Free Lines and Angles 01 Practice Test - 9th Grade 

Question 1

If one angle of a triangle is equal to the sum of the other two angles which are equal, then the triangle is a/an _________ .

A.

Acute angled triangle

B.

Obtuse angled triangle

C.

Right angled triangle

D.

Equilateral triangle

SOLUTION

Solution : C

Let the two equal angles be x.

Then the third angle y = 2x

Also we know,
y + 2x = 180°  (Sum of the angles of a triangle is 180°)

2x + 2x = 180°                              

4x = 180°

x = 45°

Hence the two equal angles are 45° each and the other angle is 90° which makes this triangle a right angled triangle.

Question 2

The angle between the bisectors of two adjacent supplementary angles is a right angle.

A.

True

B.

False

SOLUTION

Solution : A

Let A and B be adjacent supplementary angles as shown in the figure.


Then, A+B=180
The bisectors of the angles are drawn. We have to find 1+2.
1+2=(A2+B2)
=1802=90
Hence, the angle between the bisectors of adjacent supplementary angles is 90.

Question 3

Which of the following are true?

A.

A triangle can have two right angles.

B.

A triangle can have all angles less than 60

C.

A triangle can have two acute angles.

D.

A triangle cannot have two obtuse angles.

SOLUTION

Solution : C and D

The sum of the three angles of a triangle should be 180. Hence it cannot have two right angles. If it has all angles less than 60 then they cannot add up to form 180. A triangle can have two or three acute angles. It cannot have two obtuse angles because then the sum of the three angles would be greater than 180.

Question 4

Angles of a triangle are in the ratio 2 : 4 : 3. The smallest angle of the triangle is

A.

40

B.

30

C.

60

D.

80

SOLUTION

Solution : A

The sum of all the three angles of a triangle = 180

Since the angles are in the ratio 2 : 4 : 3, let x be a factor such that,

2x + 4x + 3x=180

=> 9x = 180

=> x = 20

Hence the three angles of the triangle are:

=> 2x = 40

=> 4x = 80

=> 3x = 60

So the smallest angle in the given triangle is 40.

Question 5

Two angles whose measures are a & b are such that 2a - 3b = 60 then 5b =

___
degrees,
 if they form a linear pair.

SOLUTION

Solution :

Given that,

 2a - 3b = 60

Also if they form a linear pair, then

2 × (180 - b) - 3b = 60      ( since  a = 180 - b)

⇒ 360 - 2b - 3b = 60

⇒ 5b =300

Question 6

An angle 'x' is 14 more than its complement. Find the angle.

A.

38

B.

52

C.

50

D.

104

SOLUTION

Solution : B

The sum of two complementary angles is 90.

If one angle is x, then another angle will be 90x.
x(90x)=14
                   2x=104
                     x=52

Question 7

Which of the following statements is not true about the figure given below?

A.

 DBC = 180

B.

 CBA = 90

C.

 CBA and DBA are supplementary.

D.

 ABD = 60

SOLUTION

Solution : D

Here ABD and  ABC are supplementary, since they lie on a straight line.
Hence DBC = 180
Given, ABC = 90
ABD = 180 - ABC = 180 - 90 = 90

Question 8

If the arms of one angle are respectively parallel to the arms of another angle, then the two angles are :

A.

Neither equal nor supplementary

B.

Not equal but supplementary

C.

Equal but not supplementary

D.

Either equal or supplementary.

SOLUTION

Solution : D

Let the two angles be ABC and PQR.
There are two possible cases:

i)

 
Here, 
1=2 [Correspoding angles] and
2=3 [Corresponding angles]
1=3
ABC=PQR

ii)

Here 1+2=180 [Co-interior angles] and
2=3 [Corresponding angles]
1+3=180
ABC+PQR=180
Hence it can be seen that the angles could either be equal or supplementary.

Question 9

In the adjoining figure, it is given that AB || CD, BAO = 108 and OCD = 120  then AOC = ________.

A.

120

B.

72

C.

132

D.

150

SOLUTION

Solution : C

Let us redraw the given figure as follows:
Now we know that interior angles on the same side of the transversal are supplementary
Therefore BAO + AOR = 180
⇒ AOR = 72
and similarly 
ROC + OCD = 180
 ⇒ ROC + 120 = 180
 ⇒ ROC = 60
Hence
AOC = 60 + 72
⇒ AOC = 132

Question 10

In the fig. below, if AB || CD, APQ = 50 and PRD = 127, find x and y.

A.

50, 67

B.

50, 77

C.

77, 87

D.

77, 97

SOLUTION

Solution : B

Since AB || CD

APQ = PQR (Alternate interior angles of transversal PQ)

⇒x = 50

APR = PRD (Alternate interior angles of transversal PR)

⇒y + 50 = 127

⇒ y = 127 - 50

= 77