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

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

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ML AGGARWAL CLASS 8 Chapter 12 Linear Equations and Inqualities In one Variable Exercise 12.1

 Exercise 12.1

Question 1

(i) $5 x-3=3 x-5$

=$5 x=3 x-5+3$

=$5 x=3 x-2$

=$5 x-3 x=-2$

=$2 x=-2$

$x=-1$

(ii) $3 x-7=3(5-x)$

=$3 x-7=15-3 x$

=$3 x=15+7-3 x$

=$3 x+3 x=22$

=$6 x=22$

=$x=\frac{22}{6}$

Question 2

(i) $4(2 x+1)=3(x-1)+7$

=$8 x+4=3 x-3+7$

=$8 x=3 x-3+7-4$

=$8 x-3 x=-3+3=0$

$5 x=0 \Rightarrow x=0$

(ii) 

$\begin{aligned} 3(2 p-1)=5-(3 p-2) & \\ 6 p-3=5 &-3 p+2 \\ 6 p=& 5-3 j+2+3 \\ 6 p+3 p &=10 \\ 9 p &=10 \\ p &=\frac{10}{9}=1 \frac{1}{9} \end{aligned}$

Question 3

(i) $5 y-2[y-3(y-5)]=6$

=$5 y-2 y+6(y-5)=6$

=$3 y+6 y-30=6$

=$9 y=6+30=36$

=$y=\frac{36}{9}=4$

(ii) $0.3(6-x)=0.4(x+8)$

=$1.8-0.3 x=0.4 x+3.2$

=$0.4 x=1.8-0.3 x-3.2$

=$0.4 x+0.3 x=-1.4$

=$0.7 x=-1.4$

=$x=-\frac{1.4}{0.7}=2$

Question 4

(i) $\frac{x-1}{3}=\frac{x+2}{6}+3$

Multiply and divide 2 on L.H.S

=$\frac{2(x-1)}{2 \times 3}=\frac{x+2}{6}+3$

=$\frac{2 x-1}{6}-\frac{x+2}{6}=3$

=$\frac{2 x-1-x-2}{6}=3$

=$\frac{x-3}{6}=3$

=$x-3=3 \times 6$

=$x-3=18$

$x=18+3$

$x=21$

 

(ii) $\frac{x+7}{3}=1+\frac{3 x-2}{5}$

=$\frac{x+7}{3}=\frac{5}{5}+\frac{3 x-2}{5}$

=$\frac{x+7}{3}=\frac{5+3 x-2}{5}$

=$\frac{x+7}{3}=\frac{3 x+3}{5}$

Cross Multiplying

=$5(x+7)=3(3 x+3)$

=$5 x+35=9 x+9$

=$9 x=5 x+35-9$

=$9 x-5 x=26$

=$4 x=26$

$x=\frac{26}{4}=\frac{13}{2}=6 \frac{1}{2}$

Question 5

(i) $\frac{y+1}{3}-\frac{y-1}{2}=\frac{1+2 y}{3}$

Multiplying both sides by 6 i.e L.C.M of $3,2,3$ we get

=$2(y+1)-3(y-1)=2(1+2 y)$

=$2 y+2-3 y+3=2+4 y$

=$5-y=4 y+2$

=$4 y+y=5-2$

$5 y=3$

$y=3 / 5$

(ii) $\frac{1}{3}+\frac{p}{4}=55-\frac{p+40}{5}$

=$\frac{4 p+3 p}{12}=\frac{55 \times 5}{5}-\frac{p+40}{5}$

=$\frac{7 p}{12}=\frac{275-p-40}{5}$

$\frac{7 p}{12}=\frac{235-P}{5}$

Cross Multiplication

$7 p \times 5=12(235-p)$

$35 p=12 \times 235-12 p$

$35 p+12 p=12 \times 235$

$47 P=12 \times 235$

$P=\frac{12 \times 2355}{47}$

$p=60$

Question 6

(i) $n-\frac{n-1}{2}=1-\frac{\eta-2}{3}$

$\frac{2 n-(n-1)}{2}=\frac{3-(n-2)}{3}$

$\frac{2 n-n+1}{2}=\frac{3-n+2}{3}$

$\frac{n+1}{2}=\frac{5-n}{3}$

Cross multplication

$3 x(n+1)=2(5-n)$

$3 n+3=10-2 n$

$3 n+2 n=10-3$

$5 n=7$

$n=7 / 5$

(ii) $\frac{3 t-2}{3}+\frac{2 t+3}{2}=t+\frac{7}{6}$

$\frac{3 t-2}{3}+\frac{2 t+3}{2}=\frac{6 t+7}{6}$

Multiplying 6 on both sides

2(3 t-2)+3(2 t+3)=1(6 t+3)

6 t-4+6 t+9=6 t+7

6 t+5=7

6 t=7-5=2

t=2 / 6=1 / 3

Question 7

(i)

 $\begin{aligned} 4(3 x+2)-5(6 x-1) &=2(x-8)-6(7 x-4) \\ 12 x+8-30 x+5 &=2 x-16-42 x+24 \\ 13-18 x=& 8-40 x \\ 40 x-18 x &+13=8 \\ 22 x &=8-13=-5 \\ x &=\frac{-5}{22} . \end{aligned}$

 

(ii) $3(5 x+7)+5(2 x-11)=3(8 x-5)-15$

=$15 x+21+10 x-55=24 x-15-15$

=$25 x-34=24 x-30$

=$25 x-24 x=34-30$

$x=4$

Question 8

(i) $\frac{3-2 x}{2 x+5}=-\frac{3}{11}$

Cross Multiplying both sides

$-11(3-2 x)=3(2 x+5)$

$22 x-3]=6 x+15$

$22 x-6 x=33+15$

$16 x=48$

x=3

(ii) $\frac{5 p+2}{8-2 p}=\frac{7}{6}$

cross multiplying on both sides

$\begin{aligned} 6(5 p+2) &=7(8-2 p) \\ 30 p+12 &=56-14 p \\ 30 p+14 p &=56-12 \\ 44 p &=44 \Rightarrow p=1 \end{aligned}$

Question 9

(i) $\frac{5}{x}=\frac{7}{x-4}$

Cross multiplying on both sides

$\begin{array}{c}5(x-4)=7 x \\ 7 x=5 x-20 \\ 2 x=-20 \\ x=-10\end{array}$

(ii) $\frac{4}{2 x+3}=\frac{5}{x+4}$

Cross multiplication on both sides

$\begin{aligned} 5(2 x+3) &=4(x+4) \\ 10 x+15 &=4 x+16 \\ 10 x-4 x &=16-15 \\ 6 x &=1 \Rightarrow x=\frac{1}{6} \end{aligned}$

(ii) $\frac{4}{2 x+3}=\frac{5}{x+4}$

Cross multhpicalion on bolh sides

$\begin{aligned} 5(2 x+3)=4(x+4) \\ 10 x+15=4 x+16 \\ 10 x-4 x &=16-15 \\ 6 x=1 & \Rightarrow x=\frac{1}{6} \end{aligned}$

Question 10

(i) $\frac{2 x+5}{2}-\frac{5 x}{x-1}=x$

=$\frac{(2 x+5)(x-1)-(51 \times 2)}{2(x-1)}=x$

=$2 x^{2}-2 x+5 x-5-10 x=2 x(x-1)$

=$2 x^{2}+3 x-5-101=2 x^{2}-2 x$

$-7 x-5=-2 x$

$-5=7 x-2 x$

$5 x=-5$

$x=-1$

(ii) $\frac{1}{5}\left(\frac{1}{3 x}-5\right)=\frac{1}{3}\left(3-\frac{1}{x}\right)$

$\frac{1}{5}\left(\frac{1-15 x}{3 x}\right)=\left(\frac{3 x-1}{3 x}\right)$

$1-15 x=5(3 x-1)$

$1-15 x=15 x-5$

$15 x+15 x=1+5$

$30 x=6$

$x=\frac{6}{30}=\frac{1}{5}$

Question 11

(i) $\frac{2 x-3}{2 x-1}=\frac{3 x-1}{3 x+1}$

Substralting ' 1' on both Sides

=$\frac{2 x-3}{2 x-1}-1=\frac{3 x-1}{3 x+1}-1$

=$\frac{2 x-3-2 x+1}{2 x-1}=\frac{3 x-1-3 x-1}{3 x+1}$

=$\frac{-2}{2 x-1}=\frac{-2}{3 x+1}$

=$3 x+1=2 x-1$

$3 x-2 x=-1-1$

$x=-2$

(ii) $\frac{2 y+3}{3 y+2}=\frac{4 y+5}{6 y+7}$

Cross multiplication

$(2 y+3)(6 y+7)=(3 y+2)(4 y+5)$

$12 y^{2}+14 y+18 y+21=12y^{2}+15 y+8 y+10$

$32 y+21=23 y+10$

32 y-23 y=10-21

9 y=-11

$y=\frac{-11}{9}$

Question 12

$x=p+1$

$\frac{5 x-30}{2}-\frac{7 p+1}{3}=\frac{1}{4}$

multiplying 12 on both sides

$6(5 x-30)-4(7 p+1)=3$

$30 x-180-28 p-4=3$

$30(p+1)-180-28 p-4-3=0$

$30 p+30-180-28 p-7=0$

$2 p-157=0$

$p=\frac{157}{2}$

Question 13

$\frac{x+3}{3}-\frac{x-2}{2}=1 \quad$ if $\frac{1}{x}+p=1$

Mulfiplying 6 on both sides

$2(x+3)-3(x-2)=6$

$2 x+6-3 x+6=8$

x=6 

$\frac{1}{x}+p=1 \Rightarrow p=1-\frac{1}{x}=1-\frac{1}{6}$

$p=\frac{5}{6}$

 Exercise 12.2

Question 1

Let the number be 'x' 

Three more than twice a number is 

= 2x + 3 -------1

Four less than the number = x -4 ----------2

① = ②

$2 x+3=x-4$

$2 x-x=-4-3$

$x=-7$

The number is -7 

Question 2

Let the four consecutive integers are $x+1, x+2, x+3$,

x + 4 

Given sum of them =46

$\begin{array}{c}x+1+x+2+x+3+x+y=46 \\4 x+10=46 \\4 x=46-10=36\end{array}$

$x=9$

The integers are 9+1,9+2,9+3,9+4

10,11,12,13

Question 3

Let the number be 'x '

Manjula substracts $\frac{7}{3}$ from it $=x-\frac{7}{3}$

The above result is multiplied by 6 i.e $6\left(x-\frac{7}{3}\right)$

Now it is equal 2 less than twice the number 'x'

$6\left(x-\frac{7}{3}\right)=2 x-2$

6 x-14=2 x-2

6 x-2 x=14-2

4 x=12

x=3

Question 4

Let the number be x , 7x

15 is added to both number then 

it becomes x + 15 , 7x + 15

Then one new numbers becomes $\frac{5}{2}$ times the other

new number.

$7 x+15=\frac{5}{2}(x+15)$

$2(7 x+15)=5(x+15)$

$14 x+30=5 x+75$

$14 x-5 x=75-30$

9 x=45

x=5

Therefore the number are 5 , 35

Question 5

Let the three conseutive even integers are x, x+2 ,x+4

Given Sum = 0

x+x+2+x+4=0 

3 x+6=0

3 x=-6

x=-2

∴ The integers are -2 , -2+2 ,-2+4

= -2, 0 , 2

Question 6

Let the two consecutive odd integers are x+ 1 , x + 3 

Given two fifth of smaller exceeds two ninth of greater by 4

$\frac{2}{5}(x+1)+4=\frac{2}{9}(x+3)$

$\frac{2 x+2+20}{5}=\frac{2 x+6}{9}$

Cross Multiplication.

(2 x+22) 9=5(2 x+6)

18 x+198=10 x+30

18 x-10 x=30-198

8 x=-168

x=-21 

x+1=-21+1=-20

x+3=-21+3=-18

The consecutive odd integers are -20,-18

Question 7

Given the denominator of a fraction is 1 more than 

twice its numerator

It is written as $\frac{x}{2 x+1}$

Given numerator and denominator are both increased by 5 

it becomes $\frac{3}{5}$

$\frac{x+5}{2 x+1+5}=\frac{3}{5}$

$\frac{x+5}{2 x+6}=\frac{3}{5}$

Cross mulhpication.

$5(x+5)=3(2 x+6)$

$5 x+25=6 x+18$

$6 x-5 x=25-18$

$x=7$ 

$\begin{aligned} \therefore \frac{x}{2 x+1}=\frac{7}{2(7)+1} &=\frac{7}{14+1} \\ &=\frac{7}{15} . \end{aligned}$

Therefore the original fraction is $\frac{7}{15}$

Question 8

Let the two number are 2x , 5x 

Given their difference is 15 

5 x-2 x=15

3 x=15

$x=15 / 3=5$

x=5

Therefore the two positive numbers are 10 , 25

Question 9

Let the number added to each be x 

on adding 'x '' to each of the numbers it becomes 

 

12 + x , 22 + x , 42 + x , 72 + x

for the number to be in proportion

$\frac{12+x}{22+x}=\frac{42+x}{72+x}$

Cross Multiplication

 $(x+12)(x+72)=(x+22)(x+42)$

$x^{2}+72 x+12 x+864=x^{2}+22 x+42 x+924$

$84 x+864=64 x+924$

$84 x-64 x=924-864$

$20 x=60$

x=3

So the number that must be added is x = 3

Question 10

Let the unit's digit be x

As the difference of both digits is 3 , the ten's digit is x + 3

∴ The number is 10x (x + 3) 

By adding both number , We get 143

10(x+3)+x+10 x+(x+3)=143

10 x+30+x+10 x+x+3=143

22 x+33=143

22 x=143-33

22x=110

x=5

∴ $10(x+3)+x=10(5+3)+5$

=80+5 = 85

∴ The two digit number is 85

Question 11

Let the unit's digits be ' x ' 

then ,the ten's digits number be 11 - x 

The two digit  number = 10 (11- x ) + x

When we intercharge the digits , the resulting new number 

is greater than original number by 63 

$\begin{aligned} 10(x)+(11-x) &=10(11-x)+x+63 \\ 10 x+11-x &=110-10 x+x+63 \\ 9 x+11 &=173-9 x \\ 9 x+9 x &=173-11 \\ 18 x &=162 \\ x &=9 \end{aligned}$

$\begin{aligned} \therefore 10(11-x)+x &=10(11-9)+9 \\ &=10(2)+9=20+9=29 . \end{aligned}$

∴ The two digit number is 29

Question 12

Let the Raju'S present age be ' x' year 

 then , Ritu's present age be 4x years

In four times , Ritu's age will be twice of raju's age 

4 x+4=2(x+4)

4 x+4=2 x+8

4 x-2 x=8-4

2 x=4

x=2

∴ The present ages of Raju's , Ritu are 2 , 8 year

Question 13

Let the son's age be ' x ' years

then the father age be 7x years'

Two years ago , father was 13 times as old as his Son'

7 x-2=13(x-2)

7 x-2=13 x-26

13 x-7 x=26-2

6x =24

x=4

Their present age of Son, Father are 4 , 28

Question 14 

Let the ages of sona and sonali are 5x , 3x 

five years hence , the ratio of their ages were 10 : 7

$\frac{5 x+5}{3 x+5}=\frac{10}{7}$

Cross Multiplicatim

$\begin{array}{r}7(5 x+5)=10(3 x+5) \\ 35 x+35=30 x+50 \\ 35 x-30 x=50-35 \\ 5 x=15 \\ x-3\end{array}$

There fire their present ages are 15 ,9

Question 15

An employee works on a contract of 30 days 

for that he will receive Rs 200 for each days and 

 he will be find Rs 20 fo each day he is absent 

Let the number remain absent be 'x ' days

then the no of days he worked are 30 -x days 

∴ $200(30-x)-20 x=3,800$

$6,000-200 x-20 x=3,800$

$220 x=6,000-3,80^{\circ}$

$220 x=2,200$

$x=10$

∴The no. of days he remained absent are 10 days 

Question 16

Let the no of Rs 5 coins be x 

no of Rs 2 coins be 3x

no of Rs 1 coins be 160 -4x

Total Rs 300 in coins of denominaton

∴ $5 x(x)+2(3 x)+1(160-4 x)=300$

$5 x+6 x+160-4 x=300$

$7 x=300-160$

$7 x=140$

$x=20$

Coins of each denominaton are 

Rs Coins = 20

Rs 2 Coins = 60

Rs 1 coins = 80

Question 17

Let the no of passenger with Rs 5 tickets be 'X'

The no of passenger with Rs 75 tickets be 40 - x

Total receipts from passenger is Rs 230

$\begin{aligned} 5 x+2.5(40-x) &=230 . \\ 5 x+300-7 \cdot 5 x &=230 \\ 75 x-5 x &=300-230 \\ 2.5 x &=70 \\ x &=\frac{70}{2.5} \\ x &=28 \end{aligned}$

∴ Number of passenger with Rs 5 tickets are 28

Question 18

Let the no of students in the group be 'x ' 

 

They paid equally for use a of a ful boat and 

pay Rs 10 each i.e 10$\times$x---------1

If there are 3 more students in group , each would have 

paid Rs 2 less i.e Rs 8 i.e = 8 $\times$(x +3)

1 = 2

10 x=8(x+3)

10 x=8 x+24

2 x=24

x=12

Question 19

Let the number of deer in the herd be ' x ' 

half of a herd of deer are grazing in field i,e = $\frac{1}{2} x$

Three fourth of remaining are playing i,e = $\frac{3}{4}\left(\frac{1}{2} x\right)$

$=\frac{3}{8} x$

The rest 9 are drinking water 

$x-\left\{\frac{x}{2}+\frac{3 x}{8}\right\}=9$

$x-\frac{4 x+3 x}{8}=9$

$\frac{8 x-7 x}{8}=9$

$\frac{x}{8}=9$

$x=72$

Number of deer in the herd are 72

Question 20

Let the no. of flower in the beginning be 'x'

At 1st temple she offer = $\frac{1}{2} \times x=\frac{x}{2}$

2nd temple she offer =$\frac{1}{2} \times \frac{x}{2}=\frac{x}{4} .$

3rd temple she offer = $\frac{1}{2} \times \frac{1}{4}=\frac{x}{8}$

Now she is left with 6 flower at end

$x-\left\{\frac{x}{2}+\frac{x}{4}+\frac{x}{8}\right\}=6 .$

$x-\frac{4 x+2 x+x}{8}=6$

$\frac{8 x-7 x}{8}=6$

$\frac{x}{8}=6$

$x=48$

ஃNo of flower in the beginning are 48

Question 21

Let the two supplementary angles be $x, 90-x$

These angles differ by $50^{\circ}$.

i.e $90-x-x=50$

$90-2 x=50$

$2 x=90-50$

$2 x=40$

$x=20^{\circ}$

∴ The two supplementary angles are 20 ,70

Question 22

Let the angles of triangle are 5x , 6x , 7x 

Sum of angles of triangle= 180

i .e  5x + 6x + 7 x =180

18x = 180

x = 10

∴ The angles of triangles are $50^{\circ}, 60^{\circ}, 70^{\circ}$

Question 23

Two equal sides of an isosceles triangles are $3 x-1,2 x+2$

$3 x-1=2 x+2$

$3 x-2 x=2+1$

$x=3$

The third side is 2x = $2 \times 3=6$ Units

The two sides are $3 x-1=3 \times 3-1=9-1=8$ unith

$2 x+2=2 \times 3+2=6+2=8$ unih

∴ Perimeter of triangle = 6+8+8=22 Units

Question 24

Let the perimeter of given triangle be x cm 

As each Side is increased by 4cm , So the 

perimeter is increased by $3 \times 4=12 \mathrm{~cm}$

According to given information

$\frac{x+12}{x}=\frac{7}{5}$

Cross Multiplication

$7 x=5(x+12)$

$7 x=5 x+60$

$7 x-5 x=60$

$2 x=60$

x=30

∴ Perimeter of given triangle = 30cm

Question 25

Length of a rectangle is 5cm less than twice its breadth

i .e  $l=2 b-5 \Rightarrow$ 2 b=l+5-------(1)

length is decreased by 3cm i .e (l-3)cm

breadth is increased by 2cm i .e (b+2)cm

Resulting perimeter of rectangle is 72cm

$2(1-3+b+2)=72$

$2 l+2 b-2=72$

Frameq (1) we get

$2 l+1+5-2=72$

$3 x+3=72$

$3 x=72-3$

$3 x=69$

$1=23 \mathrm{~cm}$

Length of rectangle $=23 \mathrm{~cm}$

From eq (1) $\quad 2 b=l+5=23+5$

$2 b=28 \Rightarrow b=14 \mathrm{~cm}$

∴Breadth of rectangle = 14 cm

∴Area of rectangle = l $\times$ b = $23 \times 14=322\mathrm{cm}^{2}$

Question 26

Length of rectangle l = 10cm

breadth of rectangle b = 8cm

Each side of rectangle is increased by x cm , its 

perimeter is doubled 

perimeter of rectangle = $\begin{aligned} 2(10+8) &=2 \times 18 \\ &=36 \mathrm{~cm} . \end{aligned}$

i.e $2[(10+x)+(8+x)]=2 \times 36,[\because$ perimeter is doubled]

$18+2 x=36-0$

$2 x=36-18=18$

$x=18 / 2=9$

=9 cm

Area of new rectangle i .e $\begin{aligned} &(10+9) \times(8+9) \\=& 19 \times 17=323 \mathrm{~cm}^{2} \end{aligned}$

Question 27

Let the speed of streamer in still water be x km\h 

Given the speed of stream= 5km\h

The speed of streamer down steam = (x+5) km/h

Speed of streamer upwardstream= ( x-5)km/h

Both upstream and down stream takes same time 

time taken by streams for down stream is $\frac{90}{x+5}$ hr

Time taken by streamer for upstream is $\frac{60}{x-5} h r$

i.e $\frac{90}{x+5}=\frac{60}{x-5}$

Cross Multiplication

$3(x-5)=2(x+5)$

$3 x-15=2 x+10$

$3 x-2 x=10+15$

$x=25$

∴ Speed of streamer in still water = 25km/h

Question 28

Let the speed of streamer in still water be 'x'

Given the speed of stream = 1km/h

Speed of streamer down stream= (x+1)km/h

Speed of streamer upstream = (x-1)km/h

Distance Covered by streamer , downstream= $5(x+1) \mathrm{km}$

Distance Covered by streamer upstream = $6(x-1) \mathrm{km}$

According to given information

$5(x+1)=6(x-1)$

$5 x+5=6 x-6$

$6 x-5 x=5+6$

$x=11$

∴ Speed of streamer in Still water is 11 km/h

Distance between two parts =$\begin{aligned} 5(x+1) &=5(11+1) \\ &=60 \mathrm{~km} . \end{aligned}$

Question 29

Let the speed of faster car be x km/hr 

then the speed of other car be (x - 8)km/hr

Let the faster car starts from place A and the slower

car starts from place B 

Let P and Q be their position after 4hours

( diagram should be added)

 $A p=4 x \mathrm{~km}, \quad B Q=4(x-8) \mathrm{km}, p Q=62 \mathrm{~km}$

According to the given p b

$\begin{aligned} A P &+P Q+B Q=350 \\ 4 x+62+4(x-8)=350 \end{aligned}$

$4 x+62+4 x-32=350$

$8 x+30=350$

$8 x=350-30=320$

$x=320 / 8$

x=40 km/hr

∴ Speed of faster car is 40 km/hr and the speed of slower

car is (40-8) i.e 32km/h

 

 Exercise 12.3

Question 1

(i) $x>-2$

Solution set $=\{-1,0,1,3\}$

(ii) $x<-2$

Solation et $=\{-7,-5,-3\}$

(iii) $x>2$

Solution set $=\{3\}$

(iv)

 $\begin{aligned}-5<x & \leq 5 \\ & \text { Solution set }=\{-3,-1,0,1,3\} \end{aligned}$

(v) 

$\begin{aligned}-8<x &<1 \\ & \text { Solution set }=\{-7,-5,-3,-1,0\} \end{aligned}$

(vi) 

$\begin{aligned} 0 \leq x & \leq 4 \\ & \text { Solution set }=\{0,1,3\} \end{aligned}$

Question 2

It shown by thick dots on number line 

(i) $x \leq 4, x \in N$

(diagram should be added)

(ii) $x<5, x \in W$

(diagram should be added)

(iii) $-3 \leq x<3, x \in \mathbb{L}$

(diagram should be added)

Question 3

Replacement Set =$\{-6,-4,-2,0,2,4,6\}$

$-4 \leq x<4$

(diagram should be added)

Question 4

(i) $\{1,2,3 \ldots \ldots .10\}$

(diagram should be added)

(ii) $\{-1,0,1,2,5,8\}$

(diagram should be added)

(iii) $\{-5,10\}$

(diagram should be added)

(iv) $\{5,6,7,8,9,10\}$

(diagram should be added)              Solution Set=$\phi$

Question 5

(i) 2 x-3>7

2 x>7+3

2 x>10

x>5

Solution set $=\{6,9,12\}$

(ii) $\begin{aligned} 3 x+8 \leq & 2 \\ 3 x \leq & 2-8 \\ 3 x & \leq-6 \\ x & \leq-2 \end{aligned}$

Solution set $=\{-6,-3\}$

(iii) $-3<1-2 x$

$3>2 x-1$

$2 x-1<3$

$2 x<3+1$

$2 x<4$

$x<2$

Solution set $=\{-6,-3,0\}$

Question 6

(i) $\begin{aligned} 4 x+1 &<17, x \in N 1 \\ 4 x &<17-1 \\ 4 x &<16 \\ x &<4, x \in N \end{aligned}$

As $x \in N$, the Solution set is $\{1,2,3\}$

(ii) $\begin{aligned} 4 x+1 & \leq 17, x \in W \\ 4 x & \leq 17-1 \\ 4 x & \leq 16 \\ x & \leq 4 \end{aligned}$

As $x \in W$, the solution set is $\{0,1,2,3,4\}$

(iii) $\begin{array}{rl}4>3 x-11 & , x \in N \\ 3 x-11 & <4 \\ 3 x & <4+11 \\ 3 & 3 x<15 \\ & x<5\end{array}$

As $x \in N$, the solution set is $\{1,2,3,4\}$

(iv)$-17 \leq 9 x-8, x \in z$

$9 x-8 \geq-17$

$9 x \geq-17+8$

$9 x \geq-9$

$x \geq-1$

As $x+z$, the solution set is $\{-1,0,1,2,3 \ldots .\}$

Question 7

(i) $\frac{2 y-1}{5} \leq 2, y \in N$

$2 y-1 \leq 10$

$2 y \leq 10+1$

$2 y \leq 11$

$y \leq \frac{11}{2}$

As $y \in N$, the solution set is $\{1,2,3,4,5\}$

(ii) $\frac{2 y+1}{3}+1 \leq 3, y \in w$

$\frac{2 y+1}{3} \leq 3-1$

$\frac{2 y+1}{3} \leq 2$

$2 y+1 \leq 6$

$2 y+6-1$

$2 y \leq 5$

$y \leq \frac{5}{2}$

As $y \in W$, the solution set is $\{0,1,2\}$

(iii) $\frac{2}{3} p+s<9, p \in W .$

$\frac{2}{3} p<9-5$

$\frac{21}{3}<4$

$2 p<12$

$p<6$

As $p \in W$, the solution set is $\{0,1,2,3,4,5\}$

(iv) $-2(p+3)>5 \quad p \in I$

Mulhpliying '- ' on both sides

2(p+3)<-5

$2 p+6<-5$

$2 p<-5-6$

$2 p<-11$

$p<-\frac{11}{2}$

As $p \in I$, the solution set is $\{\ldots-9,-8,-7,-6\}$

Question 8

(i) $2 x-3<x+2, x \in N$

$2 x<x+2+3$

$2 x-x<5$

$x<5$

As $x \in N$, the Solufion set is $\{1,2,3,4\}$

(ii) $3-x \leq 5-3 x, \quad x \in W$

$3-x+3 x \leq 5$

$2 x+3 \leq 5$

$2 x \leq 5-3$

$2 x \leq 2$

$x \leq 1$

As $x \in W$, the solution set is $\{0,1\}$

(iv) $\frac{3}{2}-\frac{x}{2}>-1, \quad x \in N$

$\frac{3-x}{2}>-1$

$3-x>-2$

Mulhplying with '- ' on son sides

$x-3<2$

$x<2+3$

$x<5$

As $x \in N$, the solution set is $\{1,2,34\}$

Question 9

$\{-3,-2,-1,0,1,2,3\}$

$\begin{aligned} \frac{3 x-1}{2} &<2 \\ 3 x-1 &<4 \\ 3 x-4 &<4+1 \\ 3 x &<5 \end{aligned}$

$x<5 / 3$

As $x$ should be in replacement set, the solution

set is $\{-3,-2,-1,0,1\}$

(diagram should be added)

Question 10

$\frac{x}{3}+\frac{1}{4}<\frac{x}{6}+\frac{1}{2}, x \in w$

$\frac{x}{3}-\frac{x}{6}+\frac{1}{4}<\frac{1}{2}$

$\frac{x}{3}-\frac{x}{6}<\frac{1}{2}-\frac{1}{4}$

$\frac{2 x-x}{6}<\frac{2-1}{4}$

$\frac{x}{6}<\frac{1}{4}$

$x<\frac{6}{4}$

$x<\frac{3}{2}$

As $x \in W$, the solution set is $\{0,1\}$

Question 11

(i) 

$\begin{aligned}-4 \leq 4 x<14, & x \in N \\ 4 x \geqslant-4 ; & 4 x<14 \\ x \geqslant-1 ; & x<14 / 4 \\ & x<7 / 2 \end{aligned}$

As $x \in N$, the solution set is $\{-1,0,1,2,3\}$

(diagram should be added)

(ii) $-1<\frac{x}{2}+1 \leq 3, \quad x \in I$

$\frac{x}{2}+1>-1 \quad ; \quad \frac{1}{2}+1 \leq 3$

$\frac{x}{2}>-1-1 \quad ; \quad \frac{x}{2} \leq 3-1$

$\frac{x}{2}>-2 ; \quad \frac{x}{2} \leq 2$

$x>-4 ; \quad x \leq 4$

i.e $-4<x \leq 4$

As $x \in I$, the solution set is $\{-3,-2,-1,0,1,2,3,4\}$

(diagram should be added)

Continue Reading →

ML AGGARWAL CLASS 8 CHAPTER 11 Factorisation Excercise 11.1

 Excercise 11.1

Solution 1

(i) $8 x y^{3}+12 x^{2} y^{2}$

H.C.F of $8 x y^{3}$ and $12 x^{2} y^{2}$ is $4 x y^{2}$

$\therefore$ Divide each expression by $4 x y^{2}$ and keep $4 x y^{2}$

outside the bracket.

$\Rightarrow \quad 4 x y^{2}(2 y+3 x)$

$\therefore 8 x y^{3}+12 x^{2} y^{2}=4 x y^{2}(2 y+3 x)$

(ii) $15 a x^{3}-9 a x^{2}$

H.C.F of exprassion $15 a x^{3}$ and $9 a x^{2}$ is $3 a x^{2}$

$\quad 15 a x^{3}-9 a x^{2}=3 a x^{2}(5 x-3)$

Solution 2

(i) $21 p y^{2}-56 p y$

$H \cdot C \cdot F$ of $21 p y^{2}$ and $56 p y$ is $7 p y$

$\therefore 21 p y^{2}-56 p y \Rightarrow 7 p y(3 y-8)$

(ii) $4 x^{3}-6 x^{2}$

H.c.F of $4 x^{3}$ and $6 x^{2}$ is $2 x^{2}$

$\therefore 4 x^{3}-6 x^{2} \Rightarrow 2 x^{2}(2 x-3)$

Solution 3

(i) $25 a b c^{2}-15 a^{2} b^{2} c$

H.c.F of $25 a b c^{2}$ and $15 a^{2} b^{2} c$ is $5 a b c$

$\therefore 25 a b c^{2}-15 a^{2} b^{2} c \Rightarrow 5 a b c(5 c-3 a b)$

(ii)

 $x^{2} y z+x y^{2} z+x y z^{2}$

HC.F of $x^{2} y z$, $x y^{2} z$ and $x y z^{2}$ is $x y z$

$\therefore \quad x y z(x+y+z)$

Solution 4

(i) $8 x^{3}-6 x^{2}+10 x$

H.C.F of $8 x^{3}, 6 x^{2}$ and $10 x$ is $2 x$

$\Rightarrow \quad 2 x\left(4 x^{2}-3 x+5\right)$

(ii) $14 \mathrm{mn}+22 \mathrm{~m}-6 \mathrm{p}$

H.C.F of $14 \mathrm{mn}, 22 \mathrm{~m}$ and $62 \mathrm{P}$ is 2

$\Rightarrow 2(7 \mathrm{mn}+11 \mathrm{n}-31 \mathrm{P})$

Solution 5

(i) $18 p^{2} q^{2}-24 p q^{2}+30 p^{2} q$

$H \cdot c \cdot F$ of $18 p^{2} q^{2}, 24 p a^{2}$ and $30 p^{2} q$ is $6 p q$

$\Rightarrow \quad 6 p q(3 p q-4 q+5 p)$

(ii) $27 a^{3} b^{3}-18 a^{2} b^{3}+75 a^{3} b^{2}$

H. C .F of $27 a^{3} b^{3}, 18 a^{2} b^{2}$ and $7 s a^{3} b^{2}$ is $3 a^{2} b^{2}$

$\Rightarrow 3 a^{2} b^{2}(9 a b-6 b+25 a)

Solution 6

(i) $15 a(2 p-3 q)-10 b(2 p-3 q)$

$H \cdot c \cdot F \quad$ of $15 a(2 p-3 q)$ and $10 b(2 p-3 q)$ is

$5(2 p-3 q)$

$\Rightarrow 5(2 p-3 q)(3 a-2 b)$

(ii) $3 a\left(x^{2}+y^{2}\right)+6 b\left(x^{2}+y^{2}\right)$

$H \cdot C \cdot F$ of $3 a\left(x^{2}+y^{2}\right)$ and $6 b\left(x^{2}+y^{2}\right)$ is $3\left(x^{2}+y^{2}\right)$

$\therefore \Rightarrow 3\left(x^{2}+y^{2}\right)(a+2 b)$

Solution 7

(i) $6(x+2 y)^{3}+8(x+2 y)^{2}$

H.C.F of $6(x+2 y)^{3}$ and $8(x+2 y)^{2}$ is $2(x+2 y)^{2}$

$\therefore \quad 2(x+2 y)^{2} \quad(3(x+2 y)+4)$

$\Rightarrow \quad 2(x+2 y)^{2}(3 x+6 y+4)$

(ii)$14(a-3 b)^{3}-21 p(a-3 b)$

H.C.F of $14(a-3 b)^{3}$ and $21 p(a-3 b)$ is

$\quad F(a-3 b)$

$\rightarrow \quad 7(a-3 b)\left[2(a-3 b)^{2}-3 p\right]$

Solution 8

(i) $\quad 10 a(2 p+q)^{3}-15 b(2 p+q)^{2}+35(2 p+q)$

$H \cdot C \cdot F$ of $10 a(2 p+q)^{3}, 15 b(2 p+q)^{2}$ and

$35(2 p+q)$ is $5(2 p+q)$

$\Rightarrow 5(2 p+q)(2 a-3 b+7)$

 Excercise 11.2

Solution 1

(i)

 $\begin{aligned} & x^{2}+x y-x-y \\ \Rightarrow & x(x+y)-1(x+y) \\ \Rightarrow &(x+y)(x-1) \end{aligned}$

(ii)

 $\begin{aligned} & y^{2}-y z-5 y+5 z \\ \Rightarrow & y(y-z)-5(y-z) \\ \Rightarrow &(y-z)(y-5) \end{aligned}$

Solution 2

(i) 

$\begin{aligned} & 5 x y+7 y-5 y^{2}-7 x \\ \Rightarrow & 5 x y-5 y^{2}+7 y-7 x . \\ \Rightarrow & 5 x y-5 y^{2}-7 x+7 y \\ \Rightarrow & 5 y(x-y)-7(x-y) \\ \rightarrow &(x-y)(5 y-7) \end{aligned}$

(ii) $5 p^{2}-8 p q-10 p+16 q$

$5 p^{2}-10 p-8 p q+16 q$

$5 p(p-2)-8 q(p-2)$

$(p-2)(5 p-8 q)$

Solution 3

(i) 

$\begin{aligned} & a^{2} b-a b^{2}+3 a-3 b \\ \Rightarrow & a b(a-b)+3(a-b) \\ \Rightarrow &(a-b)(a b+3) \end{aligned}$

(ii)

 $\begin{aligned} & x^{3}-3 x^{2}+x-3 \\=& x^{2}(x-3)+1(x-3) \\ \Rightarrow &(x-3)\left(x^{2}+1\right) \end{aligned}$

Solution 4

(i) 

$\begin{aligned} & 6 x y^{2}-3 x y-10 y+5 \\ \Rightarrow & 3 x y(2 y-1)-5(2 y-1) \\ \Rightarrow &(2 y-1)(3 x y-5) \end{aligned}$

(ii) $3 a x-6 a y-8 b y+4 b x$

$ \quad 3 a(x-2 y)+2 b(-2 y+x)$

$\Rightarrow 3 a(x-2 y)+2 b(x-2 y)$

$\Rightarrow \quad(x-2 y)(3 a+2 b)$

Solution 5

(i) $x^{2}+x y(1+y)+y^{3}$\

$\Rightarrow x^{2}+x y+x y^{2}+y^{3}$

$\Rightarrow \quad x(x+y)+y^{2}(x+y)$

$\Rightarrow \quad(x+y)\left(x+y^{2}\right)$

(ii) $y^{2}-x y(1-x)-x^{3}$

$\Rightarrow y^{2}-x y+x^{2} y-x^{3}$

$\Rightarrow y(y-x)+x^{2}(y-x)$

$\Rightarrow(y-x)+\left(y+x^{2}\right)$

Solution 6

(i) $a b^{2}+(a-1) \cdot b-1$

$a b^{2}+a b-b-1$

$a b(b+1)-(b+1)$

$(b+1)(a b-1)$

(ii) $2 a-4 b-x a+2 b x$

=$2(a-2 b)-x(a-2 b)$

=$(a-2 b)(2-x)$

Solution 7

(i) $\quad 5 p h-10qk+2 r p h-4 q r k$

$=5(p h-2 q k)+2 r(p h-2 q k)$

$=\left(p h-2 qk\right)(5+2 r)$

(ii) $x^{2}-x(a+2 b)+2 a b$

=$x^{2}-x a-2 b x+2 a b$

=$x(x-a)-2 b(x-a)$

=$(x-a)(x-2 b)$

  

Solution 8

(i) $a b\left(x^{2}+y^{2}\right)-x y\left(a^{2}+b^{2}\right)$

=$a b x^{2}+a b y^{2}-x y a^{2}-x y b^{2}$

=$a x(b x-a y)+b y(a y-b x)$

=$a x(b x-a y)-b y(b x-a y)$

=$(b x-a y)(a x-b y)$

(ii) $(a x+b y)^{2}+(b x-a y)^{2}$

=$a^{2} x^{2}+b^{2} y^{2}+2 a x \cdot b y+b^{2} x^{2}+a^{2} y^{2}-2 b x .a y$

=$x^{2}\left(a^{2}+b^{2}\right)+y^{2}\left(a^{2}+b^{2}\right)$

=$\left(a^{2}+b^{2}\right)\left(x^{2}+y^{2}\right)$

Solution 9

 

(i) (i) $a^{3}+a b(1-2 a)-2 b^{2}$

$= a^{3}+a b-2 a b-2 b^{2}$

$= a\left(a^{2}+b\right)-2 b\left(a^{2}+b\right)$

= $\left(a^{2}+b\right)(a-2 b)$

 

(ii) $3 x^{2} y-3 x y+12 x-12$

=$3 x y(x-1)+12(x-1)$

=$(x-1)(3 x y+12)$

=$(x-1) \cdot 3 \cdot(x y+4)$

=$\therefore \quad 3(x-1)(x y+4)$

Solution 10

(i) $a^{2} b+a b^{2}-a b c-b^{2} c+a x y+b x y$

=$\left(a^{2} b+a b^{2}\right)-\left(a b c+b^{2} c\right)+(a x y+b x y)$

=$a b(a+b)-b c(a+b)+x y(a+b)$

=$(a+b)(a b-b c+x y)$

 

(ii) $a x^{2}-b x^{2}+a y^{2}-b y^{2}+a z^{2}-b z^{2}$

=$x^{2}(a-b)+y^{2}(a-b)+x^{2}(a-b)$

=$(a-b)\left(x^{2}+y^{2}+z^{2}\right)$

Solution 11

(i) $x-1-(x-1)^{2}+a x-a$

=$1(x-1)-(x-1)^{2}+a(x-1)$

=$(x-1) \quad(1-(x-1)+a)$

=$(x-1)(1-x+1+a)$

=$(x-1)(2-x+a)$

(ii) $a x+a^{2} x+a b y+b y-(a x+b y)^{2}$

=$a x+b y+a^{2} x+a b y-(a x+b y)^{2}$

=$1(a x+b y)+a(a x+b y)-(a x+b y)^{2}$

=$(a+b y)(1+a-a x-b y)$

 

 Excercise 11.3

Solution 1

(i) $\quad x^{2}-12 x+36$

$\Rightarrow(x)^{2}-2 \cdot 6 \cdot x+(6)^{2}$

by using $a^{2}-2 a b+b^{2}=(a-b)^{2}$

$\therefore \quad(x-6)^{2}$

 

(ii) $36 p^{2}-60 p q+25 q^{2}$

=$(6 p)^{2}-2 \cdot 6 p \cdot 5 q+(5 q)^{2}$

=$(6 p-5 q)^{2}$

(iii) $9 x^{2}+66 x y+121 y^{2}$

$(3 x)^{2}+23 x \cdot 11 y+(11 y)^{2}$

$=(3 x+11 y)^{2}$

 

(iv) $a^{4}+6 a^{2} b^{2}+9 b^{4}$

=$\left(a^{2}\right)^{2}+2 \cdot a^{2} \cdot 3 b^{2}+\left(3 b^{2}\right)^{2}$

=$\left(a^{2}+3 b^{2}\right)^{2}$

(v) $x^{2}+\frac{1}{x^{2}}+2$

=$(x)^{2}+2 \cdot x \cdot \frac{1}{x}+\left(\frac{1}{x}\right)^{2}$

$=\left(x+\frac{1}{x}\right)^{2}$

Solution 2

(i) $4 p^{2}-9$

=$(2 p)^{2}-3^{2}$

by using $a^{2}-b^{2}=(a+b)(a-b)$

=(2 p+3)(2 p-3)

(ii) $4 x^{2}-169 y^{2}$

=$(2 x)^{2}-(13 y)^{2}$

=$(2 x+13 y)(2 x-13 y)$

Solution 3

(i) $9 x^{2} y^{2}-25$

=$(3 x y)^{2}-5^{2}$

=$(3 x y+5)(3 x y-5)$

(ii) $16 x^{2}-\frac{1}{144}$

=$(4 x)^{2}-\left(\frac{1}{12}\right)^{2}$

=$\left(4 x^{2}+\frac{1}{12}\right)\left(4 x-\frac{1}{12}\right)$

Solution 4

(i) $20 x^{2}-45 y^{2}$

=$5\left(4 x^{2}-9 y^{2}\right)$

=$5\left((2 x)^{2}-(3 y)^{2}\right)$

=$5(2 x+3 y)(2 x-3 y)$

(ii) $\frac{9}{16}-25 a^{2} b^{2}$

=$\left(\frac{3}{4}\right)^{2}-(5 a b)^{2}$

=$\left(\frac{3}{4}+5 a b\right)\left(\frac{3}{4}-5 a b\right)$

Solution 5

(i) $(2 a+3 b)^{2}-16 c^{2}$

=$(2 a+3 b)^{2}-(4 c)^{2}$

=$(2 a+3 b-4 c)(2 a+3 b+4 c)$

(ii) $1-(b-c)^{2}$

=$1^{2}-(b-c)^{2}$

=$(1+b-c)(1-b-c)$

Solution 6

(i) $9(x+y)^{2}-x^{2}$

=$3^{2}(x+y)^{2}-x^{2}$

=$(3(x+y))^{2}-x^{2}$

=$(3(x+y)+x)(3(x+y)-x)$

=$(3 x+3 y+x)(3 x+3 y-x)$

=$(4 x+3 y)(2 x+3 y)$

(ii) $(2 m+3 n)^{2}-(3 m+2 n)^{2}$

$= \quad(2 m+3 n+3 m+2 n)(2 m+3 n-3 m-2 n)$

$=(5 m+5 n)(n-m)$

Soluyion 7

(i) $25(a+b)^{2}-16(a-b)^{2}$

$5^{2}(a+b)^{2}-4^{2}(a-b)^{2}$

$(5 a+5 b)^{2}-(4 a-4 b)^{2}$

$=(5 a+5 b+4 a-4 b)(5 a+5 b-4 a+4 b)$

$=(9 a-b)(a+a b)$

 

(ii) $9(3 x+2)^{2}-4(2 x-1)^{2}$

=$3^{2}(3 x+2)^{2}-2^{2}(2 x-1)^{2}$

=$(9 x+6)^{2}-(4 x-2)^{2}$

=$(9 x+6+4 x-2)(9 x+6-4 x+2)$

=$(13 x+4)(5 x+7)$

Solution 8

(i) $x^{3}-25 x$

=$x\left(x^{2}-25\right)$

=$x\left(x^{2}-5^{2}\right)$

=$x \quad((x+5)(x-5))$

(ii) $63 p^{2} q^{2}-7$

=$7\left(9 p^{2} q^{2}-1\right)$

=$7\left((3 p q)^{2}-1^{2}\right)$

=$7 \quad((3 p q+1)(3 p q-1))$

Solution 9

(i) $32 a^{2} b-72 b^{3}$

=$8 b\left(4 a^{2}-9 b^{2}\right)$

=$8 b\left((2 a)^{2}-(3 b)^{2}\right)$

=$8 b((2 a+3 b)(2 a-3 b))$

(ii) $9(a+b)^{3}-25(a+b)$

=$(a+b)\left(3^{2}\left(a+b^{2}\right)^{2}-5^{-2}\right)$

=$(a+b)\left((3 a+3 b)^{2}-5^{2}\right)$

=$(a+b)(3 a+3 b+5)(3 a+3 b-5)$

Solution 10

 

(i) $x^{2}-y^{2}-2 y-1$

=$x^{2}-\left(y^{2}+2 y+1\right)$           $a^{2}+b^{2}+2 a b=(a+b)^{2}$

                                                                     $(a+b)(a-b)=a^{2}-b^{2}$

=$1 \quad x^{2}-(y+1)^{2}$

=$x \quad(x+y+1)(x-y-1)$

(ii) $p^{2}-4 p q+4 q^{2}-r^{2}$

=$\left(p^{2}-2 \cdot p \cdot 2 q+(2 q)^{2}\right)-r^{2}$

=$\Rightarrow \quad(p+2 q)^{2}-r^{2}$

=$\Rightarrow \quad(p-2 q+r)(p-2 q-r)$

Solution 11

(i) $9 x^{2}-y^{2}+4 y-4$

=$(3 x)^{2}-\left(y^{2}-2 \cdot y \cdot 2+2^{-2}\right)$

=$(3 x)^{2}-(y-2)^{2}$

=$(3 x+y-2)(3 x-y+2)$

(ii) $4 a^{2}-4 b^{2}+4 a+1$

=$(2 a)^{2}+2 \cdot 2 a \cdot 1+(1)^{2}-(2 b)^{2}$

=$(2 a+1)^{2}-(2 b)^{2}$

=$(2 a+1+2 b) \cdot(2 a+1-2 b)$

Solution 12

(i) $625-p^{4}$

=$(25)^{2}-\left(p^{2}\right)^{2}$

=$\left(25+p^{2}\right)\left(25-p^{2}\right)$

(ii) $5 y^{5}-405 y$

=$5 y\left(y^{4}-81\right)$

=$5 y \quad\left(\left(y^{2}\right)^{2}-9^{2}\right)$

=$5 y\left(y^{2}+9\right)\left(y^{2}-9\right)$

Solution 13

(i) $x^{4}-y^{4}+x^{2}-y^{2}$

=$\left(x^{4}-y^{4}\right)+\left(x^{2}-y^{2}\right)$

=$\left(\left(x^{2}\right)^{2}-\left(y^{2}\right)^{2}\right)+\left(x^{2}-y^{2}\right)$

=$\left(x^{2}+y^{2}\right)\left(x^{2}-y^{2}\right)+\left(x^{2}-y^{2}\right)$

=$\left(x^{2}-y^{2}\right)\left(x^{2}+y^{2}+1\right)$

(ii) $64 a^{2}-9 b^{2}+42 b c-49 c^{2}$

=$(8 a)^{2}-(3 b)^{2}+7 c(6 b-7 c)$

=$(8 a+3 b)(8 a-3 b)+7 c(6 b-7 c)$

 Excercise 11.4

Solution 1

(i) $x^{2}+3 x+2$

$\Rightarrow \quad x^{2}+2 x+x+2$

$\Rightarrow \quad x(x+2)+1(x+2)$

$\Rightarrow(x+1)(x+2) .$

(ii) $z^{2}+10 z+24$

=$z^{2}+6 z+4 z+24$

=$z(z+6)+4(z+6)$

=$(z+6)(z+4)$

Solution 2

(i) $y^{2}-7 y+12$

=$y^{2}-4 y-3 y+12$

=$y(y-4)-3(y-4)$

=$(y-4)(y-3)$

(ii) $m^{2}-23 m+42$

=$m^{2}-21 m-2 m+42$

=$m(m-21)-2(m-21)$

=$m(m-21)(m-2)$

Solution 3

(i) $y^{2}-5 y-24$

=$y^{2}-8 y+3 y-24$

=$y(y-8)+3(y-8)$

=$(y-8)(y+3)$

(ii) $t^{2}+23 t-108$

=$t^{2}+27 t-4 t-108$

=$t(t+27)-4(t+27)$

=$(t+27)(t-4)$

Solution 4

(i) $3 x^{2}+14 x+8$

=$3 x^{2}+12 x+2 x+8$

=$3 x(x+4)+2(x+4)$

=$(x+4)(3 x+2)$

(ii) $3 y^{2}+10 y+8$

=$3 y^{2}+6 y+4 y+8$

=$3 y(y+2)+4(y+2)$

=$(y+2)(3 y+4)$

Solution 5

(i) $14 x^{2}-23 x+8$

=$14 x^{2}-16 x-7 x+8$

=$2 x(7 x-8)-1(7 x-8)$

=$(7 x-8)(2 x-1)$

(ii) $12 x^{2}-x-35$

=$12 x^{2}-21 x+20 x-35$

=$3 x(42 x-7 x)+5(4 x-7)$

=$(4 x-7) \quad(3 x+5)$

Solution 6

(i) $6 x^{2}+11 x-10$

=$6 x^{2}+15 x-4 x-10$

=$3 x(2 x+5)-2(2 x+5)$

=$(2 x+5)(3 x-2)$

(ii) $5-4 x-12 x^{2}$

=$5-10 x+6 x-12 x^{2}$

=$5(1-2 x)+6 x(1-2 x)$

=$(1-2 x)(5+6 x)$ 

 

Solution 7

(i) $1-18 y-63 y^{2}$

=$1-21 y+3 y-63 y^{2}$

=$1(1-21 y)+3 y(1-21 y)$

=$(1-21 y)(1+3 y)$

(ii) $3 x^{2}-5 x y-12 y^{2}$

=$3 x^{2}-9 x y+4 x y-12 y^{2}$

=$3 x(x-3 y)+4 y(x-3 y)$

=$(x-3 y)(3 x+4 y)$

  

Solution 8

(i) $x^{2}-3 x y-40 y^{2}$

=$x^{2}-8 x y+5 x y-40 y^{2}$

=$x(x-8 y)+5 y(x-8 y)$

=$(x-8 y)(x+5 y)$

 

(ii) $10 p^{2} q^{2}-21 p q+9$

=$10 p^{2} q^{2}-15 p q-6 p q+9$

=$5 p q(2 p q-3)-3(2 p q-3)$

=$(2 p q-3)(5 p q-3)$

Solution 9

(i) $2 a^{2} b^{2}+a b-45$

=$2 a^{2} b^{2}+10 a b-9 a b-45$

=$2 a b(a b+5)-9(a b+5)$

=$(a b+5)(2 a b-9)$

(ii) $x(12 x+7)-10$

=$12 x^{2}+7 x-10$

=$12 x^{2}+15 x-8 x-10$

=$3 x(4 x+5)-2(4 x+5)$

=$(4 x+5)(3 x-2)$

Solution 10

(i) $(a+b)^{2}-11(a+b)-42$

=$(a+b)^{2}-14(a+b)+3(a+b)-42$

=$(a+b)(a+b-14)+3(a+b-14)$

=$(a+b-14)(a+b+3)$

(ii) $8+6(p+q)-5(p+q)^{2} \quad$

=$8+10(p+q)-4(p+q)-5(p+a)^{2}$

=$2(4+5(p+q))-(p+q)(4+5(p+a))$

=

$(4+5(p+q))(2-(p+q))$

Solution 11

(i) $(x-2 y)^{2}-6(x-2 y)+5-1 \times 5$

=$(x-2 y)^{2}-5(x-2 y)-(x-2 y)+5$

=$(x-2 y)(x-2 y-5)-1(x-2 y-5)$

=$(x-2 y-5)(x-2 y-1)$

(ii) $7+10(2 x-3 y)-8(2 x-3 y)^{2} \quad 0 \times 8=-56$

$7+14(2 x-3 y)-4(2 x-3 y)-8(2 x-3 y)^{2}$

$7(1+2(2 x-3 y))-4(2 x-3 y)(1+2(2 x-3 y))$

$(1+2(2 x-3 y))(7-4(2 x-3 y))$

$(1+4 x-6 y)(7-8 x+12 y)$

 Excercise 11.5

Solution 1

(i) $(35 x+28) \div(5 x+4)$

$\frac{35 x+28}{5 x+4}$

$⇒\frac{7(5 x+4)}{5 x+4}$

$⇒\quad 7$

(ii) $7 p^{2} q^{2}(9 r-21) \div 63 p q(r-3)$

⇒ $\frac{7 p^{2} q^{2}(9 r-27)}{63 p q(r-3)}$

⇒ $\frac{7 \cdot p^{2} q^{ 2} \cdot q(r- 3)}{63 pq(x - 3)}$

⇒ pq

Solution 2

(i) $6(2 x+7)(5 x-3) \div 3(5 x-3)$

⇒$\frac{6(2 x+7)(5 x-3)}{ 3(5 x-3)}$

⇒$2(2 x+7)$

⇒$\quad 4 x+14$

(ii) $33 p q(p+3)(2 q-5) \div 11 p(2 q-5)$

⇒$\frac{38 p q(p+3)(2 q-8)}{11-pq (2 q-5)}$

⇒3(p+3)

⇒3 p+9

Solution 3

(i) $\left(7 x^{3}-63 x\right) \div 7(x-3)$

⇒ $\frac{7 x^{3}-6 x}{7(x-3)}$

⇒$\frac{7 x\left(x^{2}-9\right)}{7(x-3)}$

⇒$\frac{x\left(x^{2}-3^{2}\right)}{x-3}$

⇒$\frac{x(x+3)(x-3)}{(x-3)}$

⇒ $x(x+3)$

⇒$x^{2}+3x$

(ii) $\left(3 p^{2}+17 p+10\right) \div(p+5)$

⇒$\frac{3 p^{2}+17 p+10}{p+5}$

⇒$\frac{3 p^{2}+15 p+2 p+10}{p+5}$

⇒$\frac{3 P(p+5)+2(p+5)}{p+5}$

⇒$\frac{(p+5)(3 p+2)}{(p+5)}$

⇒ $3 p+2$

(iii) $10 x y\left(14 y^{2}+43 y-21\right) \div 5 x(7 y-3)$

⇒ $\frac{10 y\left(14 y^{2}+43 y-21\right)}{5x(7 y-3)}$

⇒$\frac{2 y\left(14 y^{2}-6 y+49 y-21\right)}{7 y-3}$

⇒$\frac{2 y(2 y(7 y-3)+7(7 y-3))}{7 y-3}$

⇒$\frac{2 y(7 y-3)(2 y+7)}{(7 y-3)}$

⇒2 y(2 y+7)

(iv) $12 p q r\left(6 p^{2}-13 p q+6 q^{2}\right) \div 6 p q(2 p-3 q)$

⇒ $\frac{12pq r\left(6 p^{2}-13 p q+6 q^{2}\right)}{6 pq(2 p-3 q)}$

⇒ $\frac{2r\left(6 p^{2}-9 p q-4 p q+6 q^{2}\right)}{2 p-3 q}$

⇒ $\frac{2 r(3 p(2 p-3 q)-2 q(2 p-3 q))}{2 p-3 q}$

⇒ $\frac{2 r(2 p-3 q)(3 p-2 q)}{(2 p-3 q)}$

⇒ 2(3 p-2 q)

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