ML AGGARWAL CLASS 8 Chapter 13 Understanding Quadrilaterals Excercise 13.1

 Excercise 13.1

Question 1

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

(a) A convex quadrilateral has two diagonal

(b) A regular hexgon has 9 diagonals

 

Question 3

(i) 8

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

given noiof sidesof polygon(n) $=8$

Som 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

(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 $=72^{\circ}$

Sumof 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

(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 isides 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 

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

givew 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.ot sides of polygon= 24 

Question 6

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

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

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

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

 Given 

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

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

(diagram should be added)

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

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

(i) Sum of angles in quadrilater = 360

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

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

(daigram should be added)

Sum interior angles

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

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

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

x=140

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

(daigram should be added)

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

(iv) Sum of interior angles of a quadrilater = 360'

(daigram should be added)

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

Question 13

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

(daigram should be added)

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

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

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

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

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

(daigram should be added)

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

$a+w=180$ (∵Forms straignt 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

Given

The petagon 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

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

Given 

 { Each interior angle of polygon }=2 \times \text { Exferior angle }

Interior angle $+$ Exterior angle $=180$

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

$3 \times$ Exterior angle $=180$

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

Exterior angle $=60^{\circ}$

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

No. of sides of polygon= 6

 Excercise 13.2

Question 1

(i) 6cm (∵opposite sides are equal )

(ii) 9cm (∵opposite sides are equal )

(iii) $\angle D C B=60^{\circ} \quad(\because \angle DCB) \angle C B A$ are Sumpplementary $)$

(iv) $\angle A D C=120^{\circ} \quad(\because$ opposite angles are equal)

(v) $\angle D A B=60^{\prime} \quad \because$ Adjacent angles are Supplementary $)$

(vi) OC = 7cm (∵ 'O' bisecs Ac )

(vii) OB = 5cm (∵ 'O' bisecs Db)

Question 2

(i) Given parallelogram 

Let the unknown angle = a

(diagram should be added)

∴ a + 120 = 180 (∵Adjacent angles are supplementary)

a= 180 - 120

a= 60

∴ a + y = 180  (∵Adjacent angles are supplementary)

$\begin{aligned} 60+y &=180^{\circ} \\ y &=180-60 \\ y &=120^{\circ} \end{aligned}$

x = 60  (∵ opposite angles are equal in parallogram)

z = 120

(ii) $A B C D$ is a parallogrom

$z=40^{\circ}(\because A B \| C D)$

(diagram should be added)

At 'O'

$100+\angle C O D=180^{\circ}$ (∵ Forms straight line)

$\angle CO D=180-100^{\circ}$

$\angle C O D=80^{\circ}$

In $\triangle C O D$

$\begin{aligned} Z+\angle C O D+y &=180^{\circ} \\ 40+80+y=& 180 \\ y+120 &=180^{\circ} \\ y &=180-120 \end{aligned}$

y = 60'

In $\triangle B O C$

$\begin{aligned} L A C B=30^{\circ} \quad(\because B C \| A D) & \\ \therefore \quad & x+\angle A C B+\angle B O C=180^{\circ} \\ & x+30+100=180 \\ x+130 &=180 \\ x &=180-130 \\ & x=50^{'} \end{aligned}$

$x=50^{\circ}, y=60^{\circ}, z=40^{\circ}$

(iii) ABCD is a parallogram 

Y= $120^{\circ}$ (∵Opposite angles are equal in parallogram)

(diagram should be added)

$Z+35^{\circ}+120=180$ (∵Adjacent angles are supplementry)

$\begin{aligned} z+155 &=180 \\ z &=180-155 \\ z &=25^{\circ} \end{aligned}$

$z=x(\because \quad A B \| C D)$

$x=25^{\circ}$

$\therefore \quad x=25^{\circ}, \quad y=120 ; \quad z=25^{\circ}$

(iv) ABCD is a parallogram 

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

(diagram should be added)

(∵Adjacent angles are supplementry)

$\begin{aligned} 67+x+y 0 &=180 \\ x+137 &=180 \\ x &=180-137 \\ x &=43^{\circ} \end{aligned}$

z = 70' (∵ opposite angles are equal in parallogram )

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

$\angle B C A=x$

$L B C A=43^{\circ}$

At 'C'

$\begin{aligned} \angle B C A+y &=180^{\circ}(\because \text { Forms a straight line }) \\ 43+y &=180 \\ y &=180-43 \\ y & \quad y=137^{\circ} \end{aligned}$

$\therefore \quad x=43^{\circ}, \quad y=137^{\circ}, z=70^{\circ}$

Question 3

Let x , y are length of adjacent sides of parallogram

Given

Perimeter $=72 \mathrm{~cm}$

$x: y=5: 7 \Rightarrow \frac{x}{y}=\frac{5}{7} \Rightarrow x=\frac{5}{7} \cdot y$

x+y+x+y=72 (∵opposite sides are equal in length)

$2(x+y)=72$

$2\left(\frac{5}{7} \cdot y+y\right)=72$

$\frac{12}{7} \cdot y=36$

$y=\frac{36 \times 7}{12}$

$y=21 \mathrm{~cm}$

 

$\begin{aligned} x &=\frac{5}{7} \cdot y \\ x &=\frac{5}{7} \times 21 \\ x &=15 \mathrm{~cm} \\ \therefore \quad x=15 \mathrm{~cm}, & y=21 \mathrm{~cm} . \end{aligned}$

ஃ $15 \mathrm{~cm}, 21 \mathrm{~cm}$ are lengths sides of parallogram

Question 4

Given

Angles of parlogram are in the ratio of $4: 5$

Let the Anglebe $4 x, 5 x$

$\begin{array}{rl}4 x+5 x & =180(\because \text { Adjacent angles are Supplementry }] \\ 9 x= & 180 \\ x & =20\end{array}$

Angles   $4 x=4 \times 20=80^{\circ}$

   $5 x=5 \times 20=100^{\circ}$

∴  Four angles of parallogram = $80^{\circ}, 100^{\circ}, 80^{\circ}, 100^{\circ}$

(∵Opposite are equal in a parallogram)

Question 5

(i) $\angle A+\angle C=180^{\circ} ?$

may (Or) may not be

∵( $\angle A = \angle C = 90'$ )

(ii) $\begin{aligned} A D &=B C=6cm, A B=5cm, D C=4.5 \mathrm{Cm} \\ & No(\because A D \neq B C) \end{aligned}$

 

(iii) $\angle B=80^{\circ}, \angle D=70^{\circ}$ ?

No, opposite angles must be equal in parallogram

(iv) $\angle B+\angle C=180^{\circ} ?$

Yes (∵Adjacent angles are supplementry )

Question 6

(daigram should be added)

$y=40^{\circ}$

$(\because \quad H E \| O P)$

At 'o '

$\angle H O P+70^{\circ}=180^{\circ} \quad(\because$ forms straight line $)$

$\angle H O P=180^{\circ}-70^{\circ}$

$\angle H O P=110^{\circ}$

∴ $\angle HOP = 110' (∵Opposite angles are equal )

$\begin{aligned} 40+z+110^{\circ} &=180^{\prime}(\because \text { ∵Adjacent angles are supplementry }) \\ z+150 &=180 \\ z &=180-150 . \\ z &=30^{\circ} \end{aligned}$

 

(ii) (daigram should be added)

At ' O'

$x+60+80=180^{\circ}$ (Forms straight line)

$\begin{aligned} x+140 &=180^{\circ} \\ x &=180-140 \\ x &=40^{\circ} \end{aligned}$

$z=x=40^{\circ}(∵\quad R 0 \| E P)$

$\angle O+\angle P=180^{\circ}$  (∵Adjacent angles are supplementry)

$\begin{aligned} x+60+y=& 180^{\circ} \\ 40+60+y=& 180 \\ y+100=& 180 \\ y &=150-100 \\y=80^{\circ} \end{aligned}$

Question 7

(daigram should be added)

Opposite sides are equal 

$\begin{array}{ll}5 y-1=24 & 4 x+2=28 \\ 5 y=24+1 & 4 x=28-2 \\ 5 y=25 & 4 x=26 \\ y=5] & x=\frac{26}{4}=\frac{13}{2}\end{array}$

$\therefore \quad x=6.5, y=5$

(daigram should be added)

'O' Bisects the $\overline{B R}$

$\begin{aligned} \overline{B O} &=\overline{O R} \\ x+y &=20 \rightarrow(1) \end{aligned}$

'o' Bisects the $\overline{N v}$

$\overline{N O}=\overline{O V}$

x+3=18----------②

x=18-3

x=15

Substituted x value in ①

$\begin{aligned} 15+y &=20 \\ y &=20-15 \\ y &=5 \end{aligned}$

Question 8

In A B C D Parallogram.

$\angle A+\angle B=180$   (∵Adjacent angles are supplementry)

$120+\angle B=180$

$\angle B=180-120$

$\angle B=60^{\circ}$

In PQRS parallelogram 

$\angle P \doteq \angle R(\because$ opposite angles are equal)

$\angle P=50^{\circ}$

In $\triangle P B X$

$\begin{aligned} \angle P+\angle B+x &=180^{\prime} \text { (- Sum of angles in triangle) } \\ 50+60+x &=180 \\ 110+x &=180 \\ \quad x &=180-110 \\ \quad x &=70^{\circ} \end{aligned}$

Question 9

(i) $\angle C A D=?$

(daigram should be added)

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

$\angle A D B=46$

In $\triangle A D O$

$\angle C A D+\angle A D B \neq 68=180$

$\angle C A D+46+68=180$

$\angle C A D+114=180$

$\quad[\angle C A D=66$

(ii) $\angle A C D=?$

$\begin{aligned} \angle D O A+\angle D O C &=180^{\circ}(\because \text { Straight line }) \\ 68+& \angle D O C=180^{\circ} \\ & \angle D O C=180-68 \\ & \angle D O C=112^{\circ} \end{aligned}$

In $\triangle$ DOC

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

$112+30+\angle A C D=180$

$\angle A C D+142=180$

$\angle A C D=38^{\prime}$

(iii) $\begin{aligned} \angle A D C &=\angle A D O+\angle B D C \\ &=46+30 \\ \angle A D C &=76^{\prime} \end{aligned}$

Question 10

(i) (daigram should be added)

$\overline{A D}=\overrightarrow{B C}$ (sides of $\|$ gm $)$

$\angle A N D=\angle C P B=90^{\circ}$

$\angle D A N=\angle B C P(: B C \| A D)$

From SAA Congruence 

$\triangle A N D \cong \triangle B P C$

(ii) As $\Delta D N \cong \triangle B A P$

$\therefore \quad \overline{A N}=\overline{C P}$

Question 11

In parallolegram ABCR

$\overrightarrow{A B}=\overline{R C}(:$ opposite sides $) \rightarrow(1)$

In parallolegram ABPC

$\overline{A B}=\overline{C P}$ ------② (∵ opposite sides )

(1) + (2)

$\begin{aligned} \overline{A B}+A \bar{B} &=\overline{R C}+\overline{C P} \\ 2 \overline{A B} &=\overline{R P} \rightarrow(3) \\ 2 \overline{A C} &=\overline{P O} \rightarrow(4) \\ 2 \overline{B C} &=Q \bar{R} \rightarrow(5) \end{aligned}$

(3) $+(\mathbb{4})+(-5)$

$2(A \bar{B}+A \bar{C}+\overline{B C})=\overline{P O}+Q K+R P$

 Excercise 13.3

Question 1

(i) Square, Rhombus

(ii) Square, Rectangle

Question 2

(i) I. opposite sides are equal and opposite sides are parallel

    II. Adjacent angles are Supplementry

(ii) I. It has four sides

     II. Sum of all interior angles in $360^{\circ}$

(iii) I. All the sides are equal

II. All interior angles are $90^{\circ}$

III. Diagnals cut perpendicularly.

(iv) I. opposite sides are equal

II. All the interior angles are $90^{\circ}$

Question 3

(i) Parallelogram, square, rectangle, rhombus

(ii) square,rhombus

(iii) Square, rectangle

Question 4 

 (daigram should be added)

Given 

side of Rhombus = one diagnal of rhoms 

$A D=D C=A C$

$\therefore \triangle A D C$ is a equilateral triangle.

$\therefore \quad \angle A D C=\angle D A C=\angle D C A=60^{\circ}$

$\triangle^{\operatorname{le}} A C B$ is also a equilateral triangle

$\therefore \angle C A B=\angle A B C=\angle B C A=60^{\circ}$

$\angle A D B=\angle D A C+\angle C A B=60^{\prime}+60^{\prime}=120^{\circ}$

$\angle D C B=\angle D C A+\angle A C B=60+60=120^{\circ}$

$\therefore$ Angles of rhombus $60^{\circ}, 120^{\circ}, 60^{\circ}, 120^{\circ}$

Question 5

 (daigram should be added)

ABCD is a rhombus diagnals rhombus bisects each other 

$\begin{aligned} \therefore \quad & x=8 \\ & y=6 . \end{aligned}$

Diagnals of rhombus cuts orthogonally

In $\Delta^{\operatorname{le}} A O B$

$\overline{D A}^{2}+\overline{O B}^{2}=\overline{A B}^{2}(\because$ Pythagorus Theorem $)$

$x^{2}+y^{2}=z^{2}$

$8^{2}+6^{2}=z^{2}$

$z^{2}=64+36$

$z^{2}=100$

z = 10

Question 6

 (daigram should be added)

$A B C D$ is a trapezium

$\angle A: \angle D=5: 7$

$\angle B=(3 x+11)^{\circ}$

$\angle C=(5 x-31)^{\circ}$

From the property of trapezium

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

Let $\angle A, \angle D=5 y, 7 y$

$\begin{aligned} \therefore \quad 5 y+7 y &=180^{\circ} \\ 12 y &=180^{\circ} \\ y &=15^{\circ} \\ \angle A=5 y &=5 \times 15=75^{\circ} \\ \angle D=7 y &=7 \times 15=105^{\circ} \end{aligned}$

$\angle B+\angle C=180^{\circ}$

$3 x+11+5 x-31=180^{\circ}$

$8 x-20=180$

$8 x=180+20$

$8 x=200$

$x=\frac{200}{8}$

$x=25^{\circ}$

$\begin{aligned} \angle B &=3 x+11 . \\ &=3 \times 25+11 \\ &=75+11 \\ \angle B &=86^{\circ} \\ \angle C &=5 x-31 \\ &=5 \times 25-31 \\ &=125-31 \\ \angle C &=94^{\circ} \end{aligned}$

∴ $\angle A=75^{\circ}, \quad \angle B=86^{\circ}, \angle C=94^{\circ}, \angle D=105^{\circ}$

Question 7

 (daigram should be added)

$\angle C E B: \angle E C B=3: 2$

$\angle C B E=90^{\circ}(\because$ anglin rectangle $)$

∴ In $\Delta$ ECB

Let

$\angle C E B=3 x, \quad \angle E C B=2 x .$

$\angle C E B+\angle E C B+\angle C B E=180^{\prime}$

$3 x+2 x+90=180$

$5 x+90=180$

$5 x=980-90$

$5 x=90$

$x=\frac{90}{5}$

$x=18^{\circ}$

(ii) $\angle C E B=3 x=3 \times 18=54^{\circ}$

At ' $c^{\prime}$

$\angle C E B+\angle D C E=\angle D C B$

$54^{\circ}+\angle D C E=90^{\circ}$ (∵Angle in rectangle)

$\angle D C E=90-54$

$\angle D C E=36^{\circ}$

$\begin{aligned} \angle D C E+\angle D C F &=180^{\circ} \quad(\because \text { Forms Straightine }) \\ 36^{\circ}+\angle D C F &=180^{\circ} \\ \angle D C F &=180-36^{\circ} \\ \angle D C F &=144^{\circ} \end{aligned}$

Question 8

 (daigram should be added)

Given $A B C D$ is a rectangle

$\overrightarrow{A O}=\overline{O B}[\because$ intersect at 'O']

$\therefore \angle D A B=\angle O B A=x .$

(i)  In $\Delta$ AOB

$\begin{aligned} \angle A O B+\angle A B O &+\angle O A B=180^{\prime} \\ 118+x+x &=180^{\circ} \\ 2 x &=180-118 \\ 2 x &=62 \\ x &=\frac{62}{2} \\ x &=31^{'} \end{aligned}$

$\therefore \angle A B O=31^{\circ}$

(ii) $\begin{aligned} \angle A O B+\angle A O D &=180^{\circ}(\because \text { Forms straightline }\\ 118+\angle A O D &=180^{\circ} . \\ \angle A O D &=180-118 \\ \angle A O D &=62^{\circ} \end{aligned}$

$\overrightarrow{O D}=\overline{O A}(\because$ diagnals bisect each other)

$\angle D A O=\angle A D O=y$

  In $\Delta$ AOD

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

$y+y-62=180$

$\begin{aligned} 2 y+62 &=180 \\ 2 y &=180-62 \\ 2 y &=118 \\ y &=\frac{118}{2} \\ y &=59^{\circ} \\ \therefore \quad \angle A D O &=59^{\circ} \end{aligned}$

(iii) Simillary by taking $\Delta$  BOC

We can prove $\angle O C B=59^{\circ}$

Question 9

 (daigram should be added)

Give $A B C D$ is a rhombus

$\angle A B D=50^{\circ}$

In $\triangle^{\prime \prime}$ AOB (∵In rhombus, diagonals cut perpendicularly )

$\therefore \quad \angle A O B+\angle O A B+\angle O B A=180^{\circ}$

$\begin{aligned} 90+\angle O A B+50^{\circ}=180 \\ & \angle O A B-140^{\circ}=180 \\ & \angle O A B=180-140 \\ & \angle O A B=40^{\circ} \end{aligned}$

$\angle O A B=\angle C A B=40^{\circ}$

(ii) $\begin{aligned} \angle B C D &=? \\ A B & \| D C \end{aligned}$

$\therefore \angle C A B=\angle A C D=40^{\circ}$

$\angle B C D=2 \times \angle A C D$ (∵fdiagonal in rhombus, bisects the angle )

$\angle B C D=2 \times 40^{\circ}$

$\angle B C D=80^{\circ}$

(iii) $\angle A D C$

$\angle A D C=\angle A B C$ (∵opposite angles are equal in rhombus )

$\angle A D C=2 \times \angle A B D$

$\angle A D C=2 \times 50^{\circ}$

$\angle A D C=100^{\circ}$

Question 10

 (daigram should be added)

In a trapezium

$\begin{aligned} \angle C+\angle B &=180^{\circ} \\ 112^{\circ}+\angle B &=180^{\circ} \\ \angle B &=180-102 \\ \angle B &=78^{\circ} \end{aligned}$

Given 

$\begin{aligned} & \overline{A D}=\overline{C B} \\ \therefore & \angle C=\angle D=102^{\circ} \\ \therefore A &=\angle B=78^{\circ} \end{aligned}$

∴ Angles in trapezium $78 ; 78^{\circ}, 102^{\circ}, 102^{\circ}$

Question 11

 (daigram should be added)

Given PQRS is a kite

$\begin{aligned} \therefore \quad \angle Q=\angle S &=120^{\circ} \\ & x=120^{\circ} \end{aligned}$

In a quadtitateral

$\begin{array}{rl}\angle p+\angle Q+\angle R+\angle S & =360 \\ y+120+50+120 & 2360 \\ y+290 & =360 \\ y & =360-290 \\ y & =70^{\circ}\end{array}$