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Triangles- 2 - Free MCQ Practice Test with solutions, Class 9 Maths


MCQ Practice Test & Solutions: Test: Triangles- 2 (25 Questions)

You can prepare effectively for Class 9 Mathematics (Maths) Class 9 with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Triangles- 2". These 25 questions have been designed by the experts with the latest curriculum of Class 9 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 25 minutes
  • - Number of Questions: 25

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Test: Triangles- 2 - Question 1

In ΔABC, AB = 2.5 cm and BC = 6 cm. Then, the length of AC cannot be

Detailed Solution: Question 1

in triangle ABC

AB = 2.5 cm

BC = 6 cm

AC =  ?

in any triangle sum of two sides >  third side

=> AB +  BC > AC

=> 2.5 + 6 > AC

=> AC < 8.5

AB + AC > BC

=> 2.5 + AC > 6

=> AC > 3.5

BC + AC > AB

=> 6 + AC > 2.5

=> AC > -3.5

Taking all together

 3.5 < AC  < 8.5

3.6 ,  3.8 & 4  lies betwenn them

but not 3.4

Hence Length of AC can not be 3.4 cm

Test: Triangles- 2 - Question 2

In figure, ABCD is a quadrilateral in which AB = BC and AD = DC. Measure of ∠BCD is:

Detailed Solution: Question 2

In ΔABC ,as it is isoceles so,
∠BCA=36°  [ (180°-108°)/2 ]
In ΔADC ,as it is isoceles so,
∠DCA=69°  [ (180°-42°)/2 ]
∠BCD=∠BCA+∠DCA
∠BCD=36°+69°
∠BCD=105°

Test: Triangles- 2 - Question 3

In the adjoining figure, △ABC ≅ △ADC. If ∠BAC = 30 and ∠ABC = 100 then ∠ACD is equal to

Detailed Solution: Question 3

Given:

  • Triangles △ABC ≅ △ADC
  • ∠BAC = 30°
  • ∠ABC = 100°

To find: ∠ACD

Step 1: Since △ABC ≅ △ADC, their corresponding angles are equal. Therefore, corresponding to:

  • ∠BAC in △ABC, we have ∠DAC in △ADC, so ∠DAC = 30°
  • ∠ABC in △ABC, we have ∠ADC in △ADC, so ∠ADC = 100°

Step 2: Use the fact that the sum of angles in a triangle is 180°.

In triangle ADC, the angles are ∠DAC, ∠ADC, and ∠ACD. So,

∠DAC + ∠ADC + ∠ACD = 180°

Substitute the known values:

30° + 100° + ∠ACD = 180°

Step 3: Solve for ∠ACD:

∠ACD = 180° - 30° - 100° = 50°

Answer: ∠ACD = 50°, which corresponds to option D.

Test: Triangles- 2 - Question 4

It is not possible to construct a triangle when the lengths of its sides are

Detailed Solution: Question 4

To check if a triangle can be formed, the sum of any two sides must be greater than the third side.

a) 4, 6, 6
4 + 6 = 10 > 6
6 + 6 = 12 > 4
4 + 6 = 10 > 6
Triangle possible

b) 9.3, 5.2, 7.4
9.3 + 5.2 = 14.5 > 7.4
9.3 + 7.4 = 16.7 > 5.2
5.2 + 7.4 = 12.6 > 9.3
Triangle possible

c) 6, 7, 8
6 + 7 = 13 > 8
7 + 8 = 15 > 6
6 + 8 = 14 > 7
Triangle possible

d) 5.3, 2.2, 3.1
5.3 + 2.2 = 7.5 > 3.1
5.3 + 3.1 = 8.4 > 2.2
2.2 + 3.1 = 5.3 = 5.3 (not greater)
Triangle not possible

Answer: d) 5.3 cm, 2.2 cm, 3.1 cm

Test: Triangles- 2 - Question 5

In fig, AC = BC and ∠ACY = 140. Find X and Y:

Detailed Solution: Question 5

∠ACB = 40ο
as ABC is a isoceles trinagle
so ∠BAC = ∠ABC = a
a + a + 40 = 180ο
2a = 140ο 
a = 70ο
x + a =180ο   and  y +a = 180ο
x = 110ο  and y = 110ο

Test: Triangles- 2 - Question 6

In triangles ABC and DEF, AB = FD and ∠A=∠D. The two triangles will be congruent by SAS axiom if :

Detailed Solution: Question 6


The triangles ABC and DEF are

It is given that

AB = FD and ∠A = ∠D

For a triangle to be congruent from the SAS axiom two sides and included angle must be equal

For ∆ ABC ≅ ∆ DFE from SAS axiom we require AC = DE

Therefore, the two triangles will be congruent by the SAS axiom if AC = DE.

Test: Triangles- 2 - Question 7

D is a Point on the Side BC of a △ABC such that AD bisects ∠BAC then:

Detailed Solution: Question 7

Given:
In △ABC, point D lies on side BC such that AD bisects ∠BAC.

By the Angle Bisector Theorem, the angle bisector divides the opposite side in the ratio of the sides containing the angle.

So,
BD / DC = BA / CA

Checking the options:

(A) BD = CD
This is true only when BA = CA, which is not given. So this is incorrect.

(B) CD > CA
CD is only a part of BC and cannot be compared directly with side CA. So this is incorrect.

(C) BD > BA
BD is a part of side BC, while BA is a full side of the triangle. Hence BD cannot be greater than BA. This is incorrect.

(D) BA > BD
BA is a full side of the triangle and BD is only a part of another side. So BA is greater than BD. This is correct.

Correct answer: d) BA > BD

Test: Triangles- 2 - Question 8

In the adjoining figure, O is Mid – point of AB. If ∠ACO = ∠BDO, then ∠OAC is equal to

Test: Triangles- 2 - Question 9

In the adjoining fig, AD = BC and ∠BAD = ∠ABC. If ∠BAD = 120 and ∠ABD = 35, then ∠CAD is

Test: Triangles- 2 - Question 10

In fig. which of the following statement is true?

Test: Triangles- 2 - Question 11

In △ABC, ∠A = 35 and ∠B = 65, then the longest side of the triangle is:

Test: Triangles- 2 - Question 12

In △ABC, if ∠A = 45 and ∠B = 70, then the shortest and the longest sides of the triangle are respectively,

Test: Triangles- 2 - Question 13

In the adjoining figure, AB = BC and ∠ABD = ∠CBD, then another angle which measures 30 is

Detailed Solution: Question 13

The two triangles are congruent by SAS congruency.

Test: Triangles- 2 - Question 14

ABC ≅ △PQR. If AB=5 cm, and  ∠A = 60ο then which of the following is true?

Detailed Solution: Question 14

ABC ≅ △PQR
so AB = PQ and ∠A = ∠P

Test: Triangles- 2 - Question 15

In fig., △ABD ≅ △ACD, AB = AC, name the criteria by which the triangles are congruent:

Test: Triangles- 2 - Question 16

P is a point on side BC of a △ABC such that AP bisects △BAC. Then

Test: Triangles- 2 - Question 17

In the adjoining figure, AB = AC and ∠A = 70, then ∠C is 

Test: Triangles- 2 - Question 18

In the adjoining figure, AC = BD. If ∠CAB = ∠DBA, then ∠ACB is equal to

Test: Triangles- 2 - Question 19

In △ABC, if ∠B = 30 and ∠C = 70, then which of the following is the longest side?

Test: Triangles- 2 - Question 20

In fig, in △ABD, AB = AC, then the value of x is:

Test: Triangles- 2 - Question 21

In the given figure, AD is the median, then ∠BAD is:

Test: Triangles- 2 - Question 22

In the adjoining fig, PQ = PR. If ∠QPR = 48, then value of x is:

Detailed Solution: Question 22

- In triangle PQR, PQ = PR , making it an isosceles triangle.

- In an isosceles triangle, the angles opposite the equal sides are equal.

- Given angle QPR = 48 degrees ,

let angle PQR = angle PRQ = y

- The sum of angles in a triangle is  180.

- Therefore, y + y + 48 = 180

- Solving: 2y = 132 degrees

therefore y = 66 degrees

 angle y + angle x = 180 degrees

66 + angle x = 180 degree 

angle x = 180-66

angle x = 132 degree 

Test: Triangles- 2 - Question 23

In the adjoining figure, ABCD is a quadrilateral in which AD = CB and AB = CD, then ∠ACB is equal to

Test: Triangles- 2 - Question 24

In the adjoining figure, PQ > PR. If OQ and OR are bisectors of ∠Q and ∠R respectively, then

Test: Triangles- 2 - Question 25

In the adjoining fig, AD = BC and ∠BAD = ∠ABC. If ∠BAD = 120 and ∠ABD = 35, then ∠CAD is

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