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Question

Mathematics Question on Triangles

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

Answer

(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

PEEQ=3.93\frac{PE}{EQ}=\frac{3.9}{3}=1.3
PFFR\frac{PF}{FR}=3.62.6\frac{3.6}{2.6}=1.5

Hence, PEEQPFFR\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

PEEQ=44.5=89\frac{PE}{EQ}=\frac{4}{4.5}=\frac{8}{9}
PFFR=89\frac{PF}{FR}=\frac{8}{9}

Hence, PEEQ=PFFR\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

PEEQ=0.181.28=18128=964\frac{PE}{EQ}=\frac{0.18}{1.28}=\frac{18}{128}=\frac{9}{64}
PFPR=0.362.56=964\frac{PF}{PR}=\frac{0.36}{2.56}=\frac{9}{64}

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