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8/14/2019 Geometry Excercise with Solutions
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Geometry Exercise with Solutions
1. In figure 3-43, l is parallel to m and 180o x y z+ + = . The degree measure of
the angles , , x y z respectively is:
(a) 45 ,55 ,80o o o
(b) 50 ,45 ,85o o o
(c) 45 ,50 ,85o o o
(d) 35 ,60 ,85o o o
Sol: Correct option is (c)
Since lines l IS PARALLEL TO m and line l is the transversal, Therefore
a and 135o form a pair of corresponding angles.
Thus, 135oa =
Now, x and a form a linear pair.
Therefore 180oa x + =
or 135 180o ox+ =
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or 180 135o ox =
or 45ox = (1)
Further, 85o and z form a pair of vertically opposite angles.
Therefore 85o
z = (2)
It is given that 180o x y z + + =
Therefore 45 85 180o o o
y+ + =
or 180 45 85o o oy =
or 50oy = (3)
From equation (1), (2) and (3), we find that
45ox = , 50oy = and 85oz =
2. l and m are parallel lines intersected by a transversal n at P and Qrespectively. The values of angles ,x y and z respectively are:
(a) 43 ,137 ,137o o o
(b) 43 ,43 ,43o o o
(c) 137 ,43 ,137o o o
(d) 137 ,137 ,43o o o
Sol: Correct option is (b)
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Intersection of lines m and n at Q forms a linear pair.
Therefore 137 180o o
x+ =
or 180 137o ox =
or 43ox = (1)
Now, angles x and y form a pair of alternate interior angles as line l linem and n is the transversal.
Therefore x y =
or 43o y=
or 43oy = (2)
The intersection of lines l and n forms a pair of vertically opposite anglesy and z
Therefore z y =
or 43oz =
Hence 43ox = , 43oy = and 43oz =
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3. In figure 3-45, l is parallel to m. The values of the angles x andy respectively are
(a) 55o and 75o
(b) 75o and 55o
(c) 75o and 125o
(d) 55o and 125o
Sol: Correct option is (b)
Since l and m are parallel, p is the transversal, y and 125o form a pair of
consecutive interior angles.
So, 125 180o oy + =
or 180 125o oy =
or 55oy = (1)
Again Since l and m are parallel and n is the transversal.
Therefore 75o
and x form a pair of alternate interior angles.
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Hence, 75ox = (2)
From equation (1) and (2), we find that 75ox = and 55oy =
4. In figure 3-46, l and m are parallel, the measure of x is
(a) 120o
(b) 60o
(c)50o
(d) 70o
Sol: Correct option is (d)
Now, l m and n is the transversal.
Therefore Angles SPQ and RQW form a pair of corresponding angles.
Hence, SPQ RQW =
or 120oSPQ = (1)
Also, angle SPR and RPQ form a pair of adjacent angles at P.
120o
Therefore SPR RPQ + =
or 50 120o oSPR + =
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or 120 50o oSPR =
or 70oSPR = (2)
Again, angles SPR and PRQ form a pair of alternate angles.
Therefore SPR PRQ =
or 70o x=
or 70ox =
5. The supplement of an angle is96o, and then its complement is
a. 84o
b. 96o
c. 90o
d. 6o
Sol: Correct option is (d)
Out of two supplementary angles the required angle =
180o - given supplementary angles = 180 96 84o o o =
Required complement = 90o - given complement
=90 84 6o o o =
6. If a bicycle has 24 spokes, the angle between a pair of adjacent spoke is:
a. 24o
b. 20o
c. 15o
d.48
o
Sol: Correct option is (c)
Let each angle between two adjacent spoke is ox . As it has 24 spokes
24 360o otherefore x = (Sum of all adjacent angles about a point equals to 360o )
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36015
24
oo o
x = =
7. The measure of the reflex angle between the hands of a clock at 11-00hours is
a. 330o
b. 300o
c. 240o
d. 280o
Sol: Correct option is (a)
The dial of a clock is dived into 60 small divisions.
Each small division on the dial =360
660
oo
=
The acute angle formed at 11-00 hours has 5 small divisions between thehours hand and the minutes hand.
Therefore, The acute angle between their hands at 11-00 hours = 5 6 30o o =
Therefore, The reflex angle formed at 11-00 hours = 360 30 330o o o =
8. In figure 3-47, l and m are parallel. The degree measure of x is
a. 110o
b. 10o
c. 70o
d. 20o
Sol: Correct option is (c)
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8/14/2019 Geometry Excercise with Solutions
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DrawOC paralell to PB paralell to QAuuur
.
Since PB parallel to OCuuur uuur
and OP is the transversal, Therefore BPO and
POC are interior angles on the same side of the transversal OP.
Thus, 180o BPO POC + =
or 107 180o oPOC+ =
or 180 107o oPOC =
or 73oPOC =
Again QA parallel to OCuuur uuur
, Therefore Angles AQO and QOC form alternate
interior angles.
Therefore AQO QOC =
AQO QOP POC = + ( QOP and POC are adjacent angles)
or 108 73o oQOP= + ( 73oPOC = )
or 108 73 35o o oQOP = =
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10. In figure 3-49, ,PQ parallel to TU SP PQuuur uuur
and 40oS = . The degree
measure of STU is
a. 120o
b. 130o
c. 40o
d. 90o
Sol: Correct option is (b)
Through T drawTV parallel to SP .
sin ceSP parallel to TVand ST intersects them,
40oTherefore STV PST = =(Alternate angles)
Also, 90oSPR VWQ = = (Corresponding angles)
and 90oVWQ WTU = = (Corresponding angles)
Now, STU STV WTU = +
40 90 130o o oSTU = + =
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11. AB, CD and EF are three concurrent lines passing through the point O
such that OF bisects BOD . If 40oBOF = , then the degree measure of
AOD
is
a. 80o
b. 160o
c. 240o
d. 100o
Sol: Correct option is (d)
Since OF is the bisector of the angle BOD
2 2 40 80o oTherefore BOD BOF = = =
Now BOD and AOC from a pair of vertically opposite angles,
80o
Therefore AOC BOD = = ( 80oBOD =Q )
Now, therefore AOD BOC = (Vertically opposite angles)
Let o AOD BOC x = =
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360oTherefore BOD AOD AOC BOC + + + = (Sum of angles about a
point=360o )
or 80 80 360o o o o ox x+ + + =
or2 160 360
o o ox + =
or 2 360 160o o ox =
or 2 200o ox =
or 100o ox =
100oAOD =
12. In figure 3-51, AB parallel to DEuuur uuur
and 45oEDC = then the degree
measure of ABC BCD + is
a. 135o
b. 225o
c. 315o
d. 325o
Sol: Correct option is (b)
Draw CF parallel to BA parallel to DEuuur uuur uuur
Now FCD EDC = (Alternate angles)
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14. If (6 14)oy + and (4 4)oy be complementary angles, then the value of y is
a. 24o
b. 16o
c. 8o
d. 12o
Sol: Correct option is (c)
As (6 14)oy + and (4 4)oy are complementary angles
(6 14) (4 4) 90
6 14 4 4 90
10 10 90
10 90 10
10 80
8
+ + =
+ + =
+ =
=
=
=
o o o
o o o o o
o o o
o o o
o o
o o
Therefore y y
y y
y
y
y
y
15. An angle equals five times its supplement. Its measure is
a. 50o
b. 130o
c. 105o
d. 150o
Sol: Correct option is (d)
Let the measure of the required angle be ox . Then the measure of its
supplement = (180 )ox
According to the problem
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5(180 )o ox x=
900 5o o
x x=
0
5 900
6 900
900150
6
o o o
o o
o
x x
x
x
+ =
=
= =
16. The measure of an angle is half the measure of is supplement. Themeasure of the angle is
a. 115o
b. 60o
c. 150o
d. 100o
Sol: Correct option is (b)
Let the measure of the required angle be ox . Then the measure of its
supplement = (180 )ox
According to the problem
1(180 )
2
2 180
3 180
60
o o
o o o
o o
o o
x x
x x
x
x
=
=
=
=
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17. An angle is 5o less than1
4of its complement. Than its measure is
a. 14o
b. 18o
c. 17o
d. 20o
Sol: Correct option is (a)
Let the measure of the required angle be ox . Then the measure of its
complement = (90 )ox
According to the problem
1(90 ) 5
4
o o ox x=
15 (90 )
4
4 20 (90 )
4 90 20
5 70
14
o o o
o o o
o o o o
o o
o o
x x
x x
x x
x
x
+ =
+ =
+ =
=
=
18. The measure of an angle whose supplement is 4 times its complement is
a. 100o
b. 110o
c. 60o
d. 130o
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Sol: Correct option is (c)
Let the measure of the required angle be
ox
.
Then the measure of its complement = (90 )ox
and the measure of its supplement = (180 )ox
According to the problem
(180 ) 4(90 )
180 360 4
4 360 180
3 180
18060
3
o o
o o o o
o o o o
o o
oo o
x x
x x
x x
x
x
=
=
+ =
=
= =
19. The measure of an angle is twice the measure of its supplement. Themeasure of the angle is
a. 110o
b. 120o
c. 100o
d. 130o
Sol: Correct option is (b)
Let the measure of the required angle be ox .
Then the measure of its supplement = (180 )ox
According to the problem
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2(180 )
360 2
2 360
3 380
360120
3
o o
o o o
o o o
o o
oo o
x x
x x
x x
x
x
=
=
+ =
=
= =
20. An angle is 24o less than its complement. Then the measure of itssupplement is
a.147
o
b. 150o
c. 153o
d. 144o
Sol: Correct option is (a)
Let the measure of the required angle be ox .
Then the measure of its complement = (90 )o
x
According to the problem
(90 ) 24
90 24
90 24
2 66
33
o o o
o o o o
o o o o
o o
o o
x x
x x
x x
x
x
=
=
+ =
=
=
Therefore Supplement of ox is 180o ox
180 33 147o o o
= =
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21. In figure 3-52 OPuuur
bisects AOC and OQuuur
bisects BOC . The measure of
POQ is
a. 70o
b. 80o
c. 90o
d. 100o
Sol: Correct option is (c)
Since OPuuur
bisects AOC1
2Therefore POC AOC =
Also, OQ bisects BOC1
2Therefore COQ BOC =
1 1
2 2
1( )
2
+ = +
+ = +
Therefore POC COQ AOC BOC
POC COQ AOC BOC
Now 180o AOC BOC + = (Linear pair angles)
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1(180) 90
2
oPOC COQ + = =
22. In figure 3-53, the value of x for which AOB becomes a straight line is
a. 15o
b. 16o
c. 18o
d. 20o
Sol: Correct option is (d)
For AOB to be a straight line
AOC and BOC should form a linear pair.
180o
Therefore AOC BOC + =
or (4 20) 6 180o o ox x + =
4 20 6 180
10 180 20
10 200
20
o o o o
o o o
o o
o o
x x
x
x
x
+ =
= +
=
=
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23. In figure 3-54, AB parallel to CD parallel to EFand GH parallel to KL .
The magnitude of GHK is
a. 100o
b. 30o
c. 150o
d. 145o
Sol: Correct option is (c)
Draw HM parallel to KLuuuur
Now KMN CKL = (Corresponding angles)
or 65oKMN =
But 180oKMN KMH + = (Linear pair)
or 65 180o oKMH+ =
or 180 65o oKMH =
or 115oKMH =
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Now AHG KMH = (Corresponding angles)
or 115oAHG =
But GHK AHG AHK = +
or 115 35 150o o oGHK = + =
24. In the given figure 3-55, AB parallel to CDand EF parallel to DH. The
value of GDH is
a. 60o
b. 120o
c. 78o
d. 36o
Sol: Correct option is (a)
As, AB parallel to CD and EG is the transversal.
= Therefore AED CDG (Corresponding angles)
or 42oAED =
Now, 180o AED DEF FEB + + = (AEB is a straight line)
or 42 78 180o o oDEF+ + =
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or 120 180o oDEF + =
or 180 120o oDEF =
or 60oDEF =
Again, DH parallel to EFand EFG is the transversal.
= Therefore GDH DEF (Corresponding angles)
or 60oGDH =
25. In figure 3-56, the value of o o o o x y z u+ + + is
a. 135o
b. 270o
c. 305o
d. 225o
Sol: Correct option is (d)
Since all the angles form a complete angle about point O
90 45 360o o o o o o o x y z u+ + + + + =
or 135 360o o o o o o x y z u+ + + + =
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or 360 135o o o o o o x y z u+ + + =
or 225o o o o o x y z u+ + + =
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