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=180∘y=180−60y=120∘
x = 60 (∵ opposite angles are equal in parallelogram)
z = 120
(ii) ABCD is a parallelogram
z=40∘(∵AB‖CD)
At 'O'
100+∠COD=180∘ (∵ Forms straight line)
∠COD=180−100∘
∠COD=80∘
In △COD
Z+∠COD+y=180∘40+80+y=180y+120=180∘y=180−120
y = 60
In △BOC
LACB=30∘(∵BC‖AD)∴x+∠ACB+∠BOC=180∘x+30+100=180x+130=180x=180−130x=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=180−155z=25∘
z=x(∵AB‖CD)
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=180−137x=43∘
z = 70 (∵ opposite angles are equal in parallelogram )
∠BCA=∠CAD(∵AD‖BC)
∠BCA=x
LBCA=43∘
At 'C'
∠BCA+y=180∘(∵ Forms a straight line )43+y=180y=180−43yy=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=5:7⇒xy=57⇒x=57⋅y
x+y+x+y=72 (∵opposite sides are equal in length)
2(x+y)=72
2(57⋅y+y)=72
127⋅y=36
y=36×712
y=21 cm
x=57⋅yx=57×21x=15 cm∴x=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(∵AD≠BC)
(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∘
(∵HE‖OP)
At 'o '
∠HOP+70∘=180∘(∵ forms straight line )
∠HOP=180∘−70∘
∠HOP=110∘
∴ $\angle HOP = 110' (∵Opposite angles are equal )
40+z+110∘=180′(∵ ∵Adjacent angles are supplementary )z+150=180z=180−150.z=30∘
At ' O'
x+60+80=180∘ (Forms straight line)
x+140=180∘x=180−140x=40∘
z=x=40∘(∵R0‖EP)
∠O+∠P=180∘ (∵Adjacent angles are supplementary)
x+60+y=180∘40+60+y=180y+100=180y=150−100y=80∘
Question 7
Sol :
Opposite sides are equal
5y−1=244x+2=285y=24+14x=28−25y=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=20−15y=5
Question 8
Sol :
In A B C D Parallelogram
∠A+∠B=180 (∵Adjacent angles are supplementary)
120+∠B=180
∠B=180−120
∠B=60∘
In PQRS parallelogram
∠P≐∠R(∵ opposite angles are equal)
∠P=50∘
In △PBX
∠P+∠B+x=180′ (- Sum of angles in triangle) 50+60+x=180110+x=180x=180−110x=70∘
Question 9
Sol :
(i) ∠CAD=?
∠CBD=∠ADB(∵AD‖BC)
∠ADB=46
In △ADO
∠CAD+∠ADB≠68=180
∠CAD+46+68=180
∠CAD+114=180
[∠CAD=66
(ii) ∠ACD=?
∠DOA+∠DOC=180∘(∵ Straight line )68+∠DOC=180∘∠DOC=180−68∠DOC=112∘
In △ DOC
∠CDO+∠DOC+∠ACD=180∘
112+30+∠ACD=180
∠ACD+142=180
∠ACD=38′
(iii) ∠ADC=∠ADO+∠BDC=46+30∠ADC=76′
Question 10
Sol :
¯AD=→BC (sides of ‖ gm )
∠AND=∠CPB=90∘
∠DAN=∠BCP(:BC‖AD)
From SAA Congruence
△AND≅△BPC
(ii) As ΔDN≅△BAP
∴¯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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