RS AGGARWAL CLASS 9 CHAPTER 8 TRIANGLES MCQ

 MULTIPLE CHOICE QUESTIONS

PAGE NO-257

Question 1:

In ∆ABC, if 3∠A = 4∠B = 6∠C then A : B : C = ?
(a) 3 : 4 : 6
(b) 4 : 3 : 2
(c) 2 : 3 : 4
(d) 6 : 4 : 3

Answer 1:

LCM of 3, 4 and 6 = 12

3∠A = 4∠B = 6∠C      (Given)

Dividing throughout by 12, we get

$\frac{3\angle \mathrm{A}}{12}=\frac{4\angle \mathrm{B}}{12}=\frac{6\angle \mathrm{C}}{12}$

$\Rightarrow \frac{\angle \mathrm{A}}{4}=\frac{\angle \mathrm{B}}{3}=\frac{\angle \mathrm{C}}{2}$

Let $\frac{\angle \mathrm{A}}{4}=\frac{\angle \mathrm{B}}{3}=\frac{\angle \mathrm{C}}{2}=k$, where k is some constant

Then, ∠A = 4k,  ∠B = 3k, ∠C = 2k

∴ ∠A : ∠B : ∠C = 4k : 3k : 2k = 4 : 3 : 2

Hence, the correct answer is option (b).

PAGE NO-258

Question 2:

In a ∆ABC, if ∠A − ∠B = 42° and ∠B − ∠C = 21° then ∠B = ?
(a) 32°
(b) 63°
(c) 53°
(d) 95°
Figure

Answer 2:

(c) 53°

Let 
$$\begin{gathered} \angle A-\angle B=42° ...\left(i\right) \text{and} \\ \angle B-\angle C=21° ...\left(ii\right) \\ \mathrm{Adding} \left(i\right) \mathrm{and} \left(ii\right),\text{ we get:} \\ \angle A-\angle C=63° \\ \angle B=\angle A-42° \left[\text{Using} \left(i\right)\right] \\ \angle C=\angle A-63° \left[\text{Using} \left(iii\right)\right] \end{gathered}$$

$$\begin{gathered} \therefore \angle A+\angle B+\angle C=180° \left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow \angle A+\angle A-42°+\angle A-63°=180° \\ \Rightarrow 3\angle A-105°=180° \\ \Rightarrow 3\angle A=285° \\ \Rightarrow \angle A=95° \\ \therefore \angle B=\left(95-42\right)° \\ \Rightarrow \angle B=53° \end{gathered}$$

Question 3:

In ∆ABC, side BC is produced to D. If ∠ABC = 50° and ∠ACD = 110° then ∠A = ?
(a) 160°
(b) 60°
(c) 80°
(d) 30°

Answer 3:

$$\begin{gathered} \therefore \angle A+\angle B=\angle ACD \\ \Rightarrow \angle A+50°=110° \\ \Rightarrow \angle A=60° \end{gathered}$$
Hence, the correct answer is option (b).

Question 4:

Side BC of ∆ABC has been produced to D on left and to E on right hand side of BC such that ∠ABD = 125° and ∠ACE = 130°. Then ∠A = ?
(a) 50°
(b) 55°
(c) 65°
(d) 75°

Answer 4:

(d) 75°

We have :
$$\begin{gathered} \mathbf{\therefore }\angle ABD+\angle ABC=180° \left[\because DE \text{is a straight line}\right] \\ \Rightarrow 125°+\angle ABC=180° \\ \Rightarrow \angle ABC=55° \end{gathered}$$
Also,
$$\begin{gathered} \angle ACE+\angle ACB=180° \\ \Rightarrow 130°+\angle ACB=180° \\ \Rightarrow \angle ACB=50° \\ \therefore \angle BAC+\angle ABC+\angle ACB=180° \left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow \angle BAC+55°+50°=180° \\ \Rightarrow \angle BAC=75° \\ \Rightarrow \angle A=75° \end{gathered}$$

Question 5:

In the given figure, the sides CB and BA of ∆ABC have been produced to D and E, respectively, such that ∠ABD = 110° and ∠CAE = 135°. Then ∠ACB = ?
(a) 65°
(b) 45°
(c) 55°
(d) 35°

Answer 5:

(a) 65°

We have :
$$\begin{gathered} \angle ABD+\angle ABC=180° \left[\because CBD \text{is a straight line}\right] \\ \Rightarrow 100°+\angle ABC=180° \\ \Rightarrow \angle ABC=70° \end{gathered}$$
Side AB of triangle ABC is produced to E.
$$\begin{gathered} \therefore \angle CAE=\angle ABC+\angle ACB \\ \Rightarrow 135°=70°+\angle ACB \\ \Rightarrow \angle ACB=65° \end{gathered}$$

Question 6:

The sides BC, CA and AB of ∆ABC have been produced to D, E and F, respectively. ∠BAE + ∠CBF + ∠ACD = ?
(a) 240°
(b) 300°
(c) 320°
(d) 360°

Answer 6:

(d) 360°

We have :

Question 7:

The the given figure, EAD ⊥ BCD. Ray FAC cuts ray EAD at a point A such that ∠EAF = 30°. Also, in ∆BAC, ∠BAC = x° and ∠ABC = (x + 10)°. Then, the value of x is
(a) 20
(b) 25
(c) 30
(d) 35

Answer 7:

In the given figure, ∠CAD = ∠EAF            (Vertically opposite angles)

∴ ∠CAD = 30°        

In ∆ABD,

∠ABD + ∠BAD + ∠ADB = 180°          (Angle sum property)

⇒ (x + 10)° + (x° + 30°) + 90° = 180°

⇒ 2x° + 130° = 180°

⇒ 2x° = 180° − 130° = 50°

⇒ x = 25

Thus, the value of x is 25.

Hence, the correct answer is option (b).

PAGE NO-259

Question 8:

In the given figure, two rays BD and CE intersect at a point A. The side BC of ∆ABC have been produced on both sides to points F and G respectively. If ∠ABF = x°, ∠ACG = y° and ∠DAE = z° then z = ?
(a) x + y – 180
(b) x + y + 180
(c) 180 – (x + y)
(d) x + y + 360°

Answer 8:

In the given figure, ∠ABF + ∠ABC = 180°         (Linear pair of angles)

∴ x° + ∠ABC = 180°

⇒ ∠ABC = 180° − x°        .....(1)

Also, ∠ACG + ∠ACB = 180°         (Linear pair of angles)

∴ y° + ∠ACB = 180°

⇒ ∠ACB = 180° − y°        .....(2)

Also, ∠BAC = ∠DAE = z°       .....(3)         (Vertically opposite angles)

In ∆ABC,

∠BAC + ∠ABC + ∠ACB = 180°       (Angle sum property)

∴ z° + 180° − x° + 180° − y° = 180°        [Using (1), (2) and (3)]

⇒ z = x + y − 180

Hence, the correct answer is option (a).

Question 9:

In the given figure, lines AB and CD intersect at a point O. The sides CA and OB have been produced to E and F respectively such that ∠OAE = x° and ∠DBF = y°.

If ∠OCA = 80°, ∠COA = 40° and ∠BDO = 70° then x° + y° = ?
(a) 190°
(b) 230°
(c) 210°
(d) 270°

Answer 9:

In the given figure, ∠BOD = ∠COA          (Vertically opposite angles)

∴ ∠BOD = 40°        .....(1)

In ∆ACO,

∠OAE = ∠OCA + ∠COA        (Exterior angle of a triangle is equal to the sum of two opposite interior angles)

⇒ x° = 80° + 40° = 120°           .....(2)

In ∆BDO,

∠DBF = ∠BDO + ∠BOD        (Exterior angle of a triangle is equal to the sum of two opposite interior angles)

⇒ y° = 70° + 40° = 110°           [Using (1)]         .....(3)

Adding (2) and (3), we get

x° + y° = 120° + 110° = 230°

Hence, the correct answer is option (b).

Question 10:

In a ∆ABC, it is given that ∠A : ∠B : ∠C = 3 : 2 : 1 and ∠ACD = 90°. If BC is produced to E then ∠ECD = ?
(a) 60°
(b) 50°
(c) 40°
(d) 25°

Answer 10:

Let $\angle A=\left(3x\right)°, \angle B=\left(2x\right)° \text{and} \angle C=x°$
Then,
$$\begin{gathered} 3x+2x+x=180° \left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow 6x=180° \\ \Rightarrow x=30° \end{gathered}$$
Hence, the angles are
$\angle A=3\times 30°=90°, \angle B=2\times 30°=60° \text{and} \angle C=30°$
Side BC of triangle ABC is produced to E.
$$\begin{gathered} \therefore \angle ACE=\angle A+\angle B \\ \Rightarrow \angle ACD+\angle ECD=90°+60° \\ \Rightarrow 90°+\angle ECD=150° \\ \Rightarrow \angle ECD=60° \end{gathered}$$
Hence, the correct answer is option (a).

Question 11:

In the given figure, BO and CO are the bisectors of ∠B and ∠C respectively. If ∠A = 50° then ∠BOC = ?
(a) 130°
(b) 100°
(c) 115°
(d) 120°

Answer 11:

(c) 115°

In $∆ABC$, we have:
$$\begin{gathered} \angle A+\angle B+\angle C=180° \left[\text{Sum of the angles of a triangle}\right] \\ \Rightarrow 50°+\angle B+\angle C=180° \\ \Rightarrow \angle B+\angle C=130° \\ \Rightarrow \frac{1}{2}\angle B+\frac{1}{2}\angle C=65° ...\left(i\right) \end{gathered}$$
In $∆OBC$, we have:
$$\begin{gathered} \angle OBC+\angle OCB+\angle BOC=180° \\ \Rightarrow \frac{1}{2}\angle B+\frac{1}{2}\angle C+\angle BOC=180° \left[\text{Using }\left(i\right)\right] \\ \Rightarrow 65°+\angle BOC=180° \\ \Rightarrow \angle BOC=115° \end{gathered}$$

PAGE NO-260

Question 12:

In the given figure, side BC of ∆ABC has been produced to a point D. If ∠A = 3y°, ∠B = x°, ∠C = 5y° and ∠CBD = 7y°. Then, the value of x is

(a) 60
(b) 50
(c) 45
(d) 35

Answer 12:

Disclaimer: In the question ∠ACD should be 7y°.

In the given figure, ∠ACB + ∠ACD = 180°         (Linear pair of angles)

∴ 5y° + 7y° = 180°

⇒ 12y° = 180°

⇒ y = 15         .....(1)

In ∆ABC,

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

∴ 3y° + x° + 5y° = 180°

⇒ x° + 8y° = 180°

⇒ x° + 8 × 15° = 180°              [Using (1)]

⇒ x° + 120° = 180°

⇒ x° = 180° − 120° = 60°

Thus, the value of x is 60.

Hence, the correct answer is option (a).