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.

similar triangles exercise solution

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.

similar triangles exercise solution

(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)`

similar triangles exercise solution

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)`

similar triangles exercise solution

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)`

similar triangles exercise solution

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)`



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