Class 10 Maths

Triangle

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:

Triangle

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

Triangle

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

Triangle

Solution: (iii) Not similar

Triangle

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

Triangle

Solution: (v) Not similar

Triangle

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.

Triangle

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

Or, ∠DOC + 120° = 180°

Hence, ∠DOC = 180° - 120° = 60°

In Δ DOC;

∠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)`

Triangle

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

In Δ ABC and Δ EOC;

`(AE)/(EC)=(BO)/(OC)`

These are similar triangles as per BPT.

Similarly, in Δ BOD and Δ FOD;

`(BF)/(FD)=(AO)/(OD)`

In Δ ABC and Δ BAD;

`(BO)/(OC)=(AO)/(OD)`

Because diagonals of a trapezium divide each other in same ratio

From above three equations, it is clear;

`(AE)/(EC)=(BF)/(FD)`

Hence, Δ ABC ∼ Δ BAD

Using the third equation;

`(BO)/(OC)=(AO)/(OD)`

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.

Triangle

Solution: Proof:

∠1 = ∠2 (given)

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

`(QR)/(QS)=(QT)/(PR)`

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.

Triangle

Solution: In Δ RPQ and Δ RTS;

∠RPQ = ∠RTS (given)

∠PRQ = ∠TRS (common)

Hence, Δ RPQ ∼ Δ RTS (AAA Criterion)