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

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

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

(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.63 cm

Asked by Pragya Singh | 1 year ago |  79

1 Answer

Solution :-

Given, in ΔPQR, E and F are two points on side PQ and PR respectively. See the figure below;

 

Triangles Exercise 6.2 Answer 3

(i) Given, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2,4 cm

Therefore, by using Basic proportionality theorem, we get,

\( \dfrac{PE}{EQ}=\dfrac{ 3.9}{3}=\dfrac{39}{30}=\dfrac{13}{10}= 1.3\)

And \( \dfrac{PF}{FR}=\dfrac{ 3.6}{2.4}=\dfrac{36}{24}=\dfrac{3}{2}= 1.3\)

So, we get, \( \dfrac{ PE}{EQ}≠\dfrac{PF}{FR}\)

Hence, EF is not parallel to QR.

 

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

Therefore, by using Basic proportionality theorem, we get,

\( \dfrac{PE}{EQ}=\dfrac{ 4}{4.5}=\dfrac{40}{45}=\dfrac{8}{9}\)

And, \( \dfrac{PF}{RF}=\dfrac{ 8}{9}\)

So, we get here,

\( \dfrac{PE}{QE}\) = \( \dfrac{PF}{RF}\)

Hence, EF is parallel to QR.

 

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

From the figure,

EQ = PQ – PE = 1.28 – 0.18 = 1.10 cm

And, FR = PR – PF = 2.56 – 0.36 = 2.20 cm

So, \( \dfrac{PE}{EQ}\)\( \dfrac{0.18}{0.10}\)\(\dfrac{18}{110}\) = \( \dfrac{9}{55}\)…………. (i)

And, \( \dfrac{PF}{FR}\) = \( \dfrac{ 0.36}{2.20}=\dfrac{36}{220 }=\dfrac{9}{5}\)………… (ii)

So, we get here,

\( \dfrac{PE}{EQ}\) = \( \dfrac{PF}{FR}\)

Hence, EF is parallel to QR.

Answered by Abhisek | 1 year ago

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