Solved MCQs: Congruence Criteria- SAS And ASA | Mathematics (Maths) Class 9 PDF Download

Q1: In triangle ABC, if AB=BC and ∠B = 70°, ∠A will be:
(a) 70°
(b) 110°
(c) 55°
(d) 130°
Ans: (c)
Sol: Given,
AB = BC
Hence, ∠A=∠C
And ∠B = 70°
By angle sum property of triangle we know:
∠A+∠B+∠C = 180°
2∠A+∠B=180°
2∠A = 180-∠B = 180-70 = 110°
∠A = 55°

Q2:  For two triangles, if two angles and the included side of one triangle are equal to two angles and the included side of another triangle. Then the congruency rule is:
(a) SSS
(b) ASA
(c) SAS
(d) None of the above
Ans: (b)

Q3: A triangle in which two sides are equal is called:
(a) Scalene triangle
(b) Equilateral triangle
(c) Isosceles triangle
(d) None of the above
Ans: (c)

Q4: The angles opposite to equal sides of a triangle are:
(a) Equal
(b) Unequal
(c) supplementary angles
(d) Complementary angles
Ans: (a)

Q5: If E and F are the midpoints of equal sides AB and AC of a triangle ABC. Then:
(a) BF = AC
(b) BF = AF
(c) CE = AB
(d) BF = CE
Ans: (d)
Sol: AB and AC are equal sides.
AB = AC (Given)
∠A = ∠A (Common angle)
AE = AF (Halves of equal sides)
∆ ABF ≅ ∆ ACE (By SAS rule)
Hence, BF = CE (CPCT)

Q6: ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB, respectively. Then:
(a) BE > CF
(b) BE < CF
(c) BE = CF
(d) None of the above
Ans: (c)
Sol: ∠A = ∠A (common arm)
∠AEB = ∠AFC (Right angles)
AB = AC (Given)
∴ ΔAEB ≅ ΔAFC
Hence, BE = CF (by CPCT)

Q7: If ABC and DBC are two isosceles triangles on the same base BC. Then:
(a) ∠ABD = ∠ACD
(b) ∠ABD > ∠ACD
(c) ∠ABD < ∠ACD
(d) None of the above
Ans: (a)
Sol: AD = AD (Common arm)
AB = AC (Sides of isosceles triangle)
BD = CD (Sides of isosceles triangle)
So, ΔABD ≅ ΔACD.
∴ ∠ABD = ∠ACD (By CPCT)

Q8: If ABC is an equilateral triangle, then each angle equals to:
(a) 90°
(b) 180°
(c) 120°
(d) 60°
Ans: (d)
Sol: Equilateral triangle has all its sides equal and each angle measures 60°.
AB= BC = AC (All sides are equal)
Hence, ∠A = ∠B = ∠C (Opposite angles of equal sides)
Also, we know that,
∠A + ∠B + ∠C = 180°
⇒ 3∠A = 180°
⇒ ∠A = 60°\
∴ ∠A = ∠B = ∠C = 60°

Q9: If AD is an altitude of an isosceles triangle ABC in which AB = AC. Then:
(a) BD = CD
(b) BD > CD
(c) BD < CD
(d) None of the above
Ans: (a)
Sol: In ΔABD and ΔACD,
∠ADB = ∠ADC = 90°
AB = AC (Given)
AD = AD (Common)
∴ ΔABD ≅ ΔACD (By RHS congruence condition)
BD = CD (By CPCT)

Q10: In a right triangle, the longest side is:
(a) Perpendicular
(b) Hypotenuse
(c) Base
(d) None of the above
Ans: (b)
Sol: In triangle ABC, right-angled at B.
∠B = 90
By angle sum property, we know:
∠A + ∠B + ∠C = 180
Hence, ∠A + ∠C = 90
So, ∠B is the largest angle.
Therefore, the side (hypotenuse) opposite to the largest angle will be the longest one.

Q11: Which of the following is not a criterion for congruence of triangles?
(a) SAS
(b) ASA
(c) SSA
(d) SSS
Ans: (c)
Sol: SSA is not a criterion for the congruence of triangles. Whereas SAS, ASA and SSS are the criteria for the congruence of triangles. 

Q12: In triangles ABC and PQR, AB = AC, ∠C = ∠P and ∠B = ∠Q. The two triangles are
(a) Isosceles and congruent
(b) Isosceles but not congruent
(c) Congruent but not isosceles
(d) Neither congruent nor isosceles
Ans: (b)
Sol: Consider two triangles, ABC and PQR. If the sides AB = AC and ∠C = ∠P and ∠B = ∠Q, then the two triangles are said to be isosceles, but they are not congruent.

Q13: In ∆ PQR, ∠R = ∠P and QR = 4 cm and PR = 5 cm. Then the length of PQ is
(a) 2 cm
(b) 2.5 cm
(c) 4 cm
(d) 5 cm
Ans: (c)
Sol: Given that, in a triangle PQR, ∠R = ∠P.
Since, ∠R = ∠P, the sides opposite to the equal angles are also equal.
Hence, the length of PQ is 4 cm.

Q14: If AB = QR, BC = PR and CA = PQ, then
(a) ∆ PQR ≅ ∆ BCA
(b) ∆ BAC ≅ ∆ RPQ
(c) ∆ CBA ≅ ∆ PRQ
(d) ∆ ABC ≅ ∆ PQR
Ans: (c)
Sol: Consider two triangles ABC and PQR.
Given that, AB = QR, BC = PR and CA = PQ.
By using Side-Side-Side (SSS rule),
We can say, ∆ CBA ≅ ∆ PRQ.

Q15: If ∆ ABC ≅ ∆ PQR, then which of the following is not true?
(a) AC = PR 
(b) BC = PQ
(c) QR = BC
(d) AB = PQ
Ans: (b)
Sol: Given that, ∆ ABC ≅ ∆ PQR
Hence, AB = PQ
BC = QR
AC =PR
Thus, BC = PQ is not true, if ∆ ABC ≅ ∆ PQR.

Q16: In ∆ ABC, BC = AB and ∠B = 80°. Then ∠A is equal to
(a) 40°
(b) 50°
(c) 80°
(d) 100°
Ans: (b) 
Sol: In a triangle, ABC, BC = AB and ∠B = 80°.Thus, the given triangle is an isosceles triangle.
By using the angle sum property of a triangle, we get
x + 80°+ x = 180°
2x + 80°= 180°
2x = 180°- 80°
2x = 100°
x = 100°/2 = 50°
Therefore, ∠A = 50°.

Q17: Two sides of a triangle are of lengths 5 cm and 1.5 cm. The length of the third side of the triangle cannot be
(a) 3.4 cm
(b) 3.6 cm
(c) 3.8 cm
(d) 4.1 cm
Ans: (a)
Sol: If two sides of a triangle are of lengths 5 cm and 1.5 cm, then the length of the third side of the triangle cannot be 3.4 cm. Because the difference between the two sides of a triangle should be less than the third side.

Q18: In ∆ ABC, AB = AC and ∠B = 50°. Then ∠C is equal to
(a) 40°
(b) 50°
(c) 80°
(d) 130°
Ans: (b)
Sol: Given that, in a triangle ABC, AB = AC and ∠B = 50°.
Since the given triangle is an isosceles triangle, the angles opposite to the equal sides are also equal. Hence, ∠C = 50°.

Q19: In ∆ PQR, if ∠R > ∠Q, then
(a) QR < PR 
(b) PQ < PR
(c) PQ > PR
(d) QR > PR 
Ans: (c)
Sol: In a triangle PQR, if ∠R > ∠Q, then PQ > PR, because the side opposite to the greater angle is longer.

Q20: It is given that ∆ ABC ≅ ∆ FDE and AB = 5 cm, ∠B = 40° and ∠A = 80°. Then which of the following is true? 
(a) DF = 5 cm, ∠F = 60°
(b) DF = 5 cm, ∠E = 60° 
(c) DE = 5 cm, ∠E = 60°
(d) DE = 5 cm, ∠D = 40
Ans: (b)
Sol: Given that, ∆ ABC ≅ ∆ FDE and AB = 5 cm, ∠B = 40° and ∠A = 80°.By using the CPCT rule, 
DF = 5 cm, since AB = 5 cm.
Again by using the CPCT rule, ∠E = ∠C.
Therefore, ∠E = ∠C = 180° – (∠A+ ∠B) [ By using angle sum property of triangle]
∠E = 180° – (80°+ 40°)
∠E = 180° – 120°
∠E = 60°