Lines and Angles Test

Result

Lines and Angles Test - 8

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Weekly Quiz Competition

Question 1
1 / -0

In the adjoining figure, if ABC = 100°, EDC = 130° and AB || DE, then BCD is equal to

A
80°

B
60°

C
50°

D
20°

Solution

ABC = 100°
CBO = 180° - ABC (Linear Pair)
= 180° - 100°
= 80°
EDO = DOB = 130° (Alternate interior angles)
DOB = OBC + OCB (Exterior angle theorem)
130° = 80° + OCB
OCB = 50°
or, BCD = 50°
Option 3 is correct.
Question 2
1 / -0

In the adjoining figure, BO and CO are angle bisectors of external angles of ABC. Find the measure of BOC.

A
90° - A

B
90° + A

C
180° - A

D
180° + A

Solution

Let,= x, = y, = z
∠CBP = 180° - y and ∠BCQ = 180° - z
Ray BO is the bisector of ∠CBP.
Therefore, = ½= ½(180° - y) = 90° - ^y/₂ … (1)
Similarly, ray CO is the bisector of ∠BCQ.
Therefore, BCO = ½BCQ = ½(180° - z) = 90° - ^z/₂ … (2)
In triangle BOC, BOC + BCO + CBO = 180° … (3)
Substituting (1) and (2) in (3), you get
BOC + 90° - ^y/₂+ 90° - ^z/₂= 180°
So, BOC = ^y/₂+ ^z/₂
Or BOC = ½(y + z) … (4)
But x + y + z = 180° (Angle sum property of a triangle)
Therefore, y + z = 180° - x
Therefore, (4) becomes:
BOC = ½(180° - x) = 90° - ^x/₂= 90° - ½BAC = 90° - ¹/₂∠A
Question 3
1 / -0

Measure of NOP in the given figure is:

A
90⁰

B
30⁰

C
120⁰

D
80⁰

Solution

In LNML + M + LNM = 180^o
LNM = 180^o – 90^o
LNM = 90^o
LNM = ONP = 90^o (Vertically opposite angles)
In ONP
O + ONP + P = 180^o
O = 180^o – 150^o
= 30^o
Option 2 is correct.
Question 4
1 / -0

If angle p is four times its supplement, then the value of p is:

A
144^o

B
36^o

C
140^o

D
72^o

Solution

Supplement of P = 180^o – PAccording to the question,
P = 4 (180^o – P)
P = 720^o – 4P
5P = 720^o
P = 144^o
Option 1 is correct.
Question 5
1 / -0

In the adjoining figure, if 2 = 55⁰ and 5 = 55⁰, the lines m and n are

A
parallel

B
not parallel

C
cannot say

D
none of these

Solution

Given - 2 = 55^o
& 5 = 55^o
Proof - The lines m & n would beto each other only when the corresponding angles are equal. But Here 2 = 5 = 55^o(which are not the corresponding angles)So, the corresponding angles, ∠2 and ∠6 cannot equal each other.
Lines m & n are not parallel to each other.
Question 6
1 / -0

The interior opposite angles of the exterior ACD are

A
B, C

B
A, C

C
A, B

D
B, E

Solution

According to exterior angle theorem, interior opposite angles of ACD are A and B.
Question 7
1 / -0

is a line. Find y.

A
30^o

B
20^o

C
45^o

D
60^o

Solution

7y + 1.5y+ 0.5y = 180^o
9y = 180^o
y = 20^o
Option 2 is correct.
Question 8
1 / -0

In the adjoining figure, the value of A + B+ C+D+ E+ F is

A
360⁰

B
270⁰

C
540⁰

D
180⁰

Solution

In ABC:A + B + C = 180^o ………. (1)
In DEF:
D + E + F = 180^o ………. (2)
Adding equations (1) and (2), we get
A + B + C + D + E + F = 180^o + 180^o
= 360^o
Option 1 is correct.
Question 9
1 / -0

A triangle divides the plane into ____ parts.

A
4

B
2

C
3

D
6

Solution

divides the plane into 3 parts
Question 10
1 / -0

If c exceeds d by one and half right angle, then d is equal to

A
112.5^o

B
247.5^o

C
160^o

D
100^o

Solution

C = D + 1 (90^o)C = D + 135^o ……….. (1)
C + D = 360^o
Putting (1) in the above equation, we get
D + D + 135^o = 360^o
D = 112.5^o
Option 1 is correct.
Question 11
1 / -0

In the adjoining figure, AB || DE. What is the value of x?

A
35^o

B
25^o

C
45^o

D
55^o

Solution

Using exterior angle theorem of triangle
Exterior angle = Sum of interior opposite angles
In BCF,
110^o = x + BFC ..... (i)
Now as CFD is a straight line
DFB + BFC = 180^o (straight angle)
BFC = 180^o – 95^o
= 85^o........ (ii)
Using eq. (ii), eq. (i) becomes
x = 110^o – 85^o = 25^o
Question 12
1 / -0

Can 90⁰, 90⁰ and 20⁰ form a triangle?

A
Yes

B
Sometimes

C
No

D
None

Solution

Angle sum property
90⁰ + 90⁰ + 20⁰ = 200⁰
cannot form a triangle.
Question 13
1 / -0

In the given figure, if three lines intersect at a point, what is the measure of angles a and b?

A
a = 50^o, b = 40^o

B
a = 40^o, b = 50^o

C
a = 30^o, b = 60^o

D
a = 60^o, b = 30^o

Solution

130^o = 90^o + b (Vertically opposite angle)b = 40^o
a + b = 90^o (complementary angles)
a + 40^o = 90^o
a = 50^o
Option 1 is correct.
Question 14
1 / -0

In the figure below, find x if AB || CD.

A
55

B
45

C
60

D
70

Solution

AB||CD
ABC = BCD (Alternate interior angles)
So, x^o= BCD
Now, EF||CD
FEC + ECD = 180^o(Co-interior Angles)
150^o + ECD = 180^o
ECD = 180^o - 150^o = 30^o
So, x^o = BCE + ECD
So, x^o = 15^o + 30^o = 45^o
x^o = 45^o
Option 2 is correct.
Question 15
1 / -0

Can three lines measuring 7 cm, 5 cm and 3 cm form a triangle?

A
Yes

B
Sometimes

C
No

D
None of these

Solution

(6 + 5) 11 > 3
Inequality property
(5 + 3) 8 > 6
(3 + 6) 9 > 5
They can form le.
Question 16
1 / -0

In the given figure, three lines are intersecting. Equation connecting x, y and z is:

A
x^o + y^o + z^o= 180^o

B
x^o + y^o + z^o = 360^o

C
2x^o + 2y^o + z^o = 360^o

D
2x^o + 2y^o + z^o = 180^o

Solution

AOB = EOD = y^o (Vertically opposite)
x^o + AOB + z^o = 180^o
Or x^o + y^o + z^o = 180^o
Option 1 is correct.
Question 17
1 / -0

In the adjoining figure, find the value of x.

A
80⁰

B
88⁰

C
90⁰

D
96⁰

Solution

AE is transverse to AB and AB||OF
AEF + EAB = 180^o
AEF = 180^o – 130^o
AEF = 50^o
Similarly, CEF + ECD = 180^o
CEF = 180 – 134^o
= 46^o
AEF + CEF = x
x = 96^o
Option 4 is correct.
Question 18
1 / -0

In the following figure, it is given that A = 60^o, CE || BA and ECD = 70^o. ACB =

A
60^o

B
50^o

C
70^o

D
90^o

Solution

CE|| BA and AC is a transversal.
ACE = BAC = 60^o
(Alternate angles)
ACD = ACE + ECD
60^o + 70^o = 130^o
Now ACB = 180^o – ACD
(ACB and ACD form a linear pair)
= 180^o – 130^o = 50^o
Question 19
1 / -0

In the given figure if p = q, then:

A
a = b

B
b = t

C
q = t

D
p = t
Question 20
1 / -0

In the given figure, BO and CO are the bisectors of the exterior angles B and C, respectively. Find BOC.

A
46⁰

B
56⁰

C
61⁰

D
76⁰

Solution

OB and OC are angle bisectors.
PBO = OBC
^∴ OCQ =OCB
PBC = A + ACB (Exterior angle theorem)
QCB = A + ABC (Exterior angle theorem)
PBC + QCB = 2A + ACB + ABC
2OBC + 2OCB = A + 180^o (Because ∠A + ∠ACB + ∠ABC = 180^o)
2(180^o – BOC) = A + 180^o
360^o – 2BOC = 58^o + 180^o
360^o – 238^o = 2BOC
BOC = 61^o
Option (3) is correct.