In Text Questions Try These (Textbook Page No. 66,76,77,83) - Chapter 4 Geometry Term 2 7th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Text Questions :  Chapter 4 Geometry Term 2 Class 7th std Maths Guide Samacheer Kalvi Solutions
Exercise $4.1$
Try These (Text book Page No. 66)
Answer the following questions.
Question $1 .$
Triangle is formed by joining three points.
Answer:
Non collinear

Question $2 .$

A triangle has ____vertices and ____ sides.
Answer:
three, three

Question $3 .$ 
A point where two sides of a triangle meet is known as ______ of a triangle
Answer: vertese 

Question $4 .$
Each angle of an equilateral triangle is of measure.
Answer:
same

Question $5 .$
A triangle has angle measurements of $29^{\circ}, 65^{\circ}$ and $86^{\circ}$. Then it is triangle.
(i) an acute angled
(ii) a right angled
(iii) an obtuse angled
(iv) a scalene
Answer:
(i) an acute angled

Question $6 .$
A triangle has angle measurements of $30^{\circ}, 30^{\circ}$ and $120^{\circ}$. Then it is triangle.
(i) an acute angled
(ii) scalene
(iii) obtuse angled
(iv) right angled
Answer:
obtuse angled

Question 7 .
Which of the following can be the sides of a triangle?
(i) $5.9 .14$
(ii) $7,7,15$
(iii) $1,2,4$
(iv) $3,6,8$
Answer:
(iv) $3,6,8$
Solution:
(i) Here $5+9=14=$ the measure of the third side.
In a triangle the sum of the measures of any two sides must be greater than the third side. $\therefore 5,9,14$ cannot be the sides of a triangle.
(ii) $7.7 .15$
Here sum of two sides $7+7=14<$ the measures of the thrid side.
So $1,1,15$ cannot be the sides of a triangles.
(iii) $1,2,4$
Here sum of two sides $1+2=3<$ the measure of the third side. $\therefore 1,2,4$ cannot be the sides of a triangle.

(iv) $3,6,8$
Sum of two sides $3+6=9>$ the third side.
$\therefore 3,6,8$ can be the sides of a triangle.
 

Question $8 .$
Ezhil wants to fence his triangular garden. If two of the sides measure 8 feet and 14 feet then the length of the third side is
(i) $11 \mathrm{ft}$
(ii) $6 \mathrm{ft}$
(iii) $5 \mathrm{ft}$
(iv) $22 \mathrm{ft}$
Answer:
(i) $11 \mathrm{ft}$
 

Question $9 .$
Can we have more than one right angle in a triangle?
Solution:
No, we cannot have more than one right angle in a triangle.
Because the sum of three angles of a triangle is $180^{\circ}$.
But if two angles are right angles then their sum itself become $180^{\circ}$.

Question $10 .$
How many obtuse angles are possible in a triangle?
Solution:
Only one.
 

Question $11 .$
In a right triangle, what will be the sum of other two angles?
Solution:
Sum of three angles of a triangle $=180^{\circ}$
If one angle is right angle (i.e. $90^{\circ}$ ).
Sum of other two sides $=180^{\circ}-90^{\circ}=90^{\circ}$
 

Question 12.
Is it possible to form an isosceles right angled triangle? Explain.
Solution:
Yes, it is possible.
If one angle is right angle, then the other two angles will be $45^{\circ}$ and $45^{\circ}$.
 

Exercise $4.2$

Try These (Text book Page No. 76)
Question $1 .$
Measure and group the pair of congruent line segments.

Solution:
$\overline{A B}=3 \mathrm{~cm}$
$\overline{C D}=4.8 \mathrm{~cm}$
$\overline{I J}=4.8 \mathrm{~cm}$
$\overline{P Q}=3 \mathrm{~cm}$
$\overline{R S}=1.7 \mathrm{~cm}$
From the above measurement $\mathrm{S}$, we can conclude that
(i) $\overline{A B} \cong \overline{P Q}$
(ii) $\overline{C D} \cong \overline{I J}$
(iii) $\overline{R S} \cong \overline{X Y}$

Try These (Text book Page No. 77)
Question $1 .$
Find the pairs of congruent angles either by superposition method or by measuring them.

Solution:
From the given figures
$\angle \mathrm{ABC}=50^{\circ}$
$\angle \mathrm{EFG}=120^{\circ}$
$\angle \mathrm{HIJ}=120^{\circ}$
$\angle \mathrm{KLH}=90^{\circ}$
$\angle \mathrm{PON}=50^{\circ}$
$\angle \mathrm{RST}=90^{\circ}$
From the above measures, we can conclude that
(i) $\angle \mathrm{ABC}=\angle \mathrm{PON}$
(ii) $\angle \mathrm{EFG}=\angle \mathrm{HIJ}$
(iii) $\angle \mathrm{KLH} \cong \angle \mathrm{RST}$
 

Try These (Text book Page No. 83)
Question $1 .$
If $\triangle \mathrm{ABC} \cong \triangle \mathrm{XYZ}$ then list the corresponding sides and corresponding angles.

Solution:
If $\Delta \mathrm{ABC} \cong \Delta \mathrm{XYZ}$ $\overline{A B} \cong \overline{X Y}-\overline{B C} \cong \overline{Y Z}$ $\overline{A C} \cong \overline{X Z}$
$\begin{aligned}
&\overline{A B} \cong \overline{X Y}-\overline{B C} \cong \overline{Y Z} \\
&\overline{A C} \cong \overline{X Z}
\end{aligned}$
And also
$\begin{aligned}
&\angle \mathrm{A} \cong \angle \mathrm{X}-\angle \mathrm{B} \cong \angle \mathrm{Y} \\
&\angle \mathrm{C} \cong \angle \mathrm{Z}
\end{aligned}$

Question $2 .$
Given triangles are congruent. Identify the corresponding parts and write the congruent statement.

Solution:
Given the set of triangles are congruent. Also we observe from the triangles that the corresponding sides.
$\begin{aligned}
&\overline{A B}=\overline{A C} \\
&\overline{B C}=\overline{Y Z} \\
&\overline{A C}=\overline{X Z}
\end{aligned}$
Here three sides of $\triangle \mathrm{ABC}$ are equal to the corresponding sides of $\triangle \mathrm{XYZ}$.
This criterion of congruency is side $-$ side $-$ side.
 

Question $3 .$
Mention the conditions needed to conclude the congruency of the triangles with reference to the above said criterions. Give reasons for your answer.

Solution:
(i) In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{XYZ}$
if $\overline{A B}=\overline{X Y}$
$\overline{B C}=\overline{Y Z}$ $\overline{A C}=\overline{X Z}$
then $\triangle \mathrm{ABC} \cong \triangle \mathrm{XYZ}$. By the Side - Side -Side Congruency Criterion.

If $\overline{\mathrm{AB}}=\overline{\mathrm{XY}}$
$\begin{aligned}
&\overline{\mathrm{BC}}=\overline{\mathrm{YZ}} \\
&\angle \mathrm{ABC}=\angle \mathrm{XYZ} \\
&\text { then } \Delta \mathrm{AB} \cong \Delta \mathrm{XYZ} .
\end{aligned}$
$\text { then } \triangle A B \cong \triangle X Y Z \text {. }$
then $\triangle \mathrm{AB} \cong \triangle \mathrm{XYZ}$.
By Side - Angle - Side Criterion.

(or) $\overline{\mathrm{AB}}=\overline{\mathrm{XY}}$
$\overline{\mathrm{AC}}=\overline{\mathrm{XY}}$
$\angle B A C=Z X Y$
$\angle \mathrm{BAC}=\angle \mathrm{XYZ}$

(or) $\overline{\mathrm{BC}}=\overline{\mathrm{YZ}}$
$\overline{\mathrm{AC}}=\overline{\mathrm{XZ}}$
$\angle \mathrm{BCA}=\angle \mathrm{YZX}$

If $\angle \mathrm{A}=\angle \mathrm{X}$
$\begin{aligned}
&\angle \mathrm{B}=\angle \mathrm{Y} \\
&\overline{\mathrm{AB}}=\overline{\mathrm{XY}}
\end{aligned}$

\begin{aligned}
&\angle \mathrm{B}=\angle \mathrm{Y} \\
&\angle \mathrm{C}=\angle \mathrm{Z} \\
&\overline{\mathrm{BC}}=\overline{\mathrm{YZ}}
\end{aligned}

\begin{aligned}
&\angle \mathrm{A}=\angle \mathrm{X} \\
&\angle \mathrm{C}=\angle \mathrm{Z} \\
&\overline{\mathrm{AC}}=\overline{\mathrm{XZ}}
\end{aligned}

then $\Delta \mathrm{ABC} \cong \Delta \mathrm{XYZ}$.
By Angle - Side - Angle Congruency Critirion.

then by RHS criterion.
$\triangle \mathrm{ABC} \cong \triangle \mathrm{XYZ}$