RS AGGARWAL CLASS 9 CHAPTER 8 TRIANGLES MCQ

 MULTIPLE CHOICE QUESTIONS


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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

3A12=4B12=6C12

A4=B3=C2

Let A4=B3=C2=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).

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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 
AB=42°   ...(i) andBC=21°   ...(ii)Adding (i) and (ii), we get:AC=63°B=A42°   [Using (i)]C=A63°   [Using (iii)]

A+B+C=180°   [Sum of the angles of a triangle]A+A42°+A63°=180°3A105°=180°3A=285°A=95°B=(9542)°B=53°

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:

A+B=ACDA+50°=110°A=60°
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 :

Also,

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 :

Side AB of triangle ABC is produced to E.

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 :
1+BAE=180°   ...i2+CBF=180°   ...ii and3+ACD=180°   ...iii
Adding (i), (ii) and (iii), we get:1+2+3+BAE+CBF+ACD=540°
180°+BAE+CBF+ACD=540°   1+2+3=180°BAE+CBF+ACD=360°

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).

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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)]

⇒ zxy − 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 A=3x°, B=2x° and C=x°
Then,
3x+2x+x=180°   Sum of the angles of a triangle6x=180°x=30°
Hence, the angles are
A=3×30°=90°, B=2×30°=60° and C=30°
Side BC of triangle ABC is produced to E.
ACE=A+BACD+ECD=90°+60°90°+ECD=150°ECD=60°
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:
A+B+C=180°   Sum of the angles of a triangle50°+B+C=180°B+C=130°12B+12C=65°   ...i
In OBC, we have:
OBC+OCB+BOC=180°12B+12C+BOC=180°   Using i65°+BOC=180°BOC=115°

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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).

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