## E and F are points on the sides PQ and PR respectively of a ΔPQR. If EF||QR, PE = (x-1)cm, EQ=(x+1)cm, PF=(x+1)cm and FR=(2x-1)cm, then x=__

Question

E and F are points on the sides PQ and PR respectively of a ΔPQR. If EF||QR, PE = (x-1)cm, EQ=(x+1)cm, PF=(x+1)cm and FR=(2x-1)cm, then x=_________.

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2 months 2021-10-09T18:51:01+00:00 2 Answers 0 views 0

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2. Step-by-step explanation:

In ΔPQR, E and F are two points on side PQ and PR respectively.

(i) PE = 3.9 cm, EQ = 3 cm (Given)

PF = 3.6 cm, FR = 2,4 cm (Given)

∴ PE/EQ = 3.9/3 = 39/30 = 13/10 = 1.3 [By using Basic proportionality theorem]

And, PF/FR = 3.6/2.4 = 36/24 = 3/2 = 1.5

So, PE/EQ ≠ PF/FR

Hence, EF is not parallel to QR.

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

∴ PE/QE = 4/4.5 = 40/45 = 8/9 [By using Basic proportionality theorem]

And, PF/RF = 8/9

So, PE/QE = PF/RF

Hence, EF is parallel to QR.

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

Here, EQ = PQ – PE = 1.28 – 0.18 = 1.10 cm

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

So, PE/EQ = 0.18/1.10 = 18/110 = 9/55 … (i)

And, PE/FR = 0.36/2.20 = 36/220 = 9/55 … (ii)

∴ PE/EQ = PF/FR.

Hence, EF is parallel to QR.