Geometry Excercise With Solutions

Geometry Exercise with Solutions 1. In figure 3-43, l is parallel to m and x + y + z = 180o . The degree measure of the angles x, y, z respectively is: (a) 45o ,55o ,80o (b) 50o , 45o ,85o (c) 45o ,50o ,85o (d) 35o , 60o ,85o 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, ∠a = ∠135o Now, ∠x and ∠a form a linear pair.

Therefore ∠a + ∠x = 180o or 135o + ∠x = 180o

or ∠x = 180o − 135o or ∠x = 45o

… (1)

Further, ∠85o and ∠z form a pair of vertically opposite angles.

Therefore ∠z = 85o

… (2)

It is given that ∠x + ∠y + ∠z = 180o

Therefore 45o + ∠y + 85o = 180o or ∠y = 180o − 45o − 85o or ∠y = 50o

… (3)

From equation (1), (2) and (3), we find that

x = 45o , y = 50o and z = 85o 2. l and m are parallel lines intersected by a transversal n at P and Q respectively. The values of angles x, y and z respectively are: (a) 43o ,137 o ,137 o (b) 43o , 43o , 43o (c) 137 o , 43o ,137 o (d) 137 o ,137 o , 43o Sol:

Correct option is (b)

Intersection of lines m and n at Q forms a linear pair.

Therefore 137o + x = 180o or x = 180o − 137o or ∠x = 43o

… (1)

Now, angles ∠x and ∠y form a pair of alternate interior angles as line l line m and n is the transversal.

Therefore ∠x = ∠y or 43o = ∠y or ∠y = 43o

… (2)

The intersection of lines l and n forms a pair of vertically opposite angles ∠y and ∠z

Therefore ∠z = ∠y or ∠z = 43o Hence x = 43o , ∠y = 43o and ∠z = 43o

3. In figure 3-45, l is parallel to m. The values of the angles ∠x and ∠y 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, ∠y + 125o = 180o or ∠y = 180o − 125o or ∠y = 55o

… (1)

Again Since l and m are parallel and n is the transversal.

Therefore ∠75o and ∠x form a pair of alternate interior angles.

Hence, ∠x = 75o

… (2)

From equation (1) and (2), we find that ∠x = 75o and ∠y = 55o

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 ∠SPQ = 120o

… (1)

Also, angle ∠SPR and ∠RPQ form a pair of adjacent angles at P.

Therefore ∠SPR + ∠RPQ = 120o or ∠SPR + 50o = 120o

or ∠SPR = 120o − 50o or ∠SPR = 70o

… (2)

Again, angles ∠SPR and ∠PRQ form a pair of alternate angles.

Therefore ∠SPR = ∠PRQ or 70o = ∠x or x = 70o 5. The supplement of an angle is 96o , 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 = 180o − 96o = 84o Required complement = 90o - given complement = 90o − 84o = 6o 6. If a bicycle has 24 spokes, the angle between a pair of adjacent spoke is: a. 24o b. 20o c. 15o d. 48o Sol:

Correct option is (c)

Let each angle between two adjacent spoke is x o . As it has 24 spokes

therefore 24 × x o = 360o (Sum of all adjacent angles about a point equals to 360o )

xo =

360o = 15o 24

7. The measure of the reflex angle between the hands of a clock at 11-00 hours 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 =

360o = 6o 60

The acute angle formed at 11-00 hours has 5 small divisions between the hours hand and the minute’s hand. Therefore, The acute angle between their hands at 11-00 hours = 5 × 6o = 30o Therefore, The reflex angle formed at 11-00 hours = 360o − 30o = 330o 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)

Draw a line n parallel to both lines l and m . Since line l line n and PO is the transversal. Therefore Angles ∠APO and ∠ROP form a pair of alternate interior angles. Hence ∠ROP = 40o . Also angles ∠CQO and ∠ROQ form a pair of alternate interior angles.

∠ROQ = 30o Now angles ∠POR and ∠ROQ form a pair of adjacent angles.

Therefore ∠POQ = ∠ROP + ∠ROQ or Therefore ∠x = 40o + 30o = 70o

uuur uuur 9. In figure 3-48, QA parallel to PB . The degree measure of ∠QOP is a. 35o b. 72o c. 73o d. 1o Sol:

Correct option is (a)

uuur Draw OC paralell to PB paralell to QA . uuur uuur Since PB parallel to OC and OP is the transversal, Therefore ∠BPO and ∠POC are interior angles on the same side of the transversal OP. Thus, ∠BPO + ∠POC = 180o or 107 o + ∠POC = 180o or ∠POC = 180o − 107o or ∠POC = 73o

uuur uuur Again QA parallel to OC , Therefore Angles ∠AQO and ∠QOC form alternate interior angles. Therefore ∠AQO = ∠QOC

∠AQO = ∠QOP + ∠POC ( ∠QOP and ∠POC are adjacent angles) or 108o = ∠QOP + 73o ( ∠POC = 73o ) or ∠QOP = 108o − 73o = 35o

uuur uuur 10. In figure 3-49, PQ parallel to TU , SP ⊥ PQ and ∠S = 40o . The degree measure of ∠STU is a. 120o b. 130o c. 40o d. 90o Sol:

Correct option is (b)

Through T draw TV parallel to SP .

sin ce SP parallel to TV and ST intersects them, Therefore ∠STV = ∠PST = 40o (Alternate angles) Also, ∠SPR = ∠VWQ = 90o (Corresponding angles) and ∠VWQ = ∠WTU = 90o (Corresponding angles) Now, ∠STU = ∠STV + ∠WTU

∠STU = 40o + 90o = 130o

11. AB, CD and EF are three concurrent lines passing through the point O such that OF bisects ∠BOD . If ∠BOF = 40o , 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

Therefore ∠BOD = 2 × ∠BOF = 2 × 40o = 80o Now ∠BOD and ∠AOC from a pair of vertically opposite angles,

Therefore ∠AOC = ∠BOD = 80o (Q ∠BOD = 80o ) Now, therefore ∠AOD = ∠BOC (Vertically opposite angles) Let ∠AOD = ∠BOC = x o

Therefore ∠BOD + ∠AOD + ∠AOC + ∠BOC = 360o (Sum point= 360o )

of

angles

about

a

or 80o + x o + 80o + x o = 360o or 2 x o + 160o = 360o or 2 x o = 360o − 160o or 2 x o = 200o or x o = 100o

∠AOD = 100o uuur uuur 12. In figure 3-51, AB parallel to DE and ∠EDC = 45o then the degree measure of ∠ABC + ∠BCD is a. 135o b. 225o c. 315o d. 325o Sol:

Correct option is (b)

uuur uuur uuur Draw CF parallel to BA parallel to DE Now ∠FCD = ∠EDC (Alternate angles)

Therefore ∠FCD = 45o uuur uuur Since CF parallel to BA and BC is the transversal, Therefore ∠FCB + ∠CBA = 180o (Interior angles on the same side of the transversal are supplementary) Now, ∠FCB + ∠ABC + ∠FCD = 180o + 45o or (∠FCB + ∠FCD) + ∠ABC = 225o (Associative property of addition) or ∠BCD + ∠ABC = 225o or ∠ABC + ∠BCD = 225o 13. An angle is equal to four times its complement. Its magnitude is a. 72o b. 40o c. 27o d. 18o Sol:

Correct option is (a)

Let the magnitude of the angle is x o . Then its complement is (90 − x)o . According to the given condition x = 4(90 − x) x = 360 − 4 x 5 x = 360 x = 72o

14. If (6 y + 14)o and (4 y − 4)o be complementary angles, then the value of y is a. 24o b. 16o c. 8o d. 12o Sol:

Correct option is (c)

As (6 y + 14)o and (4 y − 4)o are complementary angles Therefore (6 y + 14)o + (4 y − 4)o = 90o 6 y o + 14o + 4 y o − 4o = 90o 10 y o + 10o = 90o 10 y o = 90o − 10o 10 y o = 80o y o = 8o

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 x o . Then the measure of its supplement = (180 − x)o According to the problem

x o = 5(180 − x)o x o = 900 − 5 x o

x o + 5 x o = 900o 6 x o = 900o xo =

900 = 1500 6

16. The measure of an angle is half the measure of is supplement. The measure 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 x o . Then the measure of its supplement = (180 − x)o According to the problem

1 x o = (180 − x)o 2 2 x o = 180o − x o 3 x o = 180o x o = 60o

17. An angle is 5o less than

1 of its complement. Than its measure is 4

a. 14o b. 18o c. 17 o d. 20o Sol:

Correct option is (a)

Let the measure of the required angle be x o . Then the measure of its complement = (90 − x)o According to the problem

1 x o = (90 − x)o − 5o 4

x o + 5o =

1 (90 − x)o 4

4 x o + 20o = (90 − x)o 4 x o + x o = 90o − 20o 5 x o = 70o x o = 14o

18. The measure of an angle whose supplement is 4 times its complement is a. 100o b. 110o c. 60o d. 130o

Sol:

Correct option is (c)

Let the measure of the required angle be x o . Then the measure of its complement = (90 − x)o and the measure of its supplement = (180 − x)o According to the problem

(180 − x)o = 4(90 − x)o 180o − x o = 360o − 4 x o 4 x o + x o = 360o − 180o 3x o = 180o

xo =

180o = 60o 3

19. The measure of an angle is twice the measure of its supplement. The measure 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 x o . Then the measure of its supplement = (180 − x)o According to the problem

x o = 2(180 − x)o x o = 360o − 2 x o x o + 2 x o = 360o 3 x o = 380o

xo =

360o = 120o 3

20. An angle is 24o less than its complement. Then the measure of its supplement is a. 147o b. 150o c. 153o d. 144o Sol:

Correct option is (a)

Let the measure of the required angle be x o . Then the measure of its complement = (90 − x)o According to the problem

x o = (90 − x)o − 24o x o = 90o − x o − 24o x o + x o = 90o − 24o 2 x o = 66o x o = 33o

Therefore Supplement of x o is 180o − x o = 180o − 33o = 147o

uuur uuur 21. In figure 3-52 OP bisects ∠AOC and OQ bisects ∠BOC . The measure of ∠POQ is a. 70o b. 80o c. 90o d. 100o Sol:

Correct option is (c)

uuur 1 Since OP bisects ∠AOC Therefore ∠POC = ∠AOC 2

1 Also, OQ bisects ∠BOC Therefore ∠COQ = ∠BOC 2 Therefore ∠POC + ∠COQ =

1 1 ∠AOC + ∠BOC 2 2

1 ∠POC + ∠COQ = (∠AOC + ∠BOC ) 2

Now ∠AOC + ∠BOC = 180o (Linear pair angles)

∠POC + ∠COQ =

1 (180) = 90o 2

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.

Therefore ∠AOC + ∠BOC = 180o

or (4 x − 20)o + 6 x o = 180o 4 x o − 20o + 6 x o = 180o 10 x o = 180o + 20o 10 x o = 200o x o = 20o

23. In figure 3-54, AB parallel to CD parallel to EF and GH parallel to KL . The magnitude of ∠GHK is a. 100o b. 30o c. 150o d. 145o Sol:

Correct option is (c)

uuuur Draw HM parallel to KL Now ∠KMN = ∠CKL (Corresponding angles) or ∠KMN = 65o But ∠KMN + ∠KMH = 180o (Linear pair) or 65o + ∠KMH = 180o or ∠KMH = 180o − 65o or ∠KMH = 115o

Now ∠AHG = ∠KMH (Corresponding angles) or ∠AHG = 115o But ∠GHK = ∠AHG + ∠AHK or ∠GHK = 115o + 35o = 150o 24. In the given figure 3-55, AB parallel to CD and 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 ∠AED = 42o Now, ∠AED + ∠DEF + ∠FEB = 180o (AEB is a straight line) or 42o + ∠DEF + 78o = 180o

or ∠DEF + 120o = 180o or ∠DEF = 180o − 120o or ∠DEF = 60o Again, DH parallel to EF and EFG is the transversal. Therefore ∠GDH = ∠DEF (Corresponding angles)

or ∠GDH = 60o 25. In figure 3-56, the value of x o + y o + z o + u o 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 90o + x o + y o + z o + u o + 45o = 360o

or x o + y o + z o + u o + 135o = 360o

or x o + y o + z o + u o = 360o − 135o or x o + y o + z o + u o = 225o

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