If AB = QR, BC = PR and CA = PQ, then

(A) ∆ ABC ≅ ∆ PQR

(B) ∆ CBA ≅ ∆ PRQ

(C) ∆ BAC ≅ ∆ RPQ

(D) ∆ PQR ≅ ∆ BCA

. In ∆ ABC, AB = AC and ∠B = 50°. Then ∠C is equal to

(A) 40°

(B) 50°

(C) 80°

(D) 130

In ∆ ABC, BC = AB and ∠B = 80°. Then ∠A is equal to

(A) 80°

(B) 40°

(C) 50°

(D) 100°

In ∆ PQR, ∠R = ∠P and QR = 4 cm and PR = 5 cm. Then the length of PQ is

(A) 4 cm

(B) 5 cm

(C) 2 cm

(D) 2.5 cm

D is a point on the side BC of a ∆ ABC such that AD bisects ∠BAC.

Then

(A) BD = CD

(B) BA > BD

(C) BD > BA

(D) CD > CA

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°

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) 3.6 cm

(B) 4.1 cm

(C) 3.8 cm

(D) 3.4 cm

. In ∆ PQR, if ∠R > ∠Q, then

(A) QR > PR

(B) PQ > PR

(C) PQ < PR

(D) QR < PR

. In triangles ABC and PQR, AB = AC, ∠C = ∠P and ∠B = ∠Q. The two triangles are

(A) isosceles but not congruent

(B) isosceles and congruent

(C) congruent but not isosceles

(D) neither congruent nor isosceles

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

(A) BC = EF

(B) AC = DE

(C) AC = EF

(D) BC = DE