Additional Questions - Chapter 4 - Geometry - Term 2 - 7th Maths Guide Samacheer Kalvi Solutions

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Additional Questions and Answers
Exercise 4.1
Question $1 .$
"The sum of any two angles of a triangle is always greater than the third angle". Is this statement true. Justify your answer.
Solution:
No, the sum of any two angles of a triangle is not always greater than the third angle. In an isosceles right angled triangles, the angle will be $90^{\circ}, 45^{\circ}, 45^{\circ}$.
Here sum of two angles $45^{\circ}+45^{\circ}=90^{\circ}$.

Question $2 .$
The three angles of a triangle are in the ratio $1: 2: 1$. Find all the angles of the triangle. Classify the triangle in two different ways.
Solution:
Let the angles of the triangle be $\mathrm{x}, 2 \mathrm{x}, \mathrm{x}$.
Using the angle sum property, we have
$x+2 x+x=180^{\circ}$
$4 \mathrm{x}=180^{\circ}$
$\begin{aligned}
&4 x=180^{\circ} \\
&x=\frac{180^{\circ}}{4} \\
&x=45^{\circ} \\
&2 x=2 \times 45^{\circ}=90^{\circ}
\end{aligned}$
Thus the three angles of the triangle are $45^{\circ}, 90^{\circ}, 45^{\circ}$.
Its two angles are equal. It is an isoscales triangle. Its one angle is $90^{\circ}$.
$\therefore$ It is a right angled triangle.

Question $3 .$
Find the values of the unknown $x$ and $y$ in the following figures

Solution:
(i) Since angles $y$ and $120^{\circ}$ form a linear pair.
$\begin{aligned}
&y+120^{\circ}=180^{\circ} \\
&y=180^{\circ}-120^{\circ} \\
&y=60^{\circ}
\end{aligned}$
Now using the angle sum property of a triangle, we have
$\begin{aligned}
&x+y+50^{\circ}=180^{\circ} \\
&x+60^{\circ}+50^{\circ}=180^{\circ} \\
&x+110^{\circ}=180^{\circ} \\
&x=180^{\circ}-110^{\circ}=70^{\circ} \\
&x=70^{\circ} \\
&y=60
\end{aligned}$
(ii) Using the angle sum property of triangle, we have
$\begin{aligned}
&50^{\circ}+60^{\circ}+y=180^{\circ} \\
&110^{\circ}+y=180^{\circ} \\
&y=180^{\circ}-110^{\circ} \\
&y=70^{\circ}
\end{aligned}$
Again $x$ and $y$ form a linear pair
$\begin{aligned}
&\therefore \mathrm{x}+\mathrm{y}=180^{\circ} \\
&\mathrm{x}+70^{\circ}=180^{\circ} \\
&\mathrm{x}=180^{\circ}-70^{\circ}=110^{\circ} \\
&\therefore \mathrm{x}=110^{\circ} ; \mathrm{y}=70^{\circ}
\end{aligned}$

Question $4 .$
Two angles of a triangle are $30^{\circ}$ and $80^{\circ}$. Find the third angle.
Solution:
Let the third angle be $x$.
Using the angle sum property of a triangle we have,
$\begin{aligned}
&30^{\circ}+80^{\circ}+x=180^{\circ} \\
&x+110^{\circ}=180^{\circ} \\
&x=180^{\circ}-110^{\circ}=70^{\circ}
\end{aligned}$
Third angle $=70^{\circ}$.
 

Exercise 4.2
Question $1 .$
In an isoscleles $\triangle \mathrm{ABC}, \mathrm{AB}=\mathrm{AC}$. Show that angles opposite to the equal sides are equal.

Solution:

Given: $\triangle \mathrm{ABC}$ in which $\overline{A B}=\overline{A C}$.
To Prove: $\angle B=\angle K$.
Construction: Draw $\mathrm{AD} \perp \mathrm{BC}$.
Proof: In right $\triangle \mathrm{ADB}$ and right $\triangle \mathrm{ADC}$.
we have side $\mathrm{AD}=$ side $\mathrm{AD}$ (common)
$\mathrm{AB}=\mathrm{AC}$ (Hypoteneous) (given)
$\triangle \mathrm{ADB}=\mathrm{ADC}$ (RHS criterion]
$\therefore$ Their corresponding parts are equal. $\angle \mathrm{B}=\angle \mathrm{C}$.
 

Question $2 .$
$\mathrm{ABC}$ is an isosceles triangle having side $\overline{A B}=$ side $\overline{A C}$. If $\mathrm{AD}$ is perpendicular to $\mathrm{BC}$, prove that $\mathrm{D}$ is the mid-point of $\overline{B C}$.
Solution:
In $\triangle \mathrm{ABD}$ and $\triangle \mathrm{ACD}$, we have
$\angle \mathrm{ADB}=\angle \mathrm{ADC}[\because \mathrm{AD} \perp \mathrm{BC}]$
side $\overline{A D}=$ Side $\overline{A D}$ [Common]
Side $\overline{A B}=$ Side $\overline{A C}$ [Common]
Using RHS congruency, we get
$\triangle \mathrm{ABD} \cong \triangle \mathrm{ACD} \mathrm{c}$
Their corresponding parts are equal

$\therefore \mathrm{BD}=\mathrm{CD}$
$\therefore \mathrm{O}$ is the mid point of $B C$.
 

Question $3 .$
In the figure $\mathrm{PL} \perp \mathrm{OB}$ and $\mathrm{PM} \perp \mathrm{OA}$ such that $\mathrm{PL}=\mathrm{PM}$.
Prove that $\triangle \mathrm{PLO} \cong \triangle \mathrm{PMO}$.
Solution:

In $\triangle \mathrm{PLO}$ and $\triangle \mathrm{PMO}$, we have
$\angle \mathrm{PLO}=\angle \mathrm{PMO}=90^{\circ}$ [Given]
$\overline{O P}=\overline{O P}$ [Hypotenuse]
$\mathrm{PL}=\mathrm{PM}$
Using RHS congruency, we get
$\triangle \mathrm{PLO} \cong \triangle \mathrm{PMO}$