Question:

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

Updated On: Nov 2, 2023
Hide Solution
collegedunia
Verified By Collegedunia

Solution and Explanation

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

\(\frac{PE}{EQ}=\frac{3.9}{3}\)=1.3
\(\frac{PF}{FR}\)=\(\frac{3.6}{2.6}\)=1.5

Hence, \(\frac{PE}{EQ}\neq \frac{PF}{FR}\)
Therefore, EF is not parallel to QR


(ii)PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm
Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm

\(\frac{PE}{EQ}=\frac{4}{4.5}=\frac{8}{9}\)
\(\frac{PF}{FR}=\frac{8}{9}\)

Hence, \(\frac{PE}{EQ}=\frac{PF}{FR}\)
Therefore, EF is parallel to QR.


(iii)PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.63 cm
Given that: PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm

\(\frac{PE}{EQ}=\frac{0.18}{1.28}=\frac{18}{128}=\frac{9}{64}\)
\(\frac{PF}{PR}=\frac{0.36}{2.56}=\frac{9}{64}\)

Hence, \(\frac{PE}{PQ}=\frac{PF}{PR}\)
Therefore, EF is parallel to QR.

Was this answer helpful?
0
0

Top Questions on Triangles

View More Questions