1. State which pairs of triangles in Figure, are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:
Answers
Explanation :-
(i) Given, in ΔABC and ΔPQR,
- ∠A = ∠P = 60°
- ∠B = ∠Q = 80°
- ∠C = ∠R = 40°
Therefore by AAA similarity criterion,
=> ∴ ΔABC ~ ΔPQR
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(ii) Given, in ΔABC and ΔPQR,
- AB/QR = BC/RP = CA/PQ
By SSS similarity criterion,
=> ΔABC ~ ΔQRP
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(iii) Given, in ΔLMP and ΔDEF,
- LM = 2.7, MP = 2, LP = 3, EF = 5, DE = 4, DF = 6
- MP/DE = 2/4 = 1/2
- PL/DF = 3/6 = 1/2
- LM/EF = 2.7/5 = 27/50
Here , MP/DE = PL/DF ≠ LM/EF
Therefore, ΔLMP and ΔDEF are not similar.
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(iv) In ΔMNL and ΔQPR, it is given,
- MN/QP = ML/QR = 1/2
- ∠M = ∠Q = 70°
Therefore, by SAS similarity criterion
=> ∴ ΔMNL ~ ΔQPR
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(v) In ΔABC and ΔDEF, given that,
- AB = 2.5, BC = 3, ∠A = 80°, EF = 6, DF = 5, ∠F = 80°
- Here , AB/DF = 2.5/5 = 1/2
- And, BC/EF = 3/6 = 1/2
- ∠B ≠ ∠F
Hence, ΔABC and ΔDEF are not similar.
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(vi) In ΔDEF, by sum of angles of triangles, we know that,
- ∠D + ∠E + ∠F = 180°
- 70° + 80° + ∠F = 180°
- ∠F = 180° – 70° – 80°
- ∠F = 30°
Similarly, In ΔPQR,
- ∠P + ∠Q + ∠R = 180 (Sum of angles of Δ)
- ∠P + 80° + 30° = 180°
- ∠P = 180° – 80° -30°
- ∠P = 70°
Now, comparing both the triangles, ΔDEF and ΔPQR, we have
- ∠D = ∠P = 70°
- ∠F = ∠Q = 80°
- ∠F = ∠R = 30°
Therefore, by AAA similarity criterion,
=> Hence, ΔDEF ~ ΔPQR
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(i) Given, in ΔABC and ΔPQR,
∠A = ∠P = 60°
∠B = ∠Q = 80°
∠C = ∠R = 40°
Therefore by AAA similarity criterion,
∴ ΔABC ~ ΔPQR
(ii) Given, in ΔABC and ΔPQR,
AB/QR = BC/RP = CA/PQ
By SSS similarity criterion,
ΔABC ~ ΔQRP
(iii) Given, in ΔLMP and ΔDEF,
LM = 2.7, MP = 2, LP = 3, EF = 5, DE = 4, DF = 6
MP/DE = 2/4 = 1/2
PL/DF = 3/6 = 1/2
LM/EF = 2.7/5 = 27/50
Here , MP/DE = PL/DF ≠ LM/EF
Therefore, ΔLMP and ΔDEF are not similar.
(iv) In ΔMNL and ΔQPR, it is given,
MN/QP = ML/QR = 1/2
∠M = ∠Q = 70°
Therefore, by SAS similarity criterion
∴ ΔMNL ~ ΔQPR
(v) In ΔABC and ΔDEF, given that,
AB = 2.5, BC = 3, ∠A = 80°, EF = 6, DF = 5, ∠F = 80°
Here , AB/DF = 2.5/5 = 1/2
And, BC/EF = 3/6 = 1/2
⇒ ∠B ≠ ∠F
Hence, ΔABC and ΔDEF are not similar.
(vi) In ΔDEF, by sum of angles of triangles, we know that,
∠D + ∠E + ∠F = 180°
⇒ 70° + 80° + ∠F = 180°
⇒ ∠F = 180° – 70° – 80°
⇒ ∠F = 30°
Similarly, In ΔPQR,
∠P + ∠Q + ∠R = 180 (Sum of angles of Δ)
⇒ ∠P + 80° + 30° = 180°
⇒ ∠P = 180° – 80° -30°
⇒ ∠P = 70°
Now, comparing both the triangles, ΔDEF and ΔPQR, we have
∠D = ∠P = 70°
∠F = ∠Q = 80°
∠F = ∠R = 30°
Therefore, by AAA similarity criterion,
Hence, ΔDEF ~ ΔPQR