Thales Theorem Similarity of Triangles NCERT Exercise 6.2 Class Ten Mathematics

# Triangle

## Exercise 6.2 Part 1

Question 1: In the given figures, DE || BC. Find EC in first figure and AD in second figure.

Solution: In the first figure;

Δ ADE ~ Δ ABC (Because DE || BC)

Hence;

(AD)/(DC)=(AE)/(EC)

Or, (1.5)/(3)=(1)/(EC)

Or, EC=(3)/(1.5)=2 cm

Similarly, in the second figure;

Δ ADE ~ Δ ABC (Because DE || BC)

Hence;

(AD)/(DC)=(AE)/(EC)

Or, (AD)/(7.2)=(1.8)/(5.4)

Or, AD=(1.8xx7.2)/(5.4)=2.4 cm

Question 2: E and F are points on the sides PQ and PR respectively of a Δ PQR. For each of the following cases, state whether EF || QR.

(a) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

Solution: For EF || QR, the figure should fulfill following criterion;

(PE)/(EQ)=(PF)/(FR)

In this case;

(PE)/(EQ)=(3.9)/(3)=1.3

(PF)/(FR)=(3.6)/(2.4)=3/2

It is clear that;

(PE)/(EQ)≠(PF)/(FR)

Hence; EF and QR are not parallel.

(b) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

Solution: In this case;

(PE)/(EQ)=(4)/(4.5)=8/9

(PF)/(FR)=8/9

It is clear that;

(PE)/(EQ)=(PF)/(FR)

Hence; EF || QR

(c) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

Solution: In this case;

(PE)/(EQ)=(0.18)/(1.28-0.18)=(0.18)/(1.10)=(9)/(55)

(PF)/(FR)=(0.36)/(2.56-0.36)=(0.36)/(2.20)=(9)/(55)

It is clear that;

(PE)/(EQ)=(PF)/(FR)

Hence; EF || QR

Question 3: In the given figure, if LM || CB and LN || CD, prove that (AM)/(AB)=(AN)/(AD)

Solution: In Δ ABC and Δ AML;

Δ ABC ∼ Δ AML (because ML || BC)

Hence;

(AM)/(AB)=(AL)/(AC)

Hence;

(AN)/(AD)=(AL)/(AC)

From above two equations;

(AM)/(AB)=(AN)/(AD)

Question 4: In the given figure, DE || AC and DF || AE. Prove that (BF)/(FE)=(BE)/(EC)

Solution: In Δ ABC and ΔDBE;

(BE)/(EC)=(BD)/(BA)

Because Δ ABC ∼ Δ DBE

Similarly, in Δ ABE and Δ DBF;

(BF)/(FE)=(BD)/(BA)

From above two equations, it is clear;

(BF)/(FE)=(BE)/(EC)