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)
Sol: The ASA rule states that if two angles and the included side of one triangle are equal to the corresponding two angles and the included side of another triangle, then the two triangles are congruent.

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)
Sol: A triangle with two equal sides is called an isosceles triangle. In this type of triangle, the angles opposite the equal sides are also equal.

Q4: The angles opposite to equal sides of a triangle are:
(a) Equal
(b) Unequal
(c) supplementary angles
(d) Complementary angles
Ans: (a)
Sol: In a triangle, the angles opposite to the equal sides are always equal. This property is a characteristic of isosceles triangles.

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°