PQR is a triangle, S is a point on the side QR of ΔPQR such that ∠PSR = ∠QPR. Given QP = 8 cm, PR = 6 cm and SR = 3 cm.

(i) Prove ΔPQR ∼ ΔSPR.

(ii) Find the length of QR and PS.

(iii) Find $\dfrac{\text{area of ΔPQR}}{\text{area of ΔSPR}}$.

If ΔABC and ΔDEF are similar triangles in which ∠A = 47° and ∠E = 83°, then ∠C equals :

1. 50°

2. 60°

3. 70°

4. 80°

If in ΔABC and ΔPQR, we have $\dfrac{AB}{QR} = \dfrac{BC}{PR} = \dfrac{CA}{PQ}$, then:

1. ΔPQR ∼ ΔCAB

2. ΔPQR ∼ ΔABC

3. ΔBCA ∼ ΔPQR

4. ΔCBA ∼ ΔPQR

In the given figure, AB = 24 cm, AC = 18 cm, DE = 12 cm, DF = 9 cm and ∠BAC = ∠EDF. Then ΔABC ∼ ΔEDF by the condition :

1. AAA

2. SAS

3. SSS

4. AAS