}$ and $\angle Y=90^{\circ}$
By the sum property of triangles we have
$\angle \mathrm{x}+\angle \mathrm{y}+\angle \mathrm{z}=180^{\circ}$
$29^{\circ}+90^{\circ}+(2 x+1)^{\circ}=180^{\circ}$
$119^{\circ}+(2 x+1)^{\circ}=180^{\circ}$
$(2 x+1)^{\circ}=180^{\circ}-119^{\circ}$
$2 x+1^{\circ}=61^{\circ}$
$2 x=61^{\circ}-1^{\circ}$
$2 x=60^{\circ}$
$\mathrm{x}=\frac{60^{\circ}}{2}$
$x=30^{\circ}$
(iv) Let $\angle \mathrm{J}=\mathrm{x}$ and $\angle \mathrm{L}-3 \mathrm{x}$.
By angle sum property of triangles we have

$$
\begin{aligned}
&\angle \mathrm{J}+\angle \mathrm{K}+\angle \mathrm{L}=180^{\circ} \\
&\mathrm{x}+112^{\circ}+3 \mathrm{x}=180^{\circ} \\
&4 \mathrm{x}=180^{\circ}-112^{\circ} \\
&\mathrm{x}=68^{\circ} \\
&\mathrm{x}=\frac{68^{\circ}}{4} \\
&\mathrm{x}=17^{\circ}
\end{aligned}
$$
(v) Let $\angle \mathrm{S}=3 \mathrm{x}^{\circ}$
Given $\overline{\mathrm{RS}}=$ Given $\overline{\mathrm{RT}}=4.5 \mathrm{~cm}$
Given $\angle \mathrm{S}=\angle \mathrm{T}=3 \mathrm{x}^{\circ}[\because$ Angles opposite to equal sides are equal $]$ By angle sum property of a triangle we have,
$\begin{aligned}
&\angle \mathrm{R}+\angle \mathrm{S}+\angle \mathrm{T}=180^{\circ} \\
&72^{\circ}+3 \mathrm{x}+3 \mathrm{x}=180^{\circ} \\
&72^{\circ}+6 \mathrm{x}=180^{\circ} \\
&\mathrm{x}=\frac{108^{\circ}}{6} \\
&\mathrm{x}=18^{\circ}
\end{aligned}$
(vi) Given $\angle \mathrm{X}=3 \mathrm{x} ; \angle \mathrm{Y}=2 \mathrm{x} ; \angle \mathrm{Z}=\angle 4 \mathrm{x}$
By angle sum property of a triangle we have
$\begin{aligned}
&\angle \mathrm{X}+\angle \mathrm{Y}+\angle \mathrm{Z}=180^{\circ} \\
&3 \mathrm{x}+2 \mathrm{x}+4 \mathrm{x}=180^{\circ} \\
&\therefore 9 \mathrm{x}=180^{\circ} \\
&\mathrm{x}=\frac{180^{\circ}}{9}=20^{\circ}
\end{aligned}$

(vii) Given $\angle T=(x-4)^{\circ}$
$$
\begin{aligned}
&\angle \mathrm{U}=90^{\circ} \\
&\angle \mathrm{V}=(3 \mathrm{x}-2)^{\circ}
\end{aligned}
$$
By angle sum property of a triang we have
$\begin{aligned}
&\angle \mathrm{T}+\angle \mathrm{U}+\angle \mathrm{V}=180^{\circ} \\
&(\mathrm{x}-4)^{\circ}+90^{\circ}+(3 \mathrm{x}-2)^{\circ}=180^{\circ} \\
&\mathrm{x}-4^{\circ}+90^{\circ}+3 \mathrm{x}-2^{\circ}=180^{\circ} \\
&\mathrm{x}+3 \mathrm{x}+90^{\circ}-4^{\circ}-2^{\circ}=180^{\circ} \\
&4 \mathrm{x}+84^{\circ}=180^{\circ} \\
&4 \mathrm{x}=180^{\circ}-84^{\circ} \\
&4 \mathrm{x}=96^{\circ} \\
&\mathrm{x}=\frac{96^{\circ}}{4}=24^{\circ} \\
&\mathrm{x}=24^{\circ}
\end{aligned}$
(viii) Given $\angle \mathrm{N}=(\mathrm{x}+31)^{\circ}$
$\angle \mathrm{O}=(3 \mathrm{x}-10)^{\circ}$

$\angle P=(2 x-3)^{\circ}$
By angle sum property of a triangle we have
$\begin{aligned}
&\angle \mathrm{N}+\angle \mathrm{O}+\angle \mathrm{P}=\mathrm{O} \\
&(\mathrm{x}+31)^{\circ}+(3 \mathrm{x}-10)^{\circ}+(2 \mathrm{x}-3)^{\circ}=180^{\circ} \\
&\mathrm{x}+31^{\circ}+3 \mathrm{x}-10^{\circ}+2 \mathrm{x}-3^{\circ}=180^{\circ} \\
&\mathrm{x}+3 \mathrm{x}+2 \mathrm{x}+31^{\circ}-10^{\circ}-3^{\circ}=180^{\circ} \\
&6 \mathrm{x}+18^{\circ}=180^{\circ} \\
&6 \mathrm{x}=180^{\circ}+18^{\circ} \\
&6 \mathrm{x}=162^{\circ} \\
&\mathrm{x}=\frac{162^{\circ}}{6}=27^{\circ} \\
&\mathrm{x}=27^{\circ}
\end{aligned}$
 

Question $4 .$
Two line segments $\overline{A D}$ and $\overline{B C}$ intersect at $\mathrm{O}$. Joining $\overline{A B}$ and $\overline{D C}$ we get two triangles, $\Delta \mathrm{AOB}$ and $\triangle D O C$ as shown in the figure. Find the $\angle A$ and $\angle B$.

Solution:
In $\triangle \mathrm{AOB}$ and $\triangle \mathrm{DOC}$,
$\angle \mathrm{AOB}=\angle \mathrm{DOC}[\because$ Vertically opposite angles are equal $]$
Let $\angle \mathrm{AOB}=\angle \mathrm{DOC}=\mathrm{y}$
By angle sum property of a triangle we have
$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{AOB}=\angle \mathrm{D}+\angle \mathrm{C}+\angle \mathrm{DOC}=180^{\circ}$
$3 x+2 x+y=70^{\circ}+30^{\circ}+y=180^{\circ}$
$5 x+y=100^{\circ}+y=180^{\circ}$
Here $5 x+y=100^{\circ}+y$
$5 x=100^{\circ}+y-y$
$5 x=100^{\circ}$
$x=\frac{100^{\circ}}{5}=20^{\circ}$
$\angle \mathrm{A}=3 \mathrm{x}=3 \times 20=60^{\circ}$
$\angle B=2 x=2 \times 20=40^{\circ}$
$\angle \mathrm{A}=60^{\circ}$
$\angle \mathrm{B}=40^{\circ}$

Question $5 .$
Observe the figure and find the value of
$\angle \mathrm{A}+\angle \mathrm{N}+\angle \mathrm{G}+\angle \mathrm{L}+\angle \mathrm{E}+\angle \mathrm{S} \text {. }$

Solution:
In the figure we have two triangles namely $\triangle \mathrm{AGE}$ and $\triangle \mathrm{NLS}$.
By angle sum property of triangles,
Sum of angles of $\triangle \mathrm{AGE}=\angle \mathrm{A}+\angle \mathrm{G}+\angle \mathrm{E}=180^{\circ} \ldots$ (1)
Also sum of angles of $\Delta \mathrm{NLS}=\angle \mathrm{N}+\angle \mathrm{L}+\angle \mathrm{S}=180^{\circ} \ldots$ (2)
(1) $+$ (2) $\angle \mathrm{A}+\angle \mathrm{G}+\angle \mathrm{E}+\angle \mathrm{N}+\angle \mathrm{L}+\angle \mathrm{S}=180^{\circ}+180^{\circ}$
i.e., $\angle \mathrm{A}+\angle \mathrm{N}+\angle \mathrm{G}+\angle \mathrm{L}+\angle \mathrm{E}+\angle \mathrm{S}=360^{\circ}$

Question $6 .$
If the three angles of a triangle are in the ratio $3: 5: 4$, then find them.
Solution:
Given three angles of the triangles are in the ratio $3: 5: 4$.
Let the three angle be $3 x, 5 x$ and $4 x$.
By angle sum property of a triangle, we have
$3 x+5 x+4 x=180^{\circ}$
$12 \mathrm{x}=180^{\circ}$
$\begin{aligned}
&x=\frac{180^{\circ}}{12} \\
&x=15^{\circ}
\end{aligned}$
$\therefore$ The angle are $3 \mathrm{x}=3 \times 15^{\circ}=45^{\circ}$
$5 x=5 \times 15^{\circ}=75^{\circ}$
$4 x=4 \times 15^{\circ}=60^{\circ}$

Three angles of the triangle are $45^{\circ}, 75^{\circ}, 60^{\circ}$

Question $7 .$
In $\triangle \mathrm{RST}, \angle \mathrm{S}$ is $10^{\circ}$ greater than $\angle \mathrm{R}$ and $\angle \mathrm{T}$ is $5^{\circ}$ less than $\angle \mathrm{S}$, find the three angles of the triangle.
Solution:
In $\triangle R S T$. Let $\angle R=x$.
Then given $S$ is $\angle 10^{\circ}$ greater than $\angle R$
$\therefore \angle \mathrm{S}=\mathrm{x}+10^{\circ}$
Also given $\angle T$ is $5^{\circ}$ less then $\angle S$.
So $\angle \mathrm{T}=\angle \mathrm{S}-5^{\circ}=(\mathrm{x}+10)^{\circ}-5^{\circ}=\mathrm{x}+10^{\circ}-5^{\circ}$
By angle sum property of triangles, sum of three angles $=180^{\circ}$.

$\begin{aligned}
&\angle \mathrm{R}+\angle \mathrm{S}+\angle \mathrm{T}=180^{\circ} \\
&\mathrm{x}+\mathrm{x}+10^{\circ}+\mathrm{x}+5^{\circ}=180^{\circ} \\
&3 \mathrm{x}+15^{\circ}=180^{\circ} \\
&3 \mathrm{x}=180^{\circ}-15^{\circ} \\
&\mathrm{x}=\frac{165^{\circ}}{3}=55^{\circ} \\
&\angle \mathrm{R}=\mathrm{x}=55^{\circ} \\
&\angle \mathrm{S}=\mathrm{x}+10^{\circ}=55^{\circ}+10^{\circ}=65^{\circ} \\
&\angle \mathrm{T}=\mathrm{x}+5^{\circ}=55^{\circ}+5^{\circ}=60^{\circ} \\
&\therefore \angle \mathrm{R}=55^{\circ} \\
&\angle \mathrm{S}=65^{\circ} \\
&\angle \mathrm{T}=60^{\circ}
\end{aligned}$

Question $8 .$
In $\triangle \mathrm{ABC}$, if $\angle \mathrm{B}$ is 3 times $\angle \mathrm{A}$ and $\angle \mathrm{C}$ is 2 times $\angle \mathrm{A}$, then find the angles.
Solution:
In $A B C$, Let $\angle A=x$,
then $\angle B=3$ times $\angle A=3 x$
$\angle \mathrm{C}=2$ times $\angle \mathrm{A}=2 \mathrm{x}$
By angle sum property of a triangles,
Sum of three angles of $\triangle \mathrm{ABC}=180^{\circ}$.
$\begin{aligned}
&\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180 \\
&\mathrm{x}+3 \mathrm{x}+2 \mathrm{x}=180^{\circ} \\
&\mathrm{x}(1+3+2)=180^{\circ} \\
&6 \mathrm{x}=180^{\circ}
\end{aligned}$

$\begin{aligned}
&x=\frac{180^{\circ}}{6}=30^{\circ} \\
&\angle A=x=30^{\circ} \\
&\angle B=3 x=3 \times 30^{\circ}=90^{\circ} \\
&\angle C=2 x=2 \times 30^{\circ}=60^{\circ} \\
&\therefore \angle A=30^{\circ} \\
&\angle B=90^{\circ} \\
&\angle C=60^{\circ}
\end{aligned}$

Question $9 .$
In $\triangle \mathrm{XYZ}$, if $\angle \mathrm{X}: \angle \mathrm{Z}$ is $5: 4$ and $\angle \mathrm{Y}=72^{\circ}$. Find $\angle \mathrm{X}$ and $\angle \mathrm{Z}$.
Solution:
Given in $\triangle \mathrm{XYZ}, \angle \mathrm{X}: \angle \mathrm{Z}=5: 4$
Let $\angle X=5 x$; and $\angle Z=4 x$ given $\angle Y=72^{\circ}$
By the angle sum property of triangles sum of three angles of a triangles is $180^{\circ}$.
$\begin{aligned}
&\angle \mathrm{X}+\angle \mathrm{Y}+\angle \mathrm{Z}=180^{\circ} \\
&5 \mathrm{x}+72+4 \mathrm{x}=180^{\circ} \\
&5 \mathrm{x}+4 \mathrm{x}=180^{\circ}-72^{\circ} \\
&9 \mathrm{x}=108^{\circ} \\
&\mathrm{x}=\frac{108^{\circ}}{9}=12^{\circ} \\
&\angle \mathrm{X}=5 \mathrm{x}=5 \times 12^{\circ}=60^{\circ} \\
&\angle \mathrm{Z}=4 \mathrm{x}=4 \times 12^{\circ}=48^{\circ} \\
&\therefore \angle \mathrm{X}=60^{\circ} \\
&\angle \mathrm{Z}=48^{\circ}
\end{aligned}$

Question 10 .
In a right angled triangle $\mathrm{ABC}, \angle \mathrm{B}$ is right angle, $\angle \mathrm{A}$ is $\mathrm{x}+1$ and $\angle \mathrm{C}$ is $2 \mathrm{x}+5$. Find $\angle \mathrm{A}$ and $\angle \mathrm{C}$.
Solution:
Given in $\triangle \mathrm{ABC} \angle \mathrm{B}=90^{\circ}$
$\begin{aligned}
&\angle \mathrm{A}=\mathrm{x}+1 \\
&\angle \mathrm{B}=2 \mathrm{x}+5
\end{aligned}$

By angle sum property of triangles
Sum of three angles of $\triangle \mathrm{ABC}=180^{\circ}$
$\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ}$
$(x+1)+90^{\circ}+(2 x+5)=180^{\circ}$
$x+2 x+1^{\circ}+90^{\circ}+5^{\circ}=180^{\circ}$
$3 x+96^{\circ}=180^{\circ}$
$3 x=180^{\circ}-96^{\circ}=84^{\circ}$
$\mathrm{x}=\frac{84^{\circ}}{3}=28^{\circ}$
$\angle \mathrm{A}=\mathrm{x}+1=28+1=29$
$\angle \mathrm{C}=2 \mathrm{x}+5=2(28)+5=56+5=61$
$\therefore \angle \mathrm{A}=29^{\circ}$
$\angle \mathrm{C}=61^{\circ}$
 

Question $11 .$
In a right angled triangle $\mathrm{MNO}, \angle \mathrm{N}=90^{\circ}, \mathrm{MO}$ is extended to $\mathrm{P}$. If $\angle \mathrm{NOP}=128^{\circ}$, find the other two angles of $\triangle \mathrm{MNO}$.
Solution:
Given $\angle \mathrm{N}=90^{\circ}$
$\mathrm{MO}$ is extended to $\mathrm{P}$, the exterior angle $\angle \mathrm{NOP}=128^{\circ}$
Exterior angle is equal to the sum of interior opposite angles.

$\begin{aligned}
&\therefore \angle \mathrm{M}+\angle \mathrm{N}=128^{\circ} \\
&\angle \mathrm{M}+90^{\circ}=128^{\circ} \\
&\angle \mathrm{M}=128^{\circ}-90^{\circ} \\
&\angle \mathrm{M}=38^{\circ}
\end{aligned}$
By angle sum property of triangles,
$\begin{aligned}
&\therefore \angle \mathrm{M}+\angle \mathrm{N}+\angle \mathrm{O}=180^{\circ} \\
&38^{\circ}+90^{\circ}+\angle \mathrm{O}=180^{\circ} \\
&\angle \mathrm{O}=180^{\circ}-128^{\circ} \\
&\angle \mathrm{O}=52^{\circ} \\
&\therefore \angle \mathrm{M}=38^{\circ} \text { and } \angle \mathrm{O}=52^{\circ}
\end{aligned}$
 

Question 12 .
Find the value of $x$ in each of the given triangles.

Solution:
(i) In $\triangle \mathrm{ABC}$, given $\mathrm{B}=65^{\circ}$,
$\mathrm{AC}$ is extended to $\mathrm{L}$, the exterior angle at $\mathrm{C}, \angle \mathrm{BCL}=135^{\circ}$ Exterior angle is equal to the sum of opposite interior angles.
$\begin{aligned}
&\angle \mathrm{A}+\angle \mathrm{B}=\angle \mathrm{BCL} \\
&\angle \mathrm{A}+65^{\circ}=135^{\circ} \\
&\angle \mathrm{A}=135^{\circ}-65^{\circ} \\
&\therefore \angle \mathrm{A}=70^{\circ} \\
&\mathrm{x}+\angle \mathrm{A}=180^{\circ}[\because \text { linear pair }] \\
&\mathrm{x}+70^{\circ}=180^{\circ}\left[\because \angle \mathrm{A}=70^{\circ}\right]
\end{aligned}$
$\begin{aligned}
&\mathrm{x}=180^{\circ}-70^{\circ} \\
&\therefore \mathrm{x}=110^{\circ}
\end{aligned}$

(ii) In $\triangle \mathrm{ABC}$, given $\mathrm{B}=3 \mathrm{x}-8^{\circ}$
$\angle \mathrm{XAZ}=\angle \mathrm{BAC}[\because$ vertically opposite angles $]$
$8 \mathrm{x}+7+\angle \mathrm{BAC}$
i.e., In $\triangle \mathrm{ABC}, \angle \mathrm{A}=8 \mathrm{x}+7$
Exterior angle $\angle \mathrm{XCY}=120^{\circ}$
Exterior angle is equal to the sum of the interior opposite angles.
$\angle \mathrm{A}+\angle \mathrm{B}=120^{\circ}$
$8 x+7+3 x-8=120^{\circ}$
$8 x+3 x=120^{\circ}+8-7$
$11 x=121^{\circ}$
$x=\frac{121^{\circ}}{11}=11^{\circ}$
 

Question $13 .$
In $\triangle \mathrm{LMN}, \mathrm{MN}$ is extended to $\mathrm{O} .$ If $\angle \mathrm{MLN}=100-\mathrm{x}, \angle \mathrm{LMN}=2 \mathrm{x}$ and $\angle \mathrm{LNO}=6 \mathrm{x}-5$, find the value of $\mathrm{x}$.
Solution:
Exterior angle is equal to the sum of the opposite interior angles.
$\angle \mathrm{LNO}=\angle \mathrm{MLN}+\angle \mathrm{LMN}$

$\begin{aligned}
&6 x-5=100^{\circ}-x+2 x \\
&6 x-5+x-2 x=100^{\circ} \\
&6 x+x-2 x=100^{\circ}+5^{\circ} \\
&5 x=105^{\circ} \\
&x=\frac{105^{\circ}}{5}=21^{\circ} \\
&x=21^{\circ}
\end{aligned}$

Question $14 .$
Using the given figure find the value of $x$.

Solution:
In $\Delta \mathrm{EDC}$, side $\mathrm{DE}$ is extended to $\mathrm{B}$, to form the exterior angle $\angle \mathrm{CEB}=\mathrm{x}$.
We know that the exterior angle is equal to the sum of the opposite interior angles $\angle \mathrm{CEB}=\angle \mathrm{CDE}+\angle \mathrm{ECD}$
$\begin{aligned}
&x=50^{\circ}+60^{\circ} \\
&x=110^{\circ}
\end{aligned}$

Question $15 .$
Using the diagram find the value of $x$.

Solution:
Given triangle is an equilateral triangle as the three sides are equal. For an equilateral triangle all three angles are equal and is equal to $60^{\circ}$ Also exterior angle is equal to sum of opposite interior angles.
$\begin{aligned}
&x=60^{\circ}+60^{\circ} \\
&x=120^{\circ}
\end{aligned}$

Objective Type Questions
Question $16 .$
The angles of a triangle are in the ratio $2: 3: 4$. Then the angles are
(i) $20,30,40$
(ii) $40,60,80$
(iii) $80,20,80$
(iv) $10,15,20$
Answer:
(ii) $40,60,80$
 

Question $17 .$
One of the angles of a triangle is $65^{\circ}$. If the difference of the other two angles is $45^{\circ}$, then the two angles are
(i) $85^{\circ}, 40^{\circ}$
(ii) $70^{\circ}, 25^{\circ}$
(iii) $80^{\circ}, 35^{\circ}$
(iv) $80^{\circ}, 135^{\circ}$
Answer:
(iii) $80^{\circ}, 35^{\circ}$

Question 18 .
In the given figure, $\mathrm{AB}$ is parallel to $\mathrm{CD}$. Then the value of $\mathrm{b}$ is

(ii) $68^{\circ}$
(iii) $102^{\circ}$
(iv) $62^{\circ} \mathrm{A}$
Answer:
(ii) $68^{\circ}$

Question $19 .$
In the given figure, which of the following statement is true?

(i) $x+y+z=180^{\circ}$
(ii) $x+y+z=a+b+c$
(iii) $x+y+z=2(a+b+c)$
(iv) $x+y+z=3(a+b+c)$
Ans :
(iii) $x+y+z=2(a+b+c)$ ]

Question 20 .
An exterior angle of a triangle is $70^{\circ}$ and two interior opposite angles are equal. Then measure of each of these angle will be
(i) $110^{\circ}$
(ii) $120^{\circ}$
(iii) $35^{\circ}$
(iv) $60^{\circ}$
Answer:
(iii) $35^{\circ}$

Question $21 .$
In a $\triangle \mathrm{ABC}, \mathrm{AB}=\mathrm{AC}$. The value of $\mathrm{x}$ is

(ii) $80^{\circ}$
(iii) $130^{\circ}$
(iiv) $120^{\circ}$
(iv) $120^{\circ}$
(iii) $130^{\circ}$
 

Question 22.
If an exterior angle of a triangle is $115^{\circ}$ and one of the interior opposite angles is $35^{\circ}$, then the other two angles of the triangle are
(i) $45^{\circ}, 60^{\circ}$
(ii) $65^{\circ}, 80^{\circ}$
(iii) $65^{\circ}, 70^{\circ}$
(iv) $115^{\circ}, 60^{\circ}$

Answer:
(ii) $65^{\circ}, 80^{\circ}$