Class 8 Exam > Class 8 Notes > ML Aggarwal: Understanding Quadrilaterals - 2
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Page 1
1. In the given figure, ABCD is a parallelogram. Complete each
statement along with the definition or property used.
(i) AD = ………..
(ii) DC = ………..
(iii) ?DCB = ………..
(iv) ?ADC = ………..
(v) ?DAB = ………..
(vi) OC = ………..
(vii) OB = ………..
(viii) m ?DAB + m ?CDA = ………..
Solution:-
From the given figure,
(i) AD = 6 cm … [because opposite sides of parallelogram are equal]
(ii) DC = 9 cm … [because opposite sides of parallelogram are equal]
(iii) ?DCB = 60
o
(iv) ?ADC = ?ABC = 120
o
(v) ?DAB = ?DCB = 60
o
(vi) OC = AO = 7 cm
Page 2
1. In the given figure, ABCD is a parallelogram. Complete each
statement along with the definition or property used.
(i) AD = ………..
(ii) DC = ………..
(iii) ?DCB = ………..
(iv) ?ADC = ………..
(v) ?DAB = ………..
(vi) OC = ………..
(vii) OB = ………..
(viii) m ?DAB + m ?CDA = ………..
Solution:-
From the given figure,
(i) AD = 6 cm … [because opposite sides of parallelogram are equal]
(ii) DC = 9 cm … [because opposite sides of parallelogram are equal]
(iii) ?DCB = 60
o
(iv) ?ADC = ?ABC = 120
o
(v) ?DAB = ?DCB = 60
o
(vi) OC = AO = 7 cm
(vii) OB = OD = 5 cm
(viii) m ?DAB + m ?CDA = 180
o
2. Consider the following parallelograms. Find the values of x, y, z in
each.
(i)
Solution:-
(i) Consider parallelogram MNOP
From the figure, ?POQ = 120
o
We know that, sum of angles linear pair is equal to 180
o
So, ?POQ + ?PON = 180
o
120
o
+ ?PON = 180
o
?PON = 180
o
– 120
o
Page 3
1. In the given figure, ABCD is a parallelogram. Complete each
statement along with the definition or property used.
(i) AD = ………..
(ii) DC = ………..
(iii) ?DCB = ………..
(iv) ?ADC = ………..
(v) ?DAB = ………..
(vi) OC = ………..
(vii) OB = ………..
(viii) m ?DAB + m ?CDA = ………..
Solution:-
From the given figure,
(i) AD = 6 cm … [because opposite sides of parallelogram are equal]
(ii) DC = 9 cm … [because opposite sides of parallelogram are equal]
(iii) ?DCB = 60
o
(iv) ?ADC = ?ABC = 120
o
(v) ?DAB = ?DCB = 60
o
(vi) OC = AO = 7 cm
(vii) OB = OD = 5 cm
(viii) m ?DAB + m ?CDA = 180
o
2. Consider the following parallelograms. Find the values of x, y, z in
each.
(i)
Solution:-
(i) Consider parallelogram MNOP
From the figure, ?POQ = 120
o
We know that, sum of angles linear pair is equal to 180
o
So, ?POQ + ?PON = 180
o
120
o
+ ?PON = 180
o
?PON = 180
o
– 120
o
?PON = 60
o
Then, ?M = ?O = 60
o
… [Because opposite angles of parallelogram are equal]
?POQ = ?MNO
120
o
= 120
o
… [because corresponding angels are equal]
Hence, y = 120
o
Also, z = y
120
o
= 120
o
Therefore, x = 60
o
, y = 120
o
and z = 120
o
(ii)
Solution:-
Page 4
1. In the given figure, ABCD is a parallelogram. Complete each
statement along with the definition or property used.
(i) AD = ………..
(ii) DC = ………..
(iii) ?DCB = ………..
(iv) ?ADC = ………..
(v) ?DAB = ………..
(vi) OC = ………..
(vii) OB = ………..
(viii) m ?DAB + m ?CDA = ………..
Solution:-
From the given figure,
(i) AD = 6 cm … [because opposite sides of parallelogram are equal]
(ii) DC = 9 cm … [because opposite sides of parallelogram are equal]
(iii) ?DCB = 60
o
(iv) ?ADC = ?ABC = 120
o
(v) ?DAB = ?DCB = 60
o
(vi) OC = AO = 7 cm
(vii) OB = OD = 5 cm
(viii) m ?DAB + m ?CDA = 180
o
2. Consider the following parallelograms. Find the values of x, y, z in
each.
(i)
Solution:-
(i) Consider parallelogram MNOP
From the figure, ?POQ = 120
o
We know that, sum of angles linear pair is equal to 180
o
So, ?POQ + ?PON = 180
o
120
o
+ ?PON = 180
o
?PON = 180
o
– 120
o
?PON = 60
o
Then, ?M = ?O = 60
o
… [Because opposite angles of parallelogram are equal]
?POQ = ?MNO
120
o
= 120
o
… [because corresponding angels are equal]
Hence, y = 120
o
Also, z = y
120
o
= 120
o
Therefore, x = 60
o
, y = 120
o
and z = 120
o
(ii)
Solution:-
From the figure, it is given that ?PQO = 100
o
, ?OMN = 30
o
, ?PMO =40
o
.
Then, ?NOM = ?OMP … [because alternate angles are equal]
So, z = 40
o
Now, ?NMO = ?POM … [because alternate angles are equal]
So, ?NMO = a = 30
o
Consider the triangle PQO,
We know that, sum of measures of interior angles of triangle is equal to
180
o
.
?P + ?Q + ?O = 180
o
x + 100
o
+ 30
o
= 180
o
x + 130
o
= 180
o
x = 180
o
– 130
o
x = 50
o
Then, exterior angle ?OQP = y + z
100
o
= y + 40
o
By transposing we get,
y = 100
o
– 40
o
y = 60
o
Therefore, the value of x = 50
o
, y = 60
o
and z = 40
o
.
(iii)
Page 5
1. In the given figure, ABCD is a parallelogram. Complete each
statement along with the definition or property used.
(i) AD = ………..
(ii) DC = ………..
(iii) ?DCB = ………..
(iv) ?ADC = ………..
(v) ?DAB = ………..
(vi) OC = ………..
(vii) OB = ………..
(viii) m ?DAB + m ?CDA = ………..
Solution:-
From the given figure,
(i) AD = 6 cm … [because opposite sides of parallelogram are equal]
(ii) DC = 9 cm … [because opposite sides of parallelogram are equal]
(iii) ?DCB = 60
o
(iv) ?ADC = ?ABC = 120
o
(v) ?DAB = ?DCB = 60
o
(vi) OC = AO = 7 cm
(vii) OB = OD = 5 cm
(viii) m ?DAB + m ?CDA = 180
o
2. Consider the following parallelograms. Find the values of x, y, z in
each.
(i)
Solution:-
(i) Consider parallelogram MNOP
From the figure, ?POQ = 120
o
We know that, sum of angles linear pair is equal to 180
o
So, ?POQ + ?PON = 180
o
120
o
+ ?PON = 180
o
?PON = 180
o
– 120
o
?PON = 60
o
Then, ?M = ?O = 60
o
… [Because opposite angles of parallelogram are equal]
?POQ = ?MNO
120
o
= 120
o
… [because corresponding angels are equal]
Hence, y = 120
o
Also, z = y
120
o
= 120
o
Therefore, x = 60
o
, y = 120
o
and z = 120
o
(ii)
Solution:-
From the figure, it is given that ?PQO = 100
o
, ?OMN = 30
o
, ?PMO =40
o
.
Then, ?NOM = ?OMP … [because alternate angles are equal]
So, z = 40
o
Now, ?NMO = ?POM … [because alternate angles are equal]
So, ?NMO = a = 30
o
Consider the triangle PQO,
We know that, sum of measures of interior angles of triangle is equal to
180
o
.
?P + ?Q + ?O = 180
o
x + 100
o
+ 30
o
= 180
o
x + 130
o
= 180
o
x = 180
o
– 130
o
x = 50
o
Then, exterior angle ?OQP = y + z
100
o
= y + 40
o
By transposing we get,
y = 100
o
– 40
o
y = 60
o
Therefore, the value of x = 50
o
, y = 60
o
and z = 40
o
.
(iii)
Solution:-
From the above figure,
?SPR = ?PRQ
35
o
= 35
o
… [because alternate angles are equal]
Now consider the triangle PQR,
We know that, sum of measures of interior angles of triangle is equal to
180
o
.
?RPQ + ?PQR + ?PRQ = 180
o
z + 120
o
+ 35
o
= 180
o
z + 155
o
= 180
o
z = 180
o
– 155
o
z = 25
o
Then, ?QPR = ?PRQ
Z = x
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