Lines and Angles Test

Result

Lines and Angles Test - 7

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Weekly Quiz Competition

Question 1
1 / -0

In the adjoining figure POR, QOR from a liner pair and a – b = 50^0. Find angles a & b.

A
115⁰, 65⁰

B
70⁰, 100⁰

C
80⁰, 120⁰

D
120⁰, 80⁰

Solution

As we know,
a + b = 180^o-- (1)
a - b = 50^o (given)
a = 50^o + b - (2)
Put value of a in equation (1):
50^o + b + b = 180^o
2b = 180^o- 50^o
b = 130^o / 2 = 65^o
Put value of b = 65^o in equation (2):
a = 50^o + 65^o = 115^o
115^o, 65^o
Option (1) is correct.
Question 2
1 / -0

In the given figure, line II m 1 = 60^o, 8 is equal to

A
60^o

B
120^o

C
150^o

D
70^o

Solution

1 = 60^o 2 = 120^o
(Linear pair)
2 = 4 = 120^o
(Vertically opposite angles )

= 8 = 120^o(Corresponding angles)
Question 3
1 / -0

What is the degree measure of an angle whose complement is eighty percent of half of its supplement?

A
60⁰

B
45⁰

C
30⁰

D
15⁰

Solution

Let the angle be x.
Its complement = 90 - x (complementary angle sum = 90^o)
Supplementary angle = 150 - x (supplementary angle sum = 180^o)
Acc. to the question,
90 - x = 80% of ( 180 - x)
90 - x =
90 - x = (180 - x)
450 - 5x = 360 - 2x
3x = 90^o
x = 30^o
Option 3 is correct.
Question 4
1 / -0

The supplement of an angle is half of itself. The angles of its supplement are

A
60⁰, 120⁰

B
60⁰, 180⁰

C
120⁰, 360⁰

D
65⁰, 120⁰

Solution

Let the angle be x.
Its supplement = 180 - x ( sum of supplementary angles = 180)
Acc. to question,
½ x = (180 - x)
x = 360 - 2x
3x = 360
x = 120^o
Its supplement = 180 - 120 = 60^o
Option 1 is correct.
Question 5
1 / -0

In the given figure, the number of sets of parallel lines is

A
8 sets

B
7 sets

C
4 sets

D
2 sets

Solution

3 parallel lines are indicated by arrow.
AB || CD, EF || GF, AB || GF and AC || BD
total 7 sets of parallel lines.
Question 6
1 / -0

In the given figure, AB || CD, EFC = 40° and ECF = 100°, then BAF is equal to

A
140°

B
70°

C
100°

D
80°

Solution

In ECF:
ECF = 100°, EFC = 40° (Given)
ECF + CEF + EFC = 180° (Sum of the angles of a triangle is 180°)
100° + CEF + 40° = 180°
CEF = 40°
CEF + DEF = 180° (Supplementary angles)
40° + DEF = 180°
DEF = 140°
BAF = DEF = 140° (Corresponding angles)
Option (1) is correct.
Question 7
1 / -0

In the following figure, find the value of y.

A
24°

B
22°

C
20°

D
5°

Solution

COD = AOF = 27y° (Vertically opposite angles)
AOB + AOF + FOE = 180°
5y° + 27y° + 4y°= 180°
36y° = 180°
y = 5°
Option 4 is correct.
Question 8
1 / -0

Instruments used to draw a pair of parallel lines are

A
protractor and scale

B
compass and scale

C
set square and scale

D
none of these

Solution

Setsquare and scale
Question 9
1 / -0

In the given figure, if AB || CD, BAE = 70⁰, DCE = 35⁰ and CED = m⁰, the value of m is

A
95⁰

B
85⁰

C
75⁰

D
105⁰

Solution

BAE = EDC = 70^o [Alternate interior angles]
DCE = 35^o (Given)
EDC + CED + ECD = 180^o ( Sum of all angles of triangle = 180)
70^o + m + 35^o = 180^o
m^o + 105^o = 180^o
m^o = 180^o - 105^o
m^o = 75^o
Option 3 is correct .
Question 10
1 / -0

The sides BC, CA and AB of ABC are produced in order to form exterior angles ACD, BAE and CBF. ACD+ BAE + CBF is

A
180^o

B
270^o

C
360^o

D
540^o

Solution

ACD, BAE, CBE (extension angles)
ACD = CAB + ABC (extension angle = sum of 2 interior opposite angles)
ACD = x^o + y^o
Similarly, EAB = y^o + z^o = (ii)
CBF = x^o + z^o = (ii)
Acc. to question,
ACD + BAE + CBF = (x^o + y^o) + (y^o + z^o) + (2^o + x^o)
= 2x^o + 2y^o + 2z^o = 2(x^o + y^o + z^o)
= 2(180^o) (Sum of all angles of a triangle = 180^o)
= 360^o
Option 3 is correct.
Question 11
1 / -0

Example for a pair of parallel lines is

A
corners of a room

B
railway tracks

C
sides of a triangle

D
none of these

Solution

railway tracks
Question 12
1 / -0

In the given figure, if PQ || RS, then B is equal to

A
10⁰

B
50⁰

C
25⁰

D
30⁰

Solution

In BDE,
B + D + E = 180^o (Sum of angles of a triangle = 180^o)
B + 50^o + 80^o = 180^o
B = 50^o
Option (2) is correct.
Question 13
1 / -0

Find the value of x from the adjoining figure.

A
60⁰

B
75⁰

C
90⁰

D
115⁰

Solution

ACD = A + B (ACD is extension angle)
ACD = 25^o + 30^o
ACD = 55^o
AED = ACD + D (AED is extension angle)
x = 55^o + 60^o
x = 115^o
Option 4 is correct.
Question 14
1 / -0

Among the following, find the set of measures which can form a triangle.

A
70⁰, 90⁰, 25⁰

B
65⁰, 85⁰, 40⁰

C
65⁰, 85⁰, 30⁰

D
45⁰, 45⁰, 80⁰

Solution

65⁰ + 85⁰ + 30⁰ = 180⁰
Question 15
1 / -0

In the given figure, AB || CD, ABE = 130^o and BED = 30^o. Find EDC.

A
150^o

B
100^o

C
170^o

D
130^o

Solution

ABE = 130^o
ABE = BEF + BFE (ABE is an exterior angle)
130^o = 30^o + BFE
BFE = 100^o
EFB = EDC (Corresponding angles)
EDC = 100^o
Option 2 is correct.
Question 16
1 / -0

In the given figure the value of x⁰ is

A
60⁰

B
90⁰

C
70⁰

D
80⁰

Solution

For the given figure, (Using the property where an exterior angle of a triangle is equal to the sum of the two opposite interior angles)
DEB = EDA + EAD (extension angle)
DEB = 40^o + 30^o = 70^o
Using the property where an exterior angle of a triangle is equal to the sum of the two opposite interior angles
x^o = DEB + B (extension angle)
x^o = 70^o + 10^o = 80^o
x^o = 80^o

Option 4 is correct.
Question 17
1 / -0

Find the measure of angle ACD in the given figure.

A
130°

B
120°

C
160°

D
115°

Solution

90°+ 20° + x° = 180°
x° = 180° – 110° = 70°
ACD = x° + 90°
= 70°+ 90° = 160°
Question 18
1 / -0

If two angles are complementary of each other, then each angle is

A
acute

B
obtuse

C
reflex

D
right

Solution

As we know, sum of complementary angles = 90^o

a^o + b^o = 90^o
a and b are acute angles.
Option 1 is correct .
Question 19
1 / -0

In the figure, the bisectors of B and C meet at 0. Then BOC is

A
90⁰ + A

B
90⁰ + B

C
90⁰– C

D
none of these

Solution

O = 180 - (OBC + OCB) [Sum of all angles of a triangle = 180^o]
O = 180 - (2 + 4)
2 = B (OB is angle bisector of B)
Similarly,
4 = C
O = 180 -
O = 180 - (B + C)
O = 180 - (180 - A) {(B + C = 180 - A) (Sum of all angles of a triangle = 180^o)}

O = 180 - 90 + A
O = 90 + A
Option 1 is correct.
Question 20
1 / -0

If PQ || RT, the measure of y^o in the given figure is

A
70^o

B
75^o

C
80^o

D
85^o

Solution

p = 70⁰
Alternate to TRS
y⁰ + p + Q = 180⁰
y⁰ = 180⁰ - 100⁰ = 80⁰