Exercise-6.2, Class 10th, Maths, Chapter 6, NCERT

Exercise 6.2 Solutions

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Question 1

(For Fig. 6.17 (i) and (ii))
The values of EC and AD depend on the numbers shown in your book’s diagram.

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Question 2

E and F are points on PQ and PR of ΔPQR.
In each case, state whether EF ∥ QR.

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

$$\frac{PE}{EQ}=\frac{3.9}{3}=1.3,\frac{PF}{FR}=\frac{3.6}{2.4}=1.5$$

Since the ratios are not equal,

$$EF\parallel ̸QR$$

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

$$\frac{PE}{EQ}=\frac{4}{4.5}=\frac{8}{9},\frac{PF}{FR}=\frac{8}{9}$$

Ratios equal ⇒

$$EF\parallel QR$$

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

$$\frac{PE}{PQ}=\frac{0.18}{1.28}=0.1406,\frac{PF}{PR}=\frac{0.36}{2.56}=0.1406$$

Ratios equal ⇒

$$EF\parallel QR$$

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Question 3

If LM ∥ CB and LN ∥ CD, prove that

$$\frac{AM}{AN}=\frac{AB}{AD}$$

From LM ∥ CB: ΔAML ∼ ΔABC

$$\frac{AM}{AB}=\frac{AL}{AC}(1)$$

From LN ∥ CD: ΔANL ∼ ΔADC

$$\frac{AN}{AD}=\frac{AL}{AC}(2)$$

From (1) and (2):

$$\frac{AM}{AB}=\frac{AN}{AD}\Rightarrow \frac{AM}{AN}=\frac{AB}{AD}$$

Hence proved.

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Question 4

If DE ∥ AC and DF ∥ AE, prove that

$$\frac{BF}{BE}=\frac{FE}{EC}$$

Because DE ∥ AC and DF ∥ AE,
∠BFE = ∠BEC and ∠BEF = ∠BCE.
Therefore ΔBEF ∼ ΔBEC and so

$$\frac{BF}{BE}=\frac{FE}{EC}$$

Hence proved.

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Question 5

If DE ∥ OQ and DF ∥ OR, prove that EF ∥ QR.

From parallels: ∠DEF = ∠OQR and ∠DFE = ∠ORQ.
Hence ΔDEF ∼ ΔOQR ⇒ corresponding sides ∥ ⇒

$$EF\parallel QR$$

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Question 6

In ΔOPR, A, B, C are points on OP, OQ, OR such that AB ∥ PQ and AC ∥ PR.
Show that BC ∥ QR.

From AB ∥ PQ ⇒ ΔOAB ∼ ΔOPQ
and AC ∥ PR ⇒ ΔOAC ∼ ΔOPR.
Hence corresponding angles equal ⇒

$$\mathrm{\angle }ABC=\mathrm{\angle }PQR,\mathrm{\angle }ACB=\mathrm{\angle }PRQ$$

Therefore

$$BC\parallel QR$$

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Question 7

To prove: A line drawn through the midpoint of one side of a triangle parallel to another side bisects the third side.

Let ΔABC, D is midpoint of AB.
Draw DE ∥ BC, meeting AC at E.
By Basic Proportionality Theorem:

$$\frac{AD}{DB}=\frac{AE}{EC}$$

But AD = DB ⇒ AD/DB = 1
Hence AE/EC = 1 ⇒ AE = EC.
Therefore E is midpoint of AC.
Hence proved.

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Question 8

To prove: The line joining midpoints of two sides of a triangle is parallel to the third side.

Let ΔABC with D and E midpoints of AB and AC respectively.
Then

$$\frac{AD}{DB}=1,\frac{AE}{EC}=1$$

Since the sides are divided in the same ratio, by the converse of the Basic Proportionality Theorem,

$$DE\parallel BC$$

Hence proved.

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Summary Table