NCERT Solutions 10th Maths Chapter 6 Triangles Exercise 6.3

NCERT Solutions For Class 10 Maths Chapter 6 Exercise 6.3 we will restart our exploration of the world of Triangles. Thus, we are providing you Chapter 6 Triangles NCERT Class 10 Maths Solutions that will help in achieving more marks. You don't have to wander and waste your precious time in finding best CBSE NCERT Solutions.

Exercise 6.3
1. State which pairs of triangles in Fig. 6.34 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:

Answer
(i) In △ABC and △PQR, we have
∠A = ∠P = 60° (Given)
∠B = ∠Q = 80° (Given)
∠C = ∠R = 40° (Given)
∴ ΔABC ~ ΔPQR (AAA similarity criterion)
(ii) In △ABC and △PQR, we have
AB/QR = BC/RP = CA/PQ
∴ ΔABC ~ ΔQRP (SSS similarity criterion)
(iii) In △LMP and △DEF, we have
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
Hence, △LMP and △DEF are not similar.
(iv) In △MNL and △QPR, we have
MN/QP = ML/QR = 1/2
∠M = ∠Q = 70°
∴ ΔMNL ~ ΔQPR (SAS similarity criterion)
(v) In ΔABC and ΔDEF, we have
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, △ABC and △DEF are not similar.
(vi) In △DEF,we have
∠D + ∠E + ∠F = 180° (sum of angles of a triangle)
⇒ 70° + 80° + ∠F = 180°
⇒ ∠F = 180° - 70° - 80°
⇒ ∠F = 30°
In PQR, we have
∠P + ∠Q + ∠R = 180 (Sum of angles of △)
⇒ ∠P + 80° + 30° = 180°
⇒ ∠P = 180° - 80° -30°
⇒ ∠P = 70°
In △DEF and △PQR, we have
∠D = ∠P = 70°
∠F = ∠Q = 80°
∠F = ∠R = 30°
Hence, △DEF ~ △PQR (AAA similarity criterion)
2. In the fig 6.35, ΔODC ∝ ¼ ΔOBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB.
Answer
DOB is a straight line.
Therefore, ∠DOC + ∠COB = 180°
⇒ ∠DOC = 180° - 125°
= 55°
In △DOC,
∠DCO + ∠CDO + ∠DOC = 180°
(Sum of the measures of the angles of a triangle is 180º.)
⇒ ∠DCO + 70º + 55º = 180°
⇒ ∠DCO = 55°
It is given that △ODC ~ △OBA.
∴ ∠OAB = ∠OCD [Corresponding angles are equal in similar triangles.]
⇒ ∠OAB = 55°
∴ ∠OAB = ?OCD [Corresponding angles are equal in similar triangles.]
⇒ ∠OAB = 55°
3.E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.
Answer

In △ABE and △CFB,
∠A = ∠C (Opposite angles of a parallelogram)
∠AEB = ∠CBF (Alternate interior angles as AE || BC)
∴ △ABE ~ △CFB (By AA similarity criterion)