Lines And Angles Test

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Lines And Angles Test - 7

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Question 1
1 / -0

In the given figure, find the value of ∠UOP if y is one-third of z.

A
30°

B
40°

C
50°

D
60°

Solution

Given: y =
z = 3y
In the given figure,
∠SOR = ∠UOP (Vertically opposite angles)
∠TOU + ∠UOP + ∠POQ = 180° (Sum of angles on the straight line = 180°)
z + 2y + y = 180°
3y + 2y + y = 180°
6y = 180°
y = 30°
∠UOP = 2y = 2 × 30° = 60°
Therefore, ∠UOP = 60°
Question 2
1 / -0

The difference between the first and second angles of a triangle is 30°. The first angle measures 20° less than the sum of the other two. What kind of triangle is this?

A
Scalene triangle

B
Obtuse-angled triangle

C
Isosceles triangle

D
Equilateral triangle

Solution

Let the first, second and third angles of the triangle be x, y and z, respectively,
According to the given instructions,
x - y = 30° ... (i)
x = (y + z) - 20°
x - y = z - 20°
30° = z - 20° (From (i))
z = 50°
According to the angle sum property of triangle,
x + y + z = 180°
x + y + 50° = 180°
x + y = 130° ... (ii)
Solving (i) and (ii),
2x = 160°
x = 80°
Putting this value in (i),
80° - y = 30°
y = 80° - 30° = 50°
So, x = 80°, y = 50°, z = 50°
Since two angles of the triangle are equal, this is an isosceles triangle.
Question 3
1 / -0

In the given figure, MN∥PR. Find the measure of ∠DOB if ∠BCO = ∠OCA, ∠BAM = 115° and ∠CBR = 60°.

A
65°

B
75°

C
85°

D
95°

Solution

MN∥PR and AB is a transversal.
∠BAM = 115° = ∠ABR (Alternate interior angles)
∠ABR = 115°
∠ABC + ∠CBR = 115°
∠ABC + 60° = 115°
∠ABC = 55°
Now, MN∥PR and BC is a transversal.
∠CBR = ∠BCA = 60° (Alternate interior angles)
∠BCA = 60°
∠OCB + ∠OCA = 60°
2∠OCB = 60° (∠OCB = ∠OCA (given))
∠OCB = 30°
Now, ∠DOB = ∠BCO + ∠OBC (As exterior angle of a triangle is equal to the sum of the opposite interior angles)
∠DOB = 30° + 55° = 85°
Question 4
1 / -0

In the given figure, PQ ∥ RS, RS ∥ TU, and lines m and n are parallel. Find the value of angle d if the ratio of angle b to angle c is 2 : 3.

A
108°

B
110°

C
112°

D
114°

Solution

It is given that PQ ∥ RS and RS ∥ TU.
So, PQ ∥ RS ∥ TU
a + b = 180° ... (i) (As sum of co-interior angles is 180°)
b + c = 180° ... (ii) (∠SNO = b as they are vertically opposite angles)
Also, it is given that b : c = 2 : 3.

Substituting in (ii),
= 180°

c = 108°

Substituting in (ii),
b + 108° = 180°
b = 180° - 108°
b = 72°
Substituting in (i),
a + b = 180°
a + 72° = 180°
a = 180° - 72°
a = 108°
a = ∠BAM (As m∥n and PQ is a transversal, ∠a and ∠BAM are corresponding angles)
Now, ∠BAM = d (Vertically opposite angles)
So, a = d = 108°
Question 5
1 / -0

Find the supplementary angle of the angle whose measure is 15^o more than the double of its supplementary angle.

A
55°

B
25°

C
35°

D
65°

Solution

Let the measure of the angle be x.
According to the given instructions,
x = 15°+ 2(180° - x) [∵ (180° - x) is the supplementary angle of x]
x = 15° + 360° - 2x
x = -2x + 375°
3x = 375°
x = 125°
Supplementary angle of x = 180°- x = 180° - 125° = 55°
Question 6
1 / -0

What is the value of p in the given figure if AB || DE, ∠ABC = 120° and ∠CDE = 140°?

A
50°

B
80°

C
90°

D
120°

Solution

Given: AB || DE, ∠ABC = 120° and ∠CDE = 140°
If we draw a line parallel to DE through point C, then

Now, ∠XCD + ∠CDE = 180° (As sum of co-interior angles is 180°)
∠XCD + 140° = 180°
∠XCD = 180 ° - 140°
∠XCD = 40°

Similarly, ∠YCB + ∠ABC = 180° (As sum of co-interior angles is 180°)
∠YCB + 120° = 180°
∠YCB = 180° - 120°
∠YCB = 60°

Now,
∠XCD + p + ∠YCB = 180°
p = 180° - 60° - 40°
p = 80°
Question 7
1 / -0

In the figure, if || m, p || q, DAC = 30° and ABC = 70°, then what is the measure of ACD?

A
70°

B
80°

C
110°

D
150°

Solution

According to the data given in the question, || m and AC is a transversal.
∴ DAC = ACB [Alternate interior angles]
ACB = 30°
Also, q is a transversal.
∴ ABC + BCD = 180° [Co-interior angles]
70° + BCD = 180°
BCD = 110°
Now, ACB + ACD = BCD
30° + ACD = 110°
ACD = 110° - 30° = 80°
Hence, 80° is the correct answer.
Question 8
1 / -0

In the following figure, ∠OPQ = 30°, ∠PQO = 60°, and ∠QOS = 35°. Find the double of ∠TUO if line RS and line TU are parallel to each other.

A
48°

B
110°

C
140°

D
150°

Solution

Given ∠OPQ = 30°,
∠PQO = 60°
and ∠QOS = 35°

In triangle PQO,
∠OPQ + ∠PQO + ∠POQ = 180°
30° + 60° + ∠POQ = 180°
∠POQ = 180° - 90°

∠POQ = 90°
Now,
∠POQ = ∠TOU (Vertically opposite angles)
So,
∠TOU = 90°
Since line RS and line TU are parallel,
∠QOS = ∠UTO = 35° (Corresponding angles)
In triangle TUO,
∠UTO + ∠TOU + ∠OUT = 180°
35° + 90° + ∠OUT = 180°
∠OUT = 180° - 125°
∠OUT = 55°
So,
∠TUO = 55°
Double of ∠TUO = 2 × 55° = 110°
Question 9
1 / -0

Find the value of x.

A
5

B
6

C
7

D
8

Solution

5x + 22 + 2x - 10 = 5x + 28 (As exterior angle of a triangle is equal to the sum of opposite interior angles)
7x + 12 = 5x + 28
7x - 5x = 28 - 12
2x = 16
x = 8
Question 10
1 / -0

In the figure, if RS and PQ are the bisectors of ASP and SPD respectively and AB || CD, then which of the following is false?

A
ASP = SPD

B
RS || PQ

C
RSP = SPQ

D
RSP + SPC = 180°

Solution

AB || CD and m is a transversal.
ASP = SPD (Alternate interior angles)
Hence, ASP = SPD
RSP = SPQ (RS and PQ are the bisectors)
RS || PQ
RSP + SPC = 180° is false
Hence, (4) is the correct option.
Question 11
1 / -0

In the figure below, find the value of x and y.

A
x =15°, y = 8°

B
x = 13°, y = 5°

C
x = 23°, y = 9°

D
x = 34°, y =22°

Solution

AOF = ∠BOD = 30°(Vertically opposite angles)
⇒ 3y+ 3 = 30°
⇒ 3y = 27°
⇒ y = 9°
Then,
AOC = FOB = 45° (Vertically opposite angles)
⇒ 2x - 1 = 45°
⇒ 2x = 46°
⇒ x = 23°
Question 12
1 / -0

Find the value of in the given triangle.

A
1

B
4

C
6

D
7

Solution

In the given triangle,
CAB + BCA + ABC = 180°
45° + (6x + 2)° + (4x - 7)° = 180°
45° + 10x - 5 = 180°
10x = 180° - 45° + 5
10x = 140°
x = = 14°
=
= 7
Question 13
1 / -0

In the given figure, find EBD.

A
20°

B
25°

C
30°

D
35°

Solution

In the given figure,
ABF + FBE + EBD + DBC = 180°(Angles on a straight line add upto 180°)
7x° + 2x° + 3x° + 6x° = 180°
18x° = 180°
x = = 10
Now,
EBD = 3x° = 3 × 10 = 30°
Question 14
1 / -0

In the figure, AB II CD and AC II BE. What is the measure of DCF?

A
45°

B
60°

C
67°

D
68°

Solution

AC || BE and AB is a transversal.
BAC + ABE = 180° (Co-interior angles)
45° + ABE = 180°
ABE = 135°
Now, ABC + EBC = ABE
ABC + 67° = 135°
ABC = 68°
Now, AB || CD and BF is a transversal.
ABC = DCF (Corresponding angles)
DCF = 68°
Hence, (4) is the correct option.
Question 15
1 / -0

Find the value of x if 78° and x° are supplementary angles.

A
12°

B
55°

C
102°

D
143°

Solution

When two angles add up to 180°, then they are supplementary to each other.
So, 78° + x° = 180°
x° = 180° - 78°
x° = 102°
Question 16
1 / -0

In the following figure, line p is parallel to line m. Find FAD.

A
70°

B
90°

C
98°

D
105°

Solution

In ΔBCD,
CBD + BCD + CDB = 180° (Sum of the angles of a triangle is 180°.)
CBD + 40° + 30° = 180°
CBD = 110° --- (1)
EBD + CBD = 180° (linear pair of angles)
EBD + 110° = 180° (from 1)
EBD = 70°
EBD and FAD are corresponding angles.
So, FAD = 70°
Question 17
1 / -0

In the given figure, two straight lines, LM and NP intersect each other at O. If NOT = 65°, find the value of angles a, b, and c.

A
a = 96°, b = 23°, c = 42°

B
a = 92°, b = 23°, c = 44°

C
a = 92°, b = 24°, c = 45°

D
a = 92°, b = 23°, c = 47°

Solution

LOM is a straight line.
So, LON + NOT + TOM = 180° (Angles on a straight line)
4b + 65° + b = 180°
5b = 115°
b = 23°
Since, LM and NP intersect at O,
POM = LON (vertically opposite angles)
a = 4b
a = 4 × 23°
a = 92°
Since LOM is a straight line,
LOP + POM = 180°
2c + a = 180°
2c + 92° = 180°
2c = 88°
c = 44°
Thus,
a = 92°
b = 23°
c = 44°
Question 18
1 / -0

In the given figure, if AO is the bisector of DAB and AD || BC, then what is the measure of AOB?

A
20°

B
50°

C
80°

D
100°

Solution

In ΔDOA,
AOB is an exterior angle.
Hence, AOB = OAD + ADO (Exterior angle property of a triangle)
AOB = 60° + 40°AOB = 100°
So, AOB = 50°
Hence, (2) is the correct option.
Question 19
1 / -0

In the given figure, sides QP and RQ are produced to points S and T respectively. If PQT = 135° and RPS = 110°, then find the measure of angle x.

A
45°

B
115°

C
65°

D
70°

Solution

Given: PQT = 135°

Sum of all the angles of a triangle is 180° and by linear pair axiom, sum of adjacent angles on a straight line is 180°.
Now, in ΔPQR, P + Q + R = 180°… (1)
By linear pair axiom, QPR + RPS = 180°
QPR + 110° = 180°
Or, P = 70°
And, TQP + PQR = 180° 135° + Q = 180°
Or, Q = 45°
Putting the values of P, Q and R in equation (1), we get
70° + 45° + x = 180° or x = 65°
Question 20
1 / -0

Which of the following statements is true?

A
Two acute angles can be supplementary to each other.

B
Two obtuse angles can be supplementary to each other.

C
One acute and one obtuse angle can be complementary to each other.

D
One acute and one obtuse angle can be supplementary to each other.

Solution

Let one acute angle be 35°
Then, one obtuse angle = 145°
Their sum = 180°
∴ One obtuse and one acute angle will be supplementary to each other as the sum of two supplementary angles is 180°.
Question 21
1 / -0

In the given figure, TOR = 40° and UOQ = 35°. Find the measures of QOR and POS.

A
QOR = 105°, POS = 105°

B
QOR = 105°, POS = 90°

C
QOR = 90°, POS = 90°

D
QOR = 85°, POS = 75°

Solution

PQ and RS intersect each other at point O.
This implies, POS = ROQ ... (1) (Vertically opposite angles)
Also, QOR + UOQ + TOR = 180° (As sum of angles on a straight line is 180°)
QOR + 35° + 40° = 180°
QOR + 75° = 180°
QOR = 105°
Putting QOR in (1), we get:
POS = 105°
Question 22
1 / -0

Mark True or False for following statements

(1) Parallel lines are those lines that are equidistant from each other and never intersect.
(2) If the sum of two angles is 90°, then the two angles are called supplementary angles.
(3) When there is a right angle between two lines, the lines are said to be perpendicular to each other.

A

(1) (2) (3)

True True True

B

(1) (2) (3)

True False True

C

(1) (2) (3)

False True True

D

(1) (2) (3)

False False False

Solution

(1) Parallel lines are those lines that are equidistant from each other and never intersect. - TRUE

(2) If the sum of two angles is 90°, then the two angles are called supplementary angles. This statement is FALSE; if the sum of two angles is 90°, then the two angles are called complementary angles.

COA + AOB = 90°

(3) When there is a right angle between two lines, the lines are said to be perpendicular to each other. - TRUE
Question 23
1 / -0

Which of the following statements is/are incorrect?

(A) When two lines intersect, they create four angles and form two pairs of opposite angles.
(B) Alternate exterior angles are equal when a transversal intersects two parallel lines.
(C) When two lines intersect, the angles formed opposite to each other at the point of intersection (vertex) are called vertically opposite angles.
(D) Angles that have a common arm and a common vertex are called interior angles.

A
Only B

B
Both B and D

C
Only D

D
Both A and C

Solution

According to the properties of lines and angles:
(A) When two lines intersect, they create four angles and form two pairs of opposite angles. This is correct.

(B) Alternate exterior angles are equal when a transversal intersects two parallel lines. This is correct.

(C) When two lines intersect, the angles formed opposite to each other at the point of intersection (vertex) are called vertically opposite angles. This is correct.

(D) Angles that have a common arm and a common vertex are called interior angles. This is incorrect because angles that have a common arm and a common vertex are called adjacent angles.
Question 24
1 / -0

Fill in the blanks.

(A) If one angle of a linear pair is acute, then its other angle will be __P___.
(B) If a transversal intersects a pair of lines in such a way that a pair of interior angles on the same side of the transversal is 180°, then the lines are ____Q____.
(C) When two straight lines intersect, then adjacent angles are ____R____ .

A

P Q R

obtuse perpendicular supplementary

B

P Q R

obtuse perpendicular complementary

C

P Q R

obtuse parallel complementary

D

P Q R

obtuse parallel supplementary

Solution

(P) If one angle of a linear pair is acute, then its other angle will be obtuse because:
Let us assume one of the angles of a linear pair be x, such that x < 90°, i.e. acute angle.
The other angle in the linear pair becomes (180° - x°), which is clearly not acute, i.e. it is an obtuse angle.

(Q) If a transversal intersects a pair of lines in such a way that a pair of interior angles on the same side of the transversal is 180°, then the lines are parallel.
(R) When two straight lines intersect, then adjacent angles are supplementary. This is because adjacent angles form a linear pair and they are supplementary.

Question 25

1 / -0

AB and CD are two parallel lines. A transversal "t" intersects AB at P and CD at Q.

Match the following:

( P) Vertically opposite angles	(i) 1 =5
(Q) Linear pair of angles	(ii) 2 = 6
(R) Corresponding angles	(iii) 5 + 2 = 180°

(P) - (ii), (Q) - (iii), (R) - (i)

(P) - (i), (Q) - (iii), (R) - (ii)

(P) - (i), (Q) - (ii), (R) - (iii)

(P) - (ii), (Q) - (i), (R) - (iii)

Solution

From the figure, AB and CD are two parallel lines. A transversal "t" intersects AB at P and CD at Q.
1 = 5 (Corresponding angles)
2 = 6 (Vertically opposite angles)
5 + 2 = 180° (Linear pair of angles)

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(1)	(2)	(3)
True	True	True

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Lines And Angles Test - 7

Result

Lines And Angles Test - 7

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SHARING IS CARING

Question 1

30°

40°

50°

60°

Solution

Question 2

Scalene triangle

Obtuse-angled triangle

Isosceles triangle

Equilateral triangle

Solution

Question 3

65°

75°

85°

95°

Solution

Question 4

108°

110°

112°

114°

Solution

Question 5

55°

25°

35°

65°

Solution

Question 6

50°

80°

90°

120°

Solution

Question 7

70°

80°

110°

150°

Solution

Question 8

48°

110°

140°

150°

Solution

Question 9

5

6

7

8

Solution

Question 10

ASP = SPD

RS || PQ

RSP = SPQ

RSP + SPC = 180°

Solution

Question 11

x =15°, y = 8°

x = 13°, y = 5°

x = 23°, y = 9°

x = 34°, y =22°

Solution

Question 12

1

4

6

7

Solution

(1) (2) (3)

True True True

(1) (2) (3)

True False True

(1) (2) (3)

False True True

(1) (2) (3)

False False False

P Q R

obtuse perpendicular supplementary

P Q R

obtuse perpendicular complementary

P Q R

obtuse parallel complementary