Exercise 6.3 Part 1

Question 1: State which pairs of triangles in the given 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:


Solution: (i) Δ ABC ∼ Δ PQR (AAA criterion)


Solution: (ii) Δ ABC ∼ Δ QRP (SSS criterion)


Solution: (iii) Not similar


Solution: (iv) Δ LMN ∼ Δ PQR (SAS criterion)


Solution: (v) Not similar


Solution: (vi) Δ DEF ∼ Δ PQR (AAA criterion)

Question 2: In the given figure, Δ ODC ∼ Δ OBA, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.


Solution: ∠DOC + ∠COB = 180° (Linear pair of angles)

Or, ∠DOC + 120° = 180°

Hence, ∠DOC = 180° - 120° = 60°


∠DCO + ∠CDO + ∠DOC = 180° (Angle sum of triangle)

Or, ∠DCO + 70° + 60° = 180°

Or, ∠DCO = 180° - 130° = 50°

∠OCD = ∠OAB = 50° (Because Δ ODC ∼ Δ OBA; given)

Question 3: Diagonals AC and BD of a trapezium ABCD with AB || CD intersect each other at the point O. Using a similarity criterion for two triangles, show that `(OA)/(OC)=(OB)/(OD)`


Solution: Draw a line EF || CD which is passing through O.

In Δ ABC and Δ EOC;


These are similar triangles as per BPT.

Similarly, in Δ BOD and Δ FOD;


In Δ ABC and Δ BAD;


Because diagonals of a trapezium divide each other in same ratio

From above three equations, it is clear;


Hence, Δ ABC ∼ Δ BAD

Using the third equation;


Or, `(AO)/(BO)=(CO)/(DO)` proved

Question 4: In the given figure, `(QR)/(QS)=(QT)/(PR)` and ∠1 = ∠2. Show that Δ PQS ∼ Δ TQR.


Solution: Proof:

∠1 = ∠2 (given)

Hence, PQ = PR (Sides opposite to equal angles are equal in isosceles triangle)


Or, `(QR)/(QS)=(QT)/(PQ)`

Hence, PS || TR

And; Δ PQS ∼ Δ TQR proved

Question 5: S and T are points on sides PR and QR of Δ PQR such that ∠P = ∠RTS. Show that Δ RPQ ∼ Δ RTS.


Solution: In Δ RPQ and Δ RTS;

∠RPQ = ∠RTS (given)

∠PRQ = ∠TRS (common)

Hence, Δ RPQ ∼ Δ RTS (AAA Criterion)

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