Exercise 3.2 (Revised) - Chapter 3 - Understanding Quadrilaterals - Ncert Solutions class 8 - Maths

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NCERT Solutions for Class 8 Maths Chapter 3 - Understanding Quadrilaterals

Ex 3.2 Question 1.

$\text {Find } x \text { in the following figures: }$

Answer.

(a) Here, $125^{\circ}+m=180^{\circ}$
[Linear pair]

$
\Rightarrow m=180^{\circ}-125^{\circ}=55^{\circ}
$

And $125^{\circ}+n=180^{\circ}$
[Linear pair]
$
\Rightarrow n=180^{\circ}-125^{\circ}=55^{\circ}
$
$\because$ Exterior angle $x^{\circ}=$ Sum of opposite interior angles
$
\therefore x^{\circ}=55^{\circ}+55^{\circ}=110^{\circ}
$

(b) Sum of the angles of a pentagon
$
\begin{aligned}
& =(n-2) \times 180^{\circ} \\
& =(5-2) \times 180^{\circ} \\
& =3 \times 180^{\circ}=540^{\circ}
\end{aligned}
$

By linear pairs of angles,
$
\begin{aligned}
& \angle 1+90^{\circ}=180^{\circ} \\
& \angle 2+60^{\circ}=180^{\circ} \\
& \angle 3+90^{\circ}=180^{\circ} \\
& \angle 4+70^{\circ}=180^{\circ} \\
& \angle 5+x=180^{\circ}
\end{aligned}
$

Adding eq. (i), (ii), (iii), (iv) and (v),
$
\begin{aligned}
& x+(\angle 1+\angle 2+\angle 3+\angle 4+\angle 5)+310^{\circ}=900 \\
& \Rightarrow x+540^{\circ}+310^{\circ}=900^{\circ} \\
& \Rightarrow x+850^{\circ}=900^{\circ} \\
& \Rightarrow x=900^{\circ}-850^{\circ}=50^{\circ}
\end{aligned}
$

Ex 3.2 Question 2.

Find the measure of each exterior angle of a regular polygon of:
(a) 9 sides
(b) 15 sides

Answer.

(i) Sum of angles of a regular polygon $=(n-2) \times 180^{\circ}$
$
=(9-2) \times 180^{\circ}=7 \times 180^{\circ}=1260^{\circ}
$

Each interior angle $=\frac{\text { Sum of interior angles }}{\text { Number of sides }}=\frac{1260^{\circ}}{9}=140^{\circ}$
Each exterior angle $=180^{\circ}-140^{\circ}=40^{\circ}$
(ii) Sum of exterior angles of a regular polygon $=360^{\circ}$

Each exterior angle $=360 / 15$
$=24$ degrees

Ex 3.2 Question 3.

How many sides does a regular polygon have, if the measure of an exterior angle is $24^{\circ}$ ?

Answer.

Let number of sides be $n$.

Sum of exterior angles of a regular polygon $=360^{\circ}$
$
\text { Number of sides }=\frac{\text { Sum of exterior angles }}{\text { Each interior angle }}=\frac{360^{\circ}}{24^{\circ}}=15
$

Hence, the regular polygon has 15 sides.
Ex 3.2 Question 4.

How many sides does a regular polygon have if each of its interior angles is $165^{\circ}$ ?

Answer.

Let number of sides be $n$.
Exterior angle $=180^{\circ}-165^{\circ}=15^{\circ}$

Sum of exterior angles of a regular polygon $=360^{\circ}$
$
\text { Number of sides }=\frac{\text { Sum of exterior angles }}{\text { Each interior angle }}=\frac{360^{\circ}}{15^{\circ}}=24
$

Hence, the regular polygon has 24 sides.
Ex 3.2 Question 5.

(a) Is it possible to have a regular polygon with of each exterior angle as $22^{\circ}$ ?
(b) Can it be an interior angle of a regular polygon? Why?

Answer.

(a) No. (Since 22 is not a divisor of $360^{\circ}$ )
(b) No, (Because each exterior angle is $180^{\circ}-22^{\circ}=158^{\circ}$ : which is not a divisor of $360^{\circ}$ )
Ex 3.2 Question 6.

(a) What is the minimum interior angle possible for a regular polygon? Why?
(b) What is the maximum exterior angle possible for a regular polygon?

Answer.

(a) The equilateral triangle being a regular polygon of 3 sides has the least measure of an

interior angle of $60^{\circ}$.
$\because$ Sum of all the angles of a triangle
$
=180^{\circ}
$
$
\begin{aligned}
& \therefore x+x+x=180^{\circ} \\
& \Rightarrow 3 x=180^{\circ} \\
& \Rightarrow x=60^{\circ}
\end{aligned}
$
(b) By (a), we can observe that the greatest exterior angle is $180^{\circ}-60^{\circ}$
$
=120^{\circ} \text {. }
$