Lines and Angles Test

Result

Lines and Angles Test - 9

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Question 1
1 / -0

In the adjoining figure, it is given that AB || CD. IfBAO = 108^o and OCD = 120^o, then AOC is equal to

A
120^o

B
72^o

C
132^o

D
150^o

Solution

Through O draw a line EOF parallel to AB.

Now EF|| AB and CD || AB
So, CD||EF
AB || EF and AO is a transversal
We have
AOF + OAB = 180⁰
AOF + 108⁰ = 180⁰
AOF = 72⁰
EF || CD and OC is a transversal.
So,
COF + OCD = 180⁰
COF + 120⁰ = 180⁰
COF = 60⁰
AOC = AOF + COF
= 72⁰ + 60⁰ = 132⁰
Question 2
1 / -0

If an angle is equal to one-eighth of its supplement, its measure is equal to:

A
20⁰

B
50⁰

C
60⁰

D
55⁰

Solution

Let the angle be x.
x + x = 180^o9x = 180^o x 8
x = 160^o
Required angle = 20^o
Option 1 is correct.
Question 3
1 / -0

In the given figure AB = AC, CH = CB and HK || BC. If the exterior angle CAX is 140^o, the angle HCK is

A
45^o

B
55^o

C
50^o

D
30^o

Solution

As AB = AC [Given]
B = C = x [Angles opposite to equal sides are equal]
by, CH = CK
B = BHC = x [ Angles opposite to equal sides are equal]
Now, B + C = 140^o [exterior angle = sum of interior opposite angles]
2x = 140^o
x = 70^o B = C = BHC = x = 70^o
Now, In HCB.
B + BHC +BCH = 180^o [ By angle sum property]
70^o + 70^o + BCH = 180^o – 140^o
= 40^o
CHK = C – BCH
= 70^o – 40^o
= 30^o
Question 4
1 / -0

In the adjoining figure, it is given that l is parallel to m and t is a transversal. The value of x is

A
120°

B
50°

C
130°

D
60°

Solution

ADB + BDC = 180° (Supplementary angles)
BDC = 180° – 60° = 120°BDC = DGF = 120° (Corresponding angles)
Hence, option 1 is correct.
Question 5
1 / -0

The complement of 35⁰ 20' is:

A
69⁰ 40'

B
54⁰ 40'

C
59⁰ 80'

D
159⁰ 40'

Solution

35^o 20^| = 35^o + = 35^o +=
Complement of = 90^o –
= =
= 54 = 54^o 40^|
Option 2 is correct.
Question 6
1 / -0

The angles ABC and BDC are right angles. If AD = 8 cm and DC = 32 cm, the length of BD is:

A
16 cm

B
12 cm

C
15 cm

D
25 cm

Solution

In ADB:AB² = (BD)² + (AD)² …….. (2)
In BDC:
(BC)² = BD² + DC² ……… (1)
Adding equations (1) and (2), we get
(AB)² + (BC)² = 2(BA)² + (AD)² + (DC)²
(AC)² = 2(BD)² + 64 + 1024
(1600) = 2(BD)² + 1088 (Because AC = AD + DC)
BD² = 256
BD = 16 cm
Option 1 is correct.
Question 7
1 / -0

If lines AB, AC, AD and AE are parallel to a line m, then

A
A, B, C, D, E are collinear points

B
A, B, C, D, E are non-collinear points

C
AB and AC are parallel and AD and AE are perpendicular

D
None of these

Solution

A, B, C, D, E are collinear points.
Question 8
1 / -0

In the given figure, OP bisects BOC and OQ bisects AOC. In this case, POQ is equal to:

A
90⁰

B
120⁰

C
60⁰

D
100⁰

Solution

Since OP and OQ are bisectors, AOQ = QOC
Similarly, COP = POB
AOC + COB = 180^o (Supplementary angles)
2 QOC + 2 COP = 180^o
QOC + COP = 90^o
Or POQ = 90^o
Option 1 is correct.
Question 9
1 / -0

In the adjoining figure, the value of x is:

A
30°

B
18°

C
25°

D
15°

Solution

4x + 5 + 3x = 180°7x + 5 = 180°
7x = 175°
x = 25°
Hence, option 3 is correct.
Question 10
1 / -0

In the adjoining figure, AB || CD and 1 : 2 = 2 : 7. Find the measure of 6.

A
140^o

B
136^o

C
108^o

D
144^o

Solution

2x + 7x = 180^o9x = 180^o
x = 20^o2 = 7(20^o) = 140^o2 = 6 = 140^o (Corresponding angles)
Question 11
1 / -0

An angle greater than 180⁰ but less than 360⁰ is called:

A
an obtuse angle

B
right angle

C
reflex angle

D
an acute angle

Solution

If x >180^o and x < 360^o, then the angle is known as reflex angle.
Question 12
1 / -0

If D is the mid point of the hypotenuse AC of a right triangle ABC, BD is equal to:

A
AC

B
AC

C
AC

D
2AC
Question 13
1 / -0

Two straight lines AB and CD cut each other at O. If BOD = 73⁰, then BOC is:

A
63⁰

B
107⁰

C
17⁰

D
153⁰

Solution

BOD = 73^o
BOC = 180^o – BOD (Supplementary angle)
BOC = 107^o
Option 2 is correct.
Question 14
1 / -0

In the following figure, AB is a straight line. Find the value of (x + y).

A
55°

B
65°

C
70°

D
80°

Solution

If we look angles below line AB;
150° + 3x = 180°
3x = 180° – 150°
3x = 30°
x =
x = 10°
Angles above line AB;
y + 90° + 20° = 180°
y + 110° = 180°
y = 180° – 110° = 70°
So, x + y = 10° + 70° = 80°
Question 15
1 / -0

In ABC if B = C = 30⁰, which of the following is the longest side?

A
AB

B
AC

C
BC

D
None of these

Solution

By using angle – sum property
A + B + C = 180^o
A = 180^o– 60^o
= 120^o
A = 120^o
Now, In ABC
As B = C = 30^o
AB = AC = x [because sides opp to equal angles are equal]
Now, In ABD
Cos 30^o =
BD = x
Similarly, in ADC
Cos 30^o =
DC = x
BC = BD + DC
= xx =
So, we know that AB = AC = x
But BC = x
BC is the longest side.
Question 16
1 / -0

The complement of 23⁰ 46' is:

A
157⁰ 14'

B
66⁰14'

C
147⁰ 14'

D
67⁰21'
Question 17
1 / -0

In the given figure, if x + y = w + z, then:

A
points A, O, B are collinear.

B
points O, A, C are collinear.

C
COD is a straight line.

D
BOD is a straight line.

Solution

Three points are said to be collinear if they lie on a single straight line.
So points AOB are collinear.
Question 18
1 / -0

In a ABC if A = 50⁰ and B = 70⁰, then the shortest and the largest sides of the triangle are respectively:

A
AB, BC

B
BC, AC

C
AB, AC

D
none of these

Solution

In ABC, A = 50^o, B = 70^o
By using angle – sum property
A + B + C = 180^o
C = 180^o – (70^o + 50^o)
= 60^o

As we know that in a triangle, side opp. to largest angle would be the largest.
&ly side opposite to the smallest angle would be the smallest.
As B = 70^o, which is the largest
Longest side is AC
ly A = 50^o, which is the smallest
smallest side is BC.
Question 19
1 / -0

The supplement of 120⁰50' is:

A
59⁰ 10'

B
167⁰10'

C
47⁰ 20'

D
57⁰20'
Question 20
1 / -0

In the given figure, if a is greater than or equal to one fifth of a right angle, which of the following is true?

A
b > 162⁰

B
b < 162⁰

C
b 162⁰

D
b 162⁰

Solution

a= (90^o)a18^o
Minimum value of a = 18^o
b = 180^o – 18^o
= 162^o
b 162^o
Option 3 is correct.