ML AGGARWAL CLASS 8 Chapter 13 Understanding Quadrilaterals Excercise 13.1

 Exercise 13.1

Question 1

Sol :
(a) simple curves:- (i), (ii), (iii), (v), (vi)

(b) Simple closed curves:- (iii), (v), (vi)

(c) polygon:- (iii), (vi)

(d) Convex polygon: - (iii),

(e) Concave polygon:- (v)

Question 2

Sol :
(a) A convex quadrilateral has two diagonal

(b) A regular hexagon has 9 diagonals

Question 3

Sol :
(i) 8

Sum of all interior angles $=(2 n-4) \times 90$

given no of sides of polygon(n) =8

Sum of all interior angles $=(2 \times 8-4) \times 90$

$=(16-4) \times 90$

$=12 \times 90$

=1080


(ii) Given

No . of sides of polygon (n) = 12

Sum of all interior angles = $(2 n-4) \times 90$

$=(2 \times 12-4) \times 90$

=1800

Question 4

Sol :

(i) Given

exterior angle of polygon $=24^{\circ}$

Sum of all exterior angles polygon  = 360

$\begin{aligned} \therefore \quad n \times 24 &=360 \\ n &=15 \end{aligned}$


(ii) Given
exterior angle of polygon $=60^{\circ}$

Sum of all exterior angles of polygon $=360^{\circ}$

$\therefore n \times 60=360$

∴ No. of sides of given polygon is 6


(iii) Given 

exterior angle of polygon $=72^{\circ}$

Sum of all exterior angles of polygon $=360^{\circ}$

$\therefore n \times 72^{\circ}=360$

n= 5

∴ No. of sides of given polygon is 5

Question 5

Sol :
(i) For polygon with 'n ' sides

The each interior angle of polygon is given by 

=$\frac{(2 n-4) \times 90}{n}$

given

interior angle of polygon = 90

$\frac{(2 n-4) \times 90}{n}=90$

$(2 n-4)=n

2 n-n=4

n=4

∴ no. of sides of polygon = 4


(ii) For   a polygon a with 'n' sides 

The each interior angles of polygon is given by 

$=\frac{(2 n-4) \times 90}{n}$

given interior angle = $108^{\circ}$

$\frac{(2 n-4) \times 90}{n}=108^{\circ}$

$2 n-4=\frac{6}{5} \cdot n=5(2 n-4)=6 n$

10 n-20=6 n

10 n-6 n=20

4 n=20

$n=\frac{20}{4}$

n=5

∴ no. of sides of polygon = 5


(iii) For a polygon with 'n' sides .

The each interior angle of polygon is given by 

$\frac{(2 n-4) \times 90}{n}$

given interior angle $=165^{\circ}$

$\therefore \quad \frac{(2 n-4) \times 90}{n}=165$

$\frac{(2 n-4)}{n}=\frac{11}{6}$

$6(2 n-4)=11 . \times n$

12 n-24=11 n

12 n-11 n=24

n=24

∴ no. of sides of polygon= 24 

Question 6

Sol :
Given sum of interior angles of a polygon = 1260

∴ $(2 n-4) \times 90=1260^{\circ}$

Where n - no. of sides of polygon 

$\begin{aligned}(2 n-4) &=\frac{1260}{90} \\ 2 n-4 &=14 \\ 2 n &=14+4 \\ 2 n &=18 \\ n &=18 / 2 \\ n &=9 \end{aligned}$

∴ Given polygon has nine sides 

Question 7

Sol :
Given 

Ratio of angles of pentagon =7: 8: 11: 13: 15

Let Angles of pentagon =7 x, 8 x ,11 x, 13 x, 15x

Sum of angle of polygon = $(2 n-4) \times 90$

$\begin{aligned} 7 x+8 x+11 x+13 x+15 &=(2 \times 5-4) \times 90 \\ 54 x &=6 \times 90 \\ x &=\frac{540}{54} \\ {x} &=10^{\circ} \end{aligned}$

Question 8

Sol :
Given angles of pentagon =$x^{\circ},(x-10)^{\circ},(x+20)^{\circ},(2 x-44)^{\circ}$
$\operatorname{and}(2 x-70)^{\circ}$

Sum of interior angles of polygon = $(2 n-4) \times 90$

$x+(x+10)+(x+20)+(2 x-44)+(2 x-70)=(2 \times 5-4) \times 90$

$7 x-104=6 \times 90$

7 x=540+104

$\begin{aligned} 7 x &=644 \\ x &=\frac{644}{7} \\ x &=92 \end{aligned}$

∴ Angle of pentagon = $92^{\circ},(92-10)^{\circ},(92+20)^{\circ}$
$(2 \times 92-44),(2 \times 92-70)$

=92,82,112,140,114

Question 9

Sol :
Given

Exterior angles Ratio =1: 2: 3: 4: 5

Let exterior angles = x, 2 x, 3 x, 4 x, 5 x

Sum of the exterior angles = 360

$\begin{aligned} x+2 x+3 x+4 x+5 x &=360 \\ 15 x &=360 \\ x &=\frac{360}{15} \\ x &=24 \end{aligned}$

Exterior angles of pentagon 

$24,48,72^{\circ}, 96^{\circ}, 120^{\circ}$

Internal angle = 180 - External angle

Internal angles of pentagon = $180-24,180-48,150-72,180-96,180-120$

Interior angles $=156^{\circ}, 132^{\circ}, 108^{\circ}, 84^{\circ}, 60^{\circ}$ of pentgon

Question 10

Sol :
Given 

$\angle A: \angle D=2: 3$

$L B: \angle C=7: 8$







Let $\quad \angle A=2 x,  \quad \angle$  L D=3 x

Let $\quad\angle B=7 y, \quad\angle  C=8 y$

$\quad\angle$ B+$\quad\angle$C=180$(\because A B \| D C)$

$7 y+8 y=180^{\circ}$

15 y=180

y=12

$\therefore \quad \begin{aligned} \angle B &=7 y=7 \times 12=84^{\circ} \\ \therefore \quad\angle B &=8 y=8 \times 12=96^{\circ} \end{aligned}$

$\quad\angle A+\angle D=180^{\circ}$

2 x+3 x=180

$\begin{aligned} 5 x &=180 \\ x &=36 \end{aligned}$

$\angle A=2 x=2 \times 36=72^{\circ}$

$\angle D=3 x=3 \times 36=108$

$\therefore\angle  A=72^{\circ}, \angle B=84^{\circ}, L C=96^{\circ}, \angle D=108^{\circ}$

Question 11

Sol :
From $\triangle D B C$

$\begin{aligned}\angle D B C+\angle C+\angle C D B &=180^{\circ} \\ x+5 x+8+\angle C D B &=180^{\circ} \\ 6 x+8+\angle C P B &=180 \\ \angle C D B &=180-6 x-8 \\ \angle C D B &=172-6 x \end{aligned}$

$\begin{aligned} \angle C D B+& \angle A D B=3 x+10^{\prime} \\ 172-6 x+&\angle  A D B=3 x+16\\ & \angle A D B=3 x+10+6 x-172 \\ &=9 x-162 \end{aligned}$

In $\triangle A D B$

$\angle A D B+\angle D A B+\angle A B D=180^{\circ}$

9 x-162+3 x+4+50=180

12 x-108=180

12 x=180+108=288

$x=\frac{288}{12}$

$x=24^{\circ}$


(ii) $\begin{aligned} \angle D A B &=3 x+4 \\ &=3 \times 24+4 \\ &=72+4 \\ \angle D A B &=76^{\circ} \end{aligned}$


(iii) $\begin{aligned} \angle A D B &=9 x-162 \\ &=9 \times 24-162 \\ \angle A D B &=216-162 \\ \angle A D B &=54^{\circ} \end{aligned}$

Question 12

Sol :
(i) Sum of angles in quadrilateral = 360

∴ $\begin{aligned} 40+140+100+x &=360 \\ 280+x &=360 \\ x &=360-280 \\ x &=80 \end{aligned}$


(ii) Interior angle = 180-- (exterior angle)







Sum interior angles

pentagon $=(2 \times 5-4) \times 90$

$\begin{aligned} 40+x+x+120+100 &=6 \times 90 \\ 2 x+260 &=540 \\ 2 x &=540-260 \\ 2 x &=280 \end{aligned}$

$x=\frac{280}{2}$

x=140


(iii) Sum of interior angles of a quadrilateral = 360'







$\begin{aligned} \therefore x+110+60+90 &=360 \\ x+260 &=360 \\ x &=360-260 \\ x &=100 \end{aligned}$


(iv) Sum of interior angles of a quadrilateral = 360







$\begin{aligned} 110^{\circ}+83^{\prime}+180-x+90 &=360 \\ 463-x &=360 \\ x &=463-360 \\ x &=103 \end{aligned}$

Question 13

Sol :
(i) Sum of angles in a triangle = 180'







$\begin{aligned} 90+70+180-& z=180 \\ z &=90+70 \\ z &=160 \end{aligned}$

$\begin{aligned} 90+x &=180(\because \text { Forms straight line }) \\ x &=180-90 \\ x &=90 \end{aligned}$

$\begin{aligned} 70+y &=180 \text { (-: Forms straight line) } \\ y &=180-70 \\y &=110 \end{aligned}$

$\begin{aligned} \therefore \quad & x+y+z=90+110+160 \\ \therefore \quad & x+y+z=360^{\circ} \end{aligned}$


(ii)  Sum of interior angles of a quadrilateral = 360'












$\begin{aligned} 70+80+130+a &=360^{\circ} \\ 280+a &=360 \\ a &=360-280 \\ a &=80 \end{aligned}$

a+w=180 (∵Forms straight line)

$80+\omega=180$

$\omega=180-80$

$\omega=100$

z+70=180 ($\because$ Forms straight line )

z=180-70

z=110

$\begin{aligned} 80+y &=180(\because \text { forms straight line }) \\ y &=180-80 \\ y &=100 \end{aligned}$

$\begin{aligned} 130+x &=180 \quad(\therefore \text { Forms straight line) }\\ x &=180-130 \\ x &=50 \end{aligned}$

$\therefore x+y+z+\omega=50+100+110+100=360^{\circ}$

Question 14

Sol :

Given

The pentagon has three equal angles = $120^{\circ} 120^{\circ}, 120^{\circ}$

Let remaining four equal angles = x, x,x, x

Sum of interior angles of heptagon = $(2 \times 7-4) \times 90$

=$10 \times 90$

=900

$\therefore \quad 120+120+120+x+x+x+x$

$\begin{aligned} 360+4 x &=900 \\ 4 x &=900-360 \\ 4 x &=540 \\ x &=\frac{540}{4} \\ x &=135^{\circ} \end{aligned}$

The other equal angle of heptagon = 135

Question 15

Sol :
Ratio between exterior and interior angles

=1: 5

(i) Let exterior angle =x

Interior angle =5 x

exterior angle + Interior angle =180

$\begin{aligned} x+5 x &=180 \\ 6 x &=180 \\ x &=\frac{180}{6} \\ x &=30 \end{aligned}$

Each exterior angle =x =30


(ii) Each interior angle = 5x = $5 \times 30=150^{\circ}$

(iii) $\begin{aligned} \text { no.of sides of polygon } &=\frac{360}{\text { Exlerior angle }} \\ &=\frac{360}{30}=12 \end{aligned}$

∴ No. of sides of polygon = 12

Question 16

Sol :
Given 
{ Each interior angle of polygon }=2 \times \text { Exterior angle }

Interior angle+Exterior angle =180

$2 \times$ Exterior angle + Exterior angle =180

$3 \times$ Exterior angle $=180$

Exterior angle $=\frac{180}{3}$

Exterior angle $=60^{\circ}$

$\begin{aligned} \text { no.of sides of polygon } &=\frac{360}{\text { Exterior angle }} \\ &=\frac{360}{60} \end{aligned}$

No. of sides of polygon= 6

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