Class 10 Mathematics

Triangle

Exercise 6.2 (NCERT) 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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Theorem (Part 1)

Theorem (Part 2)

Exercise 6.1

Exercise 6.2 (Part 2)

Exercise 6.3 (Part 1)

Exercise 6.3 (Part 2)

Exercise 6.3 (Part 3)

Exercise 6.4

Exercise 6.5 (Part 1)

Exercise 6.5 (Part 2)

Exercise 6.6 (Part 1)

Exercise 6.6 (Part 2)