# 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 _________ .

Acute angled triangle

Obtuse angled triangle

Right angled triangle

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.

True

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 triangle can have two right angles.

A triangle can have all angles less than 60∘

A triangle can have two acute angles.

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

40∘

30∘

60∘

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,

#### 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.

38∘

52∘

50∘

104∘

#### SOLUTION

Solution :B

The sum of two complementary angles is 90∘.

If one angle is x, then another angle will be 90∘−x.

⇒x−(90∘−x)=14∘

⇒ 2x=104∘

⇒ x=52∘

### Question 7

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

∠DBC = 180∘

∠CBA = 90∘

∠CBA and ∠DBA are supplementary.

∠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 :

Neither equal nor supplementary

Not equal but supplementary

Equal but not supplementary

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=∠PQRii)

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 = ________.

120∘

72∘

132∘

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 supplementaryTherefore ∠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.

50∘, 67∘

50∘, 77∘

77∘, 87∘

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∘