ML AGGARWAL CLASS 8 Chapter 13 Understanding Quadrilaterals Exercise 13.2

 Exercise 13.2

Question 1

Sol :
(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 Supplementary)

(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

Sol :

(i) Given parallelogram 

Let the unknown angle = a







∴ 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 parallelogram)

z = 120


(ii) ABCD is a parallelogram

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








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 parallelogram 

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








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

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

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

$x=25^{\circ}$

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


(iv) ABCD is a parallelogram 

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










(∵Adjacent angles are supplementary)

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

z = 70 (∵ opposite angles are equal in parallelogram )

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

$\angle B C A=x$

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

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

Sol :
Let x , y are length of adjacent sides of parallelogram

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 parallelogram

Question 4

Sol :
Given

Angles of parallelogram are in the ratio of 4: 5

Let the Angle be 4 x, 5 x

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

Angles   $4 x=4 \times 20=80^{\circ}$

   $5 x=5 \times 20=100^{\circ}$

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

(∵Opposite are equal in a parallelogram)


Question 5

Sol :
(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 parallelogram


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

Yes (∵Adjacent angles are supplementary )


Question 6

Sol :







$y=40^{\circ}$

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

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 )

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


 
(ii) 








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 supplementary)

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

Question 7

Sol :








Opposite sides are equal 

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

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










'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

Sol :
In A B C D Parallelogram

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

$120+\angle B=180$

$\angle B=180-120$

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

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

Sol :
(i) $\angle C A D=?$







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

$\angle A D B=46$

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

Sol :
(i) 






$\overline{A D}=\overrightarrow{B C}$ (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

Sol :
In parallelogram ABCR

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

In parallelogram 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) $+(\mathbb{4})+(-5)$

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

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