Exercise 4.3 - Chapter 4 Geometry Term 2 7th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $4.3$
Miscellaneous Practice Problems
Question $1 .$
In an isoscales triangle one angle is $76^{\circ}$. If the other two angles are equal, find them.
Solution:
In an isoscales triangle, angle opposite to equal sides are equal. Let the equal angles be $x^{\circ}$ and $x^{\circ}$.
In a triangle the sum of the three angles is $180^{\circ}$.
$x^{\circ}+x^{\circ}+76^{\circ}=180^{\circ}$
$x^{\circ}(1+1)=180^{\circ}-76^{\circ}=104^{\circ}$
$2 \mathrm{x}=104^{\circ}$
$x=\frac{104^{\circ}}{2}=52^{\circ}$
$x=52^{\circ}$ $\therefore$ Other two angles are $52^{\circ}$ and $52^{\circ}$.
 

Question $2 .$
If two angles of a triangle are $46^{\circ}$ each, how can you classify the triangle?
Solution:
Given two angles of the triangle are same and is equal to $46^{\circ}$. If two angles are equal the sides opposite to equal angles are equal. Therefore it will be an isoscales triangle.
 

Question $3 .$
If an angle of a triangle is equal to the sum of the other two angles, find the type of the triangle.
Solution:
Let $\angle \mathrm{B}$ is the greater angle then by the given condition $\angle \mathrm{B}=\angle \mathrm{A}+\angle \mathrm{C}$.
Sum of three angle of a triangle $=180^{\circ}$.

$\begin{aligned}
&\angle \mathrm{A}+\angle \mathrm{B}+\angle \mathrm{C}=180^{\circ} \\
&\angle \mathrm{A}+(\angle \mathrm{A}+\angle \mathrm{C})+\angle \mathrm{C})=180^{\circ} \\
&2 \angle \mathrm{A}+2 \angle \mathrm{C}=180^{\circ} \\
&2(\angle \mathrm{A}+\angle \mathrm{C})=180^{\circ} \\
&\angle \mathrm{A}+\angle \mathrm{C}=\frac{180^{\circ}}{2} \\
&\angle \mathrm{B}=90^{\circ}
\end{aligned}$
$\therefore$ One of the angle of the triangle $=90^{\circ}$
It will be a right angled triangle.
 

Question $4 .$
If the exterior angle of a triangle is $140^{\circ}$ and its interior opposite angles are equal, find all the interior angles of the triangle.
Solution:
Given the exterior angle $=140^{\circ}$
Interior opposite angle are equal.

Let one of the interior opposite angle be $x$.
Then $\mathrm{x}+\mathrm{x}=140^{\circ}$.
[ Exterior angle $=$ sum of interior opposite angles $]$
$\begin{aligned}
&2 \mathrm{x}=140^{\circ} \\
&\mathrm{x}=\frac{140^{\circ}}{2}=70^{\circ}
\end{aligned}$
$x=70^{\circ}$ Interior opposite angle $=70^{\circ}, 70^{\circ} .$
$\mathrm{x}=70^{\circ}$
Sum of the three angles of a triangle $=180^{\circ}$.
$70^{\circ}+70^{\circ}+$ Third angle $=180^{\circ}$
$140^{\circ}+$ Third angle $=180^{\circ}$
Third angle $=180^{\circ}-140^{\circ}=40^{\circ}$
$\therefore$ Interior angle are $40^{\circ}, 70^{\circ}, 70^{\circ}$.
 

Question $5 .$
In $\triangle \mathrm{JKL}$, if $\angle \mathrm{J}=60^{\circ}$ and $\angle \mathrm{K}=40^{\circ}$, then find the value of exterior angle formed by extending the side KL.
Solution:
When extending the side $\mathrm{KL}$, the exterior angle formed in equal to the sum of the interior opposite angles.

$\begin{aligned}
&\angle \mathrm{JLX}=\angle \mathrm{LJK}+\angle \mathrm{LKJ} \\
&=60^{\circ}+40^{\circ}=100^{\circ}
\end{aligned}$
Exterior angle formed $=100^{\circ}$

Question $6 .$
Find the value of ' $x$ ' in the given figure.

Solution:
Given $\angle \mathrm{DCB}=1000$ and $\angle \mathrm{DBA}=128^{\circ}$
In the given figure
$\angle \mathrm{CBD}+\angle \mathrm{DBA}=180^{\circ}$
$\angle \mathrm{CBD}+128^{\circ}=180^{\circ}$
$\angle C B D=52^{\circ}$
Now exterior angle $\mathrm{x}=$ Sum of interior opposite angles.
$\mathrm{x}=\angle \mathrm{DCB}+\angle \mathrm{CBD}=100^{\circ}+52^{\circ}=152^{\circ}$
$x=152^{\circ}$
 

Question $7 .$
If $\triangle \mathrm{MNO} \cong \Delta \mathrm{DEF}, \angle \mathrm{M}=60^{\circ}$ and $\angle \mathrm{E}=45^{\circ}$ then find the value of $\angle \mathrm{O}$.
Solution:
Given $\triangle \mathrm{MNO} \cong \triangle \mathrm{DEF}$

$\therefore$ Corresponding parts of conqruent triangle are congruent.
$\angle \mathrm{M}=\angle \mathrm{D}=60^{\circ}$ [given $\angle \mathrm{M}=60^{\circ}$ ]
$\angle \mathrm{N}=\angle \mathrm{E}=45^{\circ}$ [given $\angle \mathrm{E}=45^{\circ}$ ]
$\angle \mathrm{O}=\angle \mathrm{F}$
In triangle $\mathrm{MNO}$, sum of the three angle $-180^{\circ}$.
$\angle \mathrm{M}+\angle \mathrm{N}+\angle \mathrm{O}=180^{\circ}$
$60^{\circ}+45^{\circ}+\angle \mathrm{O}=180^{\circ}$
$105^{\circ}+\angle \mathrm{O}=180^{\circ}$
$\angle \mathrm{O}=180^{\circ}-105^{\circ}=75^{\circ}$
Value of $\angle O=75^{\circ}$

Question $8 .$
In the given figure ray $A Z$ bisects $\angle B A D$ and $\angle D C B$, prove that
(i) $\triangle \mathrm{BAC} \cong \triangle \mathrm{DAC}$
(ii) $\mathrm{AB}=\mathrm{AD}$

Solution:
(i) In $\triangle \mathrm{BAC}$ and $\triangle \mathrm{DAC}$
$\angle \mathrm{BAC}=\angle \mathrm{DAC}[$ Given $\overline{A Z}$ bisects $\angle \mathrm{BAD}$ ]
$\angle \mathrm{BCA}=\angle \mathrm{DCA}[\overline{A Z}$ bisects $\angle \mathrm{DCB}]$
$\mathrm{AC}=\mathrm{AC}[\because$ common side $]$
$\therefore$ Here $\mathrm{AC}$ is the included side of the angles. By ASA criterior, $\triangle \mathrm{BAC} \cong \triangle \mathrm{DAC}$.
(ii) $\mathrm{By}$ (i) $\triangle \mathrm{BAC} \cong \triangle \mathrm{DAC}$
$\mathrm{BA}=\mathrm{DA}$ [By CPCTC]
i.e., $\mathrm{AB}=\mathrm{AD}$

Question $9 .$
In the given figure $\mathrm{FG}=\mathrm{FI}$ and $\mathrm{H}$ is midpoint of $\mathrm{GI}$, prove that $\Delta \mathrm{FGH} \cong \Delta \mathrm{FHI}$
Solution:
In $\Delta \mathrm{FGH}$ and $\Delta \mathrm{FHI}$
Given $\mathrm{FG}=\mathrm{HI}$
Also, $\mathrm{GH}=\mathrm{HI}[\because \mathrm{H}$ is the midpoint of GI]

$\mathrm{FH}=\mathrm{FH}$ [Common]
$\therefore$ By S.SS congruency criteria, $\Delta \mathrm{FGH} \cong \Delta \mathrm{FIH}$. Hence proved.
 

Question $10 .$
Using the given figure, prove that the triangles are congruent. Can you conclude that $\mathrm{AC}$ is parallel to DE.
Solution:

In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{EBD}$,

$\begin{aligned}
&\mathrm{AB}=\mathrm{EB} \\
&\mathrm{BC}=\mathrm{BD} \\
&\angle \mathrm{ABC}=\angle \mathrm{EBD}[\because \text { Vertically opposite angles }]
\end{aligned}$
$\angle \mathrm{ABC}=\angle \mathrm{EBD}[\because$ Vertically opposite angles $]$ $\mathrm{By}$ SAS congruency criteria. $\triangle \mathrm{ABC} \cong \triangle \mathrm{EBD}$.
We know that corresponding parts of congruent triangles are congruent.
$\therefore \angle \mathrm{BCA} \cong \angle \mathrm{BDE}$
$\text { and } \angle \mathrm{BAC} \cong \angle \mathrm{BED}$
and $\angle B A C \cong \angle B E D$
$\angle \mathrm{BCA} \cong \angle \mathrm{BDE}$ means that alternate interior angles are equal if $\mathrm{CD}$ is the transversal to lines $\mathrm{AC}$ and DE.
Similarly, if $\mathrm{AE}$ is the transversal to $\mathrm{AC}$ and $\mathrm{DE}$, we have $\angle \mathrm{BAC} \cong \angle \mathrm{BED}$
Again interior opposite angles are equal.
We can conclude that $\mathrm{AC}$ is parallel to $\mathrm{DE}$.

Challenge Problems

Question 11.
In given figure $B D=B C$, find the value of $x$.

Solution:
Given that $B D=B C$
$\triangle \mathrm{BDC}$ is on isoscales triangle.
In isoscales triangle, angles opposite to equal sides are equal.
$\angle \mathrm{BDC}=\angle \mathrm{BCD} \ldots . . .(1)$
Also $\angle \mathrm{BCD}+\angle \mathrm{BCX}=180^{\circ}[\because$ Liner Pair $]$
$\angle \mathrm{BCD}+115^{\circ}=180^{\circ}$
$\angle B C D=180^{\circ}-115^{\circ}$
$\angle \mathrm{BCD}=65^{\circ}[\mathrm{By}(1)]$
In $\triangle \mathrm{ADB}$

$\angle \mathrm{BAD}+\angle \mathrm{ADB}=\angle \mathrm{BDC}$
[ $\because \mathrm{BDC}$ is the exterior angle and $\angle \mathrm{BAD}$ and $\angle \mathrm{ABD}$ are interior opposite angles] $35^{\circ}+\mathrm{x}=65^{\circ}$
$\begin{aligned}
&x=65^{\circ}-35^{\circ} \\
&x=30^{\circ}
\end{aligned}$

Question $12 .$
In the given figure find the value of $x$.

Solution:
For $\triangle \mathrm{LNM}, \angle \mathrm{LMK}$ is the exterior angle at $\mathrm{M}$.
Exterior angle $=$ sum of opposite interior angles
$\angle \mathrm{LMK}=\angle \mathrm{MLN}+\angle \mathrm{LNM}=26^{\circ}+30^{\circ}=56^{\circ}$
$\angle \mathrm{JMK}=56^{\circ}[\because \angle \mathrm{LMK}=\angle \mathrm{JMK}]$
$x$ is the exterior angle at $\mathrm{J}$ for $\triangle \mathrm{JKM}$.
$\therefore \mathrm{x}=\angle \mathrm{JKM}+\angle \mathrm{KMJ}[\because$ Sum of interior opposite angles]
$x=58^{\circ}+56^{\circ}\left[\because \angle \mathrm{JMK}=56^{\circ}\right]$
$\mathrm{x}=114^{\circ}$
 

Question $13 .$
In the given figure find the values of $x$ and $y$.

Solution:
In $\triangle \mathrm{BCA}, \angle \mathrm{BAX}=62^{\circ}$ is the exterior angle at $\mathrm{A}$.
Exterior angle $=$ sum of interior opposite angles.
$\angle \mathrm{ABC}+\angle \mathrm{ACB}=\angle \mathrm{BAX}$
$28^{\circ}+\mathrm{x}=62^{\circ}$ $\mathrm{x}=62^{\circ}-28^{\circ}=34^{\circ}$ Also $\angle \mathrm{BAC}+\angle \mathrm{BAX}=180^{\circ}[\because$ Linear pair $]$
$2 \mathrm{ABC}+\angle \mathrm{ACB}=\angle \mathrm{BAX}$
Also $\angle B A C+\angle B$ $y+62^{\circ}=180^{\circ}$
$\begin{aligned}
&y+62^{\circ}=180^{\circ} \\
&y=180^{\circ}-62^{\circ}=118^{\circ} \\
&x=34^{\circ} \\
&y=118^{\circ}
\end{aligned}$
 

Question $14 .$
In $\triangle \mathrm{DEF}, \angle \mathrm{F}=48^{\circ}, \angle \mathrm{E}=68^{\circ}$ and bisector of $\angle \mathrm{D}$ meets $\mathrm{FE}$ at $\mathrm{G}$. Find $\angle \mathrm{FGD}$.

Solution:
Given $\angle \mathrm{F}=48^{\circ}$
$\angle \mathrm{E}=68^{\circ}$
In $\triangle \mathrm{DEF}$,
$\angle \mathrm{D}+\angle \mathrm{F}+\angle \mathrm{E}=180^{\circ}$ [By angle sum property]
$\angle \mathrm{D}+68^{\circ}+68^{\circ}=180^{\circ}$
$\angle \mathrm{D}+116^{\circ}=180^{\circ}$
$\angle \mathrm{D}=180^{\circ}-116^{\circ}=64^{\circ}$
Since $D G$ is the angular bisector of $\angle D$.
$\angle \mathrm{FDG}=\angle \mathrm{GDE}$
Also $\angle \mathrm{FDG}+\angle \mathrm{GDE}=\angle \mathrm{D}$
$2 \angle \mathrm{FDG}=64^{\circ}$
$2 \angle \mathrm{FDG}=64^{\circ}$
$\angle \mathrm{FDG}=\frac{64^{\circ}}{2}=32^{\circ}$
$\angle \mathrm{FDG}=32^{\circ}$
In $\triangle \mathrm{FDG}$,
$\angle \mathrm{FDG}+\angle \mathrm{GFD}=180^{\circ}$ [By angle sum property of triangles]
$32^{\circ}+\angle \mathrm{FDG}+48^{\circ}=180^{\circ}$
$\angle \mathrm{FDG}+80^{\circ}=180^{\circ}$
$\angle \mathrm{FDG}=180^{\circ}-80^{\circ}$
$\angle \mathrm{FDG}=100^{\circ}$

Question $15 .$
In the figure find the value of $x$.

Solution:
Exterior angle is equal to the sum of opposite interior angles.
in $\triangle \mathrm{TSP} \angle \mathrm{TSP}+\angle \mathrm{SPT}=\angle \mathrm{UTP}$
$75^{\circ}+\angle \mathrm{SPT}=105^{\circ}$
$\angle \mathrm{SPT}=105^{\circ}-75^{\circ}$
$\angle \mathrm{SPT}=30^{\circ} \ldots \ldots(1)$
$\angle \mathrm{SPT}+\angle \mathrm{TPR}+\angle \mathrm{RPQ}=180^{\circ}\left[\because\right.$ Sum of angles at a point on a line is $\left.180^{\circ}\right]$
$30^{\circ}+90^{\circ}+\angle \mathrm{RPQ}=180^{\circ}$
$120^{\circ}+\angle \mathrm{RPQ}=180^{\circ}$
$\angle \mathrm{RPQ}=180^{\circ}-120^{\circ}$
$\angle R P Q=60^{\circ} \ldots . .(2)$
$\angle \mathrm{VRQ}+\angle \mathrm{QRP}=180^{\circ}[\because$ linear pair $]$
$145^{\circ}+\angle Q R P=180^{\circ}$
$\angle \mathrm{QRP}=180^{\circ}-145^{\circ}$
$\angle \mathrm{QRP}=35^{\circ}$
Now in $\triangle \mathrm{PQR}$
$\angle \mathrm{QRP}+\angle \mathrm{RPQ}=\mathrm{x}[\because \mathrm{x}$ in the exterior angle $]$
$35^{\circ}+60^{\circ}=x$
$95^{\circ}=\mathrm{x}$
 

Question $16 .$
From the given figure find the value of $y$.

Solution:
From the figure,
$\angle \mathrm{ACB}=\angle \mathrm{XCY}$ [Vertically opposite angles]
$\angle \mathrm{ACB}=48^{\circ} \ldots(1)$
In $\triangle \mathrm{ABC}, \angle \mathrm{CBD}$ is the exterior angle at $\mathrm{B}$.
Exterior angle $=$ Sum of interior opposite angles.
$\angle \mathrm{CBD}=\angle \mathrm{BAC}+\angle \mathrm{ACB}$
$\angle \mathrm{CBE}+\angle \mathrm{EBD}=57^{\circ}+48^{\circ}$
$65^{\circ}+\angle \mathrm{EBD}=105^{\circ}$
$\angle \mathrm{EBD}=105^{\circ}+65^{\circ}=40^{\circ} \ldots \ldots . .$ (2)
In $\triangle \mathrm{EBD}, \mathrm{y}$ is the exterior angle at D.
$\mathrm{y}=\angle \mathrm{EBD}+\angle \mathrm{BED}$
$[\because$ Exterior angle $=$ Sum of opposite interior angles $]$
$y=40^{\circ}+97^{\circ}[\because$ From (2) $]$
$y=137^{\circ}$