geometry excercise with solutions

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


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


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


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



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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


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


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


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


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


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



supplement = (180 )ox


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



complement = (90 )ox


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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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


(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


Let the measure of the required angle be ox .

Then the measure of its supplement = (180 )ox



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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


Let the measure of the required angle be ox .

Then the measure of its complement = (90 )o

x


(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


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


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


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


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


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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geometry excercise with solutions

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