Question

E and F are points on the sides PQ and PR respectively of a ∆PQR for each of the following cases, state whether EF||QR:​

Answer 1

Step-by-step explanation:

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Answer 2

Question :-

• E and F are points on the sides PQ and PR respectively of a ∆PQR for each of the following cases, state whether EF||QR:

To find :-

• E and F are points on the sides PQ and PR respectively of a ∆PQR

Solution :-

Case 1st :-

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

F2.4 cm

$\large{ \sf \underline{ ╬ \: Using \: Basic \: proportionality \: theorem \: ╬ \: \: }}$

☞$</p><p>\bold{∴ \dfrac{ PE}{EQ}} = \dfrac{3.9}{3} = \dfrac{39}{30} = \dfrac{13}{10} = 1.3$

☞$\bold{ \dfrac{ PF}{FR} } = \dfrac{3.6}{2.4} = \dfrac{36}{24} = \dfrac{3}{2} = 1.5$

☞$\bold{\dfrac{ PE}{ EQ} }≠ \bold{ \dfrac{ PF}{FR}}$

So, EF is not parallel to QR.

Case 2 :-

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

$\large{ \sf \underline{ ╬ \: Using \: Basic \: proportionality \: theorem \: ╬ \: \: }}$

☞$\bold{∴ \dfrac{ PE}{QE}} = \dfrac{4}{4.5} = \dfrac{40}{45} = \dfrac{8}{9}$

☞$\bold{ \dfrac{ PF}{FR} } = \dfrac{8}{9}$

☞$\bold{ \dfrac{ PE}{QE}} = \bold{ \dfrac{ PF}{FR}}$

So, EF is parallel to QR.

Case 3 :-

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

$\large{ \sf \underline{ ╬ \: Using \: Basic \: proportionality \: theorem \: ╬ \: \: }}$

EQ = PQ − PE = 1.28 − 0.18 = 1.10 cm

FR = PR − PF = 2.56 − 0.36 = 2.20 cm

↪$\bold{\dfrac{ PE}{ EQ} } = \dfrac{0.18}{1.10} = \dfrac{18}{110} = \dfrac{9}{55} ...(1)$

↪$\bold{ \dfrac{ P E}{FR} } = \dfrac{0.36}{2.20} = \dfrac{36}{220} = \dfrac{9}{55} ..(2)$

↪$\bold{∴\dfrac{ PE}{EQ}}= \bold{ \dfrac{ PF}{FR}}$

So, EF is parallel to QR.