*Triangles*credits:openclipart |

*(MCQs asked in CBSE Papers)***Q1: In Δ ABC, AB = AC and ∠B = 50°. Then ∠C is equal to**

(a) 40°

(b) 50°

(c) 80°

(d) 130°

**Q2(CBSE 2010): In ΔABC, ∠C = ∠A and BC = 6 cm and AC = 5 cm. Then the length of AB is:**

(a) 6 cm

(b) 5 cm

(c) 3 cm

(d) 2.5 cm

**Q3: If one angle of a triangle is equal to the sum of other two angles, then the triangle is**

(a) an isosceles triangle

(b) an obtuse triangle

(c) an equilateral triangle

(d) a right triangle

**Q4(CBSE 2011): If ΔABC, is right angled at B, then :**

(a) AB = AC

(b) AC < AB

(c) AB = BC

(d) AC > AB

**Q5(CBSE 2011): In ΔABC if AB = BC, then :**

(a) ∠B > ∠C

(b) ∠A = ∠C

(c) ∠A = ∠B

(d) ∠A < ∠B

**Q6**

**(CBSE 2010)**: In ΔPQR, ∠P = 60°, ∠Q = 50°. Which side of the triangle is the longest ?(a) PQ

(b) QR

(c) PR

(d) none

**Q7**

**: P is a point on side BC of Δ ABC, such that AP bisects ∠BAC, then :****(CBSE 2011)**(a) BP = CP

(b) BA > BP

(c) BP > BA

(d) CP < CA

**Q8**

**(CBSE 2011**

**: In the give figure q-8, AD is the median, then ∠BAD is:****)**(a) 55°

(b) 50°

(c) 100°

(d) 40°

**Q9**

**(CBSE 2010****: If ΔABC ≅ ΔDEF by SSS congruence rule then :****)**(a) AB = EF, BC = FD, CA = DE

(b) AB = FD, BC = DE, CA = EF

(c) AB = DE, BC = EF, CA = FD

(d) AB = DE, BC = EF, ∠C = ∠F

**Q10**

**(CBSE 2011****: If Δ ABC is a right angled at B, then****)**(a) AB = AC

(b) AC < AB

(c) AB = AC

(d) AC > AB

**Q11: In a Δ ABC, ∠A = ∠C. If BC = 3 and AC = 4 then the perimeter of the triangle is:**

(a) 7

(b) 10

(c) 12

(d) 14

**Q12: From the following which condition is not possible for the congruence of two triangles ?**

(a) ASA

(b) AAS

(c) AAA

(d) SSS

**Q13: In a right angled triangle, if one acute angle is half the other, then the smallest angle is:**

(a) 15°

(b) 25°

(c) 30°

(d) 35°

Answers:

1: (b) 50° [

**Hint**: Δ ABC, AB = AC is an isosceles triangle, ∠B =∠C ]

2: (a) 6 cm [

**Hint**: Δ ABC, ∠C = ∠A, an isosceles triangle, BC = AB]

3: (d) a right triangle

[

**Hint**: Δ ABC, ∠A = ∠B + ∠C, ∵ ∠A + ∠B + ∠C = 180° ⇒ 2∠A = 180° ⇒ ∠A = 90° ]

4: (d) AC > AB [

**Hint**: Δ ABC, ∠B = 90°, AC is hypotenuse]

5: (b) ∠A = ∠C [

**Hint**: Δ ABC, AB = BC is an isosceles triangle, ∠A =∠C ]

6: (a) PQ [

**Hint**: ∠R = 70° is the largest, the side opposite to the largest angle is the longest.]

7: (b) BA > BP

[

**Hint**: Consider the triangle as show in fig-q7,

Given, ∠BAP = ∠CAP = z (say)

∠APC is an exterior angle for ΔBAP

⇒ ∠APC = ∠ABP + ∠BAP

⇒ ∠APC > ∠BAP (greater angle has greater side opposite to it.)

⇒ BA > BP]

8: (b) 50°

[

**Hint**:AB = AC ⇒ ∠ABD = ∠ACD = 40°. ⇒ ∠BAC = 100°

Δ ABD ≅ Δ ACD by SSS congruence

∠BAD = ∠CAD = 100°/2 = 50°]

9: (c) AB = DE, BC = EF, CA = FD

[

**Hint**: Remember CPCT for

*corresponding parts of congruent triangles*.]

10: (d) AC > AB

[

**Hint**: ∠B = 90°, AC is hypotenuse the longest side. ]

11: (b) 10 [

**Hint**: ∠A = ∠C, ⇒AB = BC = 3.]

12: (c) AAA

13: (c) 30° [

**Hint**: x + 2x + 90° = 180° ⇒ 3x = 90° ⇒ x = 30° ]

.

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