
Given: \(\frac{QR}{QS}=\frac{QT}{PR}\) and \(\angle 1 =\angle 2\)
To Prove: ΔPQR∼ΔTQR
Proof: In ∆PQR,
\(\angle\)PQR = \(\angle\)PRQ
∴ PQ = PR ………………(i)
Using (i) we obtain
\(\frac{QR}{QS}=\frac{QT}{QP}\).............(ii)
In ΔPQS and ΔTQR,
\(\frac{QR}{QS}=\frac{QT}{QP}\) [using (ii)]
\(\angle\)Q=\(\angle\)Q
∴ ΔPQS∼ΔTQR
Hence Proved
Fill in the blanks using the correct word given in the brackets :
(i) All circles are __________. (congruent, similar)
(ii) All squares are __________. (similar, congruent)
(iii) All __________ triangles are similar. (isosceles, equilateral)
(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are __________ and (b) their corresponding sides are __________. (equal, proportional)



| Case No. | Lens | Focal Length | Object Distance |
|---|---|---|---|
| 1 | \(A\) | 50 cm | 25 cm |
| 2 | B | 20 cm | 60 cm |
| 3 | C | 15 cm | 30 cm |