RS AGGARWAL CLASS 9 CHAPTER 7 LINES AND ANGLES MCQ

 MULTIPLE CHOICE QUESTIONS

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Question 1:

If one angle of a triangle is equal to the sum of the other two angles, then the triangle is
(a) an isosceles triangle
(b) an obtuse triangle
(c) an equilateral triangle
(d) a right triangle

Answer 1:

Let ∆ABC be such that ∠A = ∠B + ∠C.

In ∆ABC,

∠A + ∠B + ∠C = 180º    (Angle sum property)

⇒ ∠A + ∠A = 180º         (∠A = ∠B + ∠C)

⇒ 2∠A = 180º

⇒ ∠A = 90º

Therefore, ∆ABC is a right triangle.

Thus, if one angle of a triangle is equal to the sum of the other two angles, then the triangle is a right triangle.

Hence, the correct answer is option (d).

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Question 2:

An exterior angle of a triangle is 110° and its two interior opposite angles are equal. Each of these equal angles is
(a) 70°
(b) 55°
(c) 35°
(d) 27 $\frac{1°}{2}$

Answer 2:

Let the measure of each of the two equal interior opposite angles of the triangle be x.

In a triangle, the exterior angle is equal to the sum of the two interior opposite angles.

∴ x + x = 110°

⇒ 2x = 110°

⇒ x = 55°

Thus, the measure of each of these equal angles is 55°.

Hence, the correct answer is option (b).

Question 3:

The angles of a triangle are in the ration 3 : 5 : 7. The triangle is
(a) acute-angled
(b) obtuse-angled
(c) right-angled
(d) an isosceles triangle

Answer 3:

(a) acute-angled

Let the angles measure $\left(3x\right)°, \left(5x\right)° \text{and} \left(7x\right)°$.
Then,
$$\begin{gathered} 3x+5x+7x=180° \\ \Rightarrow 15x=180° \\ \Rightarrow x=12° \end{gathered}$$
Therefore, the angles are $3\left(12\right)°=\mathbf{36}\mathbf{°}, 5\left(12\right)°=\mathbf{60}\mathbf{°} \text{and }7\left(12\right)°=\mathbf{84}\mathbf{°}$.
Hence, the triangle is acute-angled.

Question 4:

If one of the angles of a triangle is 130° then the angle between the bisectors of the other two angles can be
(a) 50°
(b) 65°
(c) 90°
(d) 155°

Answer 4:

Let ∆ABC be such that ∠A = 130°.



Here, BP is the bisector of ∠B and CP is the bisector of ∠C.

∴ ∠ABP = ∠PBC = $\frac{1}{2}$∠B                .....(1)

Also, ∠ACP = ∠PCB = $\frac{1}{2}$∠C           .....(2)

In ∆ABC,

∠A + ∠B + ∠C = 180°         (Angle sum property)

⇒ 130° + ∠B + ∠C = 180°

⇒ ∠B + ∠C = 180° − 130° = 50°

⇒ $\frac{1}{2}$∠B + $\frac{1}{2}$∠C = $\frac{1}{2}$ × 50° = 25°

⇒ ∠PBC + ∠PCB = 25°     .....(3)     [Using (1) and (2)]

In ∆PBC,

∠PBC + ∠PCB + ∠BPC = 180°         (Angle sum property)

⇒ 25° + ∠BPC = 180°                       [Using (3)]

⇒ ∠BPC = 180° − 25° = 155°

Thus, if one of the angles of a triangle is 130° then the angle between the bisectors of the other two angles is 155°.

Hence, the correct answer is option (d).

Question 5:

In the given figure, AOB is a straight line. The value of x is

(a) 12
(b) 15
(c) 20
(d) 25

Answer 5:

It is given that, AOB is a straight line.

∴ 60º + (5xº + 3xº) = 180º           (Linear pair)

⇒ 8xº = 180º − 60º = 120º

⇒ xº = 15º

Thus, the value of x is 15.

Hence, the correct answer is option (b).

Question 6:

The angles of a triangle are in the ratio 2 : 3 : 4. The largest angle of the triangle is
(a) 12°
(b) 100°
(c) 80°
(d) 60°
 

Answer 6:

Suppose ∆ABC be such that ∠A : ∠B : ∠C = 2 : 3 : 4.

Let ∠A = 2k, ∠B = 3k and ∠C = 4k, where k is some constant.

In ∆ABC,

∠A + ∠B + ∠C = 180º      (Angle sum property)

⇒ 2k + 3k + 4k = 180º

⇒ 9k = 180º

⇒ k = 20º

∴ Measure of the largest angle = 4k = 4 × 20º = 80º

Hence, the correct answer is option (c).

Question 7:

In the given figure, ∠OAB = 110° and ∠BCD = 130° then ∠ABC is equal to

(a) 40°
(b) 50°
(c) 60°
(d) 70°

Answer 7:

In the given figure, OA || CD.

Construction: Extend OA such that it intersects BC at E.



Now, OE || CD and BC is a transversal.

∴ ∠AEC = ∠BCD = 130°       (Pair of corresponding angles)

Also, ∠OAB + ∠BAE = 180°         (Linear pair)

∴ 110° + ∠BAE = 180°

⇒ ∠BAE = 180° − 110° = 70°

In ∆ABE,

∠AEC = ∠BAE + ∠ABE                (In a triangle, exterior angle is equal to the sum of two opposite interior angles)

∴ 130° = 70° + x°

⇒ x° = 130° − 70° = 60°

Thus, the measure of angle ∠ABC is 60°.

Hence, the correct answer is option (c).

Question 8:

If two angles are complements of each other, then each angle is
(a) an acute angle
(b) an obtuse angle
(c) a right angle
(d) a reflex angle

Answer 8:

(a) an acute angle

If two angles are complements of each other, that is, the sum of their measures is $90°$, then each angle is an acute angle.

Question 9:

An angle which measures more than 180° but less than 360°, is called
(a) an acute angle
(b) an obtuse angle
(c) a straight angle
(d) a reflex angle

Answer 9:

An angle which measures more than 180° but less than 360° is called a reflex angle.

Hence, the correct answer is option (d).

Question 10:

The measure of an angle is five times its complement. The angle measures
(a) 25°
(b) 35°
(c) 65°
(d) 75°

Answer 10:

(d) 75°

Let the measure of the required angle be .
Then, the measure of its complement will be .

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Question 11:

Two complementary angles are such that twice the measure of one is equal to three times the measure of the other. The measure of larger angle is
(a) 72°
(b) 54°
(c) 63°
(d) 36°

Answer 11:

(b) 54°

Let the measure of the required angle be $x°$.
Then, the measure of its complement will be$\left(90-x\right)°$.
$$\begin{gathered} \mathbf{\therefore }2x=3\left(90-x\right) \\ \Rightarrow 2x=270-3x \\ \Rightarrow 5x=270 \\ \Rightarrow x=54 \end{gathered}$$

Question 12:

In the given figure, AOB is a straight line. If ∠AOC = 4x° and ∠BOC = 5x°, then ∠AOC = ?
(a) 40°
(b) 60°
(c) 80°
(d) 100°

Answer 12:

(c) 80°

We have :
$$\begin{gathered} \angle AOC+\angle BOC=180° \left[\text{Since }\mathit{\text{AOB }}\text{is a straight line}\right] \\ \Rightarrow 4x+5x=180° \\ \Rightarrow 9x=180° \\ \Rightarrow x=20° \\ \therefore \angle AOC=4\times 20°=80° \end{gathered}$$

Question 13:

In the given figure, AOB is a straight line. If ∠AOC = (3x + 10)° and ∠BOC (4x − 26)°, then ∠BOC = ?
(a) 96°
(b) 86°
(c) 76°
(d) 106°

Answer 13:

(b) 86°

We have :
$$\begin{gathered} \angle AOC+\angle BOC=180° \left[\text{Since }\mathit{\text{AOB }}\text{is a straight line}\right] \\ \Rightarrow 3x+10+4x-26=180° \\ \Rightarrow 7x=196° \\ \Rightarrow x=28° \end{gathered}$$

$$\begin{gathered} \therefore \angle BOC={\left[4\times 28-26\right]}^{°} \\ \mathrm{Hence}, \angle BOC=86°. \end{gathered}$$

Question 14:

In the given figure, AOB is a straight line. If ∠AOC = (3x − 10)°, ∠COD = 50° and ∠BOD = (x + 20)°, then ∠AOC = ?
(a) 40°
(b) 60°
(c) 80°
(d) 50°

Answer 14:

(c) 80°

We have :
$$\begin{gathered} \angle AOC+\angle COD+\angle BOD=180° \left[\text{Since }\mathit{\text{AOB }}\text{is a straight line}\right] \\ \Rightarrow 3x-10+50+x+20=180 \\ \Rightarrow 4x=120 \\ \Rightarrow x=30 \end{gathered}$$

$$\begin{gathered} \therefore \angle AOC={\left[3\times 30-10\right]}^{°} \\ \Rightarrow \angle AOC=80° \end{gathered}$$

Question 15:

Which of the following statements is false?
(a) Through a given point, only one straight line can be drawn.
(b) Through two given points, it is possible to draw one and only one straight line.
(c) Two straight lines can intersect at only one point.
(d) A line segment can be produced to any desired length.

Answer 15:

(a) Through a given point, only one straight line can be drawn.

Clearly, statement (a) is false because we can draw infinitely many straight lines through a given point.

Question 16:

An angle is one-fifth of its supplement. The measure of the angle is
(a) 15°
(b) 30°
(c) 75°
(d) 150°

Answer 16:

(b) 30°

Let the measure of the required angle be 
Then, the measure of its supplement will be

Question 17:

In the adjoining figure, AOB is a straight line. If x : y : z = 4 : 5 : 6, then y = ?
(a) 60°
(b) 80°
(c) 48°
(d) 72°

Answer 17:

(a) 60°

Let  

Then, we have:

Question 18:

In the given figure, straight lines AB and CD intersect at O. If ∠AOC = ɸ, ∠BOC = θ and θ = 3ɸ, then ɸ = ?
(a) 30°
(b) 40°
(c) 45°
(d) 60°

Answer 18:

(c) 45°

We have :

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Question 19:

In the given figure, straight lines AB and CD intersect at O. If ∠AOC + ∠BOD = 130°, then ∠AOD = ?
(a) 65°
(b) 115°
(c) 110°
(d) 125°

Answer 19:

(b) 115°

We have :
$$\begin{gathered} \angle AOC=\angle BOD \left[\text{Vertically-Opposite Angles}\right] \\ \therefore \angle AOC+\angle BOD=130° \\ \Rightarrow \angle AOC+\angle AOC=130° \left[\because \angle AOC=\angle BOD\right] \\ \Rightarrow 2\angle AOC=130° \\ \Rightarrow \angle AOC=65° \end{gathered}$$
Now,
$$\begin{gathered} \angle AOC+\angle AOD=180° \left[\text{∵}\mathit{\text{COD }}\text{is a straight line}\right] \\ \Rightarrow 65°+\angle AOD=180° \\ \Rightarrow \angle AOC=115° \end{gathered}$$

Question 20:

In the given figure, AB is a mirror, PQ is the incident ray and QR is the reflected ray. If ∠PQR = 108°, find ∠AQP.
(a) 72°
(b) 18°
(c) 36°
(d) 54°

Answer 20:

(c) 36°

We know that angle of incidence = angle of reflection.
Then, let $\angle AQP=\angle BQR=x°$
Now,
$$\begin{gathered} \angle AQP+\angle PQR+\angle BQR=180° \left[\text{∵}\mathit{\text{AQB }}\text{is a straight line}\right] \\ \Rightarrow x+108+x=180° \\ \Rightarrow 2x=72° \\ \Rightarrow x=36° \\ \therefore \angle AQP=36° \end{gathered}$$

Question 21:

In the given figure, AB || CD. If ∠BAO = 60° and ∠OCD = 110°, then ∠AOC = ?
(a) 70°
(b) 60°
(c) 50°
(d) 40°

Answer 21:

(c) 50°

Draw $EOF\parallel AB\parallel CD$.
Now, $EO\parallel AB \text{and} OA$ is the transversal.
$\therefore \angle EOA=\angle OAB=60° \left[\text{Alternate Interior Angles}\right]$
Also,
$OF\parallel CD \text{and} OC$ is the transversal.
$$\begin{gathered} \therefore \angle COF+\angle OCD=180° \left[\text{Angles on the same side of a transversal line are supplementary}\right] \\ \Rightarrow \angle COF+110°=180° \\ \Rightarrow \angle COF=70° \end{gathered}$$
Now,
$$\begin{gathered} \angle EOA+\angle AOC+\angle COF=180° \left[\because EOF \text{is a straight line}\right] \\ \Rightarrow 60°+\angle AOC+70°=180° \\ \Rightarrow \angle AOC=\mathbf{50}\mathbf{°} \end{gathered}$$

Question 22:

In the given figure, AB || CD. If ∠AOC = 30° and ∠OAB = 100°, then ∠OCD = ?
(a) 130°
(b) 150°
(c) 80°
(d) 100°

Answer 22:

(a) 130°

Draw 
Now,  is the transversal.

Also, 
 is the transversal.

Question 23:

In the given figure, AB || CD. If ∠CAB = 80° and ∠EFC = 25°, then ∠CEF = ?
(a) 65°
(b) 55°
(c) 45°
(d) 75°

Answer 23:

(c) 45°

 is the transversal.

Side EC of triangle EFC is produced to D.

Question 24:

In the given figure, if AB || CD, CD || EF and y : z = 3 : 7, x = ?
(a) 108°
(b) 126°
(c) 162°
(d) 63°

Answer 24:

(b) 126°
Let 
Let the transversal intersect AB at P, CD at O and EF at Q.



Then, we have:

Also,

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Question 25:

In the given figure, AB || CD. If ∠APQ = 70° and ∠PRD = 120°, then ∠QPR = ?
(a) 50°
(b) 60°
(c) 40°
(d) 35°

Answer 25:

(a) 50°

$AB\parallel CD \text{and} PQ$ is the transversal.
$\therefore \angle PQR=\angle APQ=70° \left[\text{Alternate Interior Angles}\right]$
Side QR of traingle PQR is produced to D.
$$\begin{gathered} \therefore \angle PQR+\angle QPR=\angle PRD \\ \Rightarrow 70°+\angle QPR=120° \\ \Rightarrow \angle QPR=50° \end{gathered}$$

Question 26:

In the given figure, AB || CD. If ∠EAB = 50° and ∠ECD = 60°, then ∠AEB = ?
(a) 50°
(b) 60°
(c) 70°
(d) 55°

Answer 26:

(c) 70°

$AB\parallel CD \text{and} BC$ is the transversal.
$\therefore \angle ABE=\angle BCD=60° \left[\text{Alternate Internal Angles}\right]$
In $∆ABE$, we have:
$$\begin{gathered} \angle EAB+\angle ABE+\angle AEB=180° \left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow 50°+60°+\angle AEB=180° \\ \Rightarrow \angle AEB=70° \end{gathered}$$

Question 27:

In the given figure, ∠OAB = 75°, ∠OBA = 55° and ∠OCD = 100°. Then ∠ODC = ?
(a) 20°
(b) 25°
(c) 30°
(d) 35°

Answer 27:

(c) 30°
In $∆OAB$, we have:
$$\begin{gathered} \angle OAB+\angle OBA+\angle AOB=180° \left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow 75°+55°+\angle AOB=180° \\ \Rightarrow \angle AOB=50° \end{gathered}$$
$\therefore \angle COD=\angle AOB=50° \left[\text{Vertically-Opposite Angles}\right]$
In $∆OCD$, we have:
$$\begin{gathered} \angle COD+\angle OCD+\angle ODC=180° \left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow 50°+100°+\angle ODC=180° \\ \Rightarrow \angle ODC=30° \end{gathered}$$

Question 28:

In the adjoining figure, y = ?
(a) 36°
(b) 54°
(c) 63°
(d) 72°

Answer 28:

(b) 54°

We have:
$$\begin{gathered} 3x+72=180° \left[\because AOB \text{is a straight line}\right] \\ \Rightarrow 3x=108 \\ \Rightarrow x=36 \end{gathered}$$
Also,
$$\begin{gathered} \angle AOC+\angle COD+\angle BOD=180° \left[\because AOB \text{is a straight line}\right] \\ \Rightarrow 36°+90°+y=180° \\ \Rightarrow y=54° \end{gathered}$$