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) DCB=60(DCB)CBA are Supplementary)

(iv) ADC=120( opposite angles are equal)

(v) DAB=60 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)

60+y=180y=18060y=120

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

z = 120


(ii) ABCD is a parallelogram

z=40(ABCD)








At 'O'

100+COD=180 (∵ Forms straight line)

COD=180100

COD=80

In COD

Z+COD+y=18040+80+y=180y+120=180y=180120

y = 60


In BOC

LACB=30(BCAD)x+ACB+BOC=180x+30+100=180x+130=180x=180130x=50

x=50,y=60,z=40


(iii) ABCD is a parallelogram 

Y= 120 (∵Opposite angles are equal in parallelogram)








Z+35+120=180 (∵Adjacent angles are supplementary)

z+155=180z=180155z=25

z=x(ABCD)

x=25

x=25,y=120;z=25


(iv) ABCD is a parallelogram 

A+D=180










(∵Adjacent angles are supplementary)

67+x+y0=180x+137=180x=180137x=43

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

BCA=CAD(ADBC)

BCA=x

LBCA=43

At 'C'

BCA+y=180( Forms a straight line )43+y=180y=18043yy=137

x=43,y=137,z=70

Question 3

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

Given
Perimeter =72 cm

x:y=5:7xy=57x=57y

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

2(x+y)=72

2(57y+y)=72

127y=36

y=36×712

y=21 cm
 
x=57yx=57×21x=15 cmx=15 cm,y=21 cm.

ஃ 15 cm,21 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

4x+5x=180( Adjacent angles are Supplementary ]9x=180x=20

Angles   4x=4×20=80

   5x=5×20=100

∴  Four angles of parallelogram = 80,100,80,100

(∵Opposite are equal in a parallelogram)


Question 5

Sol :
(i) A+C=180?

may (Or) may not be

∵( A=C=90 )


(ii) AD=BC=6cm,AB=5cm,DC=4.5CmNo(ADBC)

 
(iii) B=80,D=70 ?

No, opposite angles must be equal in parallelogram


(iv) B+C=180?

Yes (∵Adjacent angles are supplementary )


Question 6

Sol :







y=40

(HEOP)

At 'o '

HOP+70=180( forms straight line )

HOP=18070

HOP=110

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

40+z+110=180( ∵Adjacent angles are supplementary )z+150=180z=180150.z=30


 
(ii) 








At ' O'

x+60+80=180 (Forms straight line)

x+140=180x=180140x=40

z=x=40(R0EP)


O+P=180  (∵Adjacent angles are supplementary)

x+60+y=18040+60+y=180y+100=180y=150100y=80

Question 7

Sol :








Opposite sides are equal 

5y1=244x+2=285y=24+14x=2825y=254x=26y=5]x=264=132

x=6.5,y=5










'O' Bisects the ¯BR

¯BO=¯ORx+y=20(1)

'o' Bisects the ¯Nv

¯NO=¯OV

x+3=18----------②

x=18-3

x=15

Substituted x value in ①

15+y=20y=2015y=5

Question 8

Sol :
In A B C D Parallelogram

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

120+B=180

B=180120

B=60

In PQRS parallelogram 

PR( opposite angles are equal)

P=50

In PBX

P+B+x=180 (- Sum of angles in triangle) 50+60+x=180110+x=180x=180110x=70

Question 9

Sol :
(i) CAD=?







CBD=ADB(ADBC)

ADB=46

In ADO

CAD+ADB68=180

CAD+46+68=180

CAD+114=180

[CAD=66


(ii) ACD=?

DOA+DOC=180( Straight line )68+DOC=180DOC=18068DOC=112

In  DOC

CDO+DOC+ACD=180

112+30+ACD=180

ACD+142=180

ACD=38


(iii) ADC=ADO+BDC=46+30ADC=76

Question 10

Sol :
(i) 






¯AD=BC (sides of gm )

AND=CPB=90

DAN=BCP(:BCAD)

From SAA Congruence 

ANDBPC


(ii) As ΔDNBAP

¯AN=¯CP

Question 11

Sol :
In parallelogram ABCR

AB=¯RC(: opposite sides )(1)

In parallelogram ABPC

¯AB=¯CP ------② (∵ opposite sides )

(1) + (2)

¯AB+AˉB=¯RC+¯CP2¯AB=¯RP(3)2¯AC=¯PO(4)2¯BC=QˉR(5)

(3) +(4)+(5)

2(AˉB+AˉC+¯BC)=¯PO+QK+RP

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