In $△$ PQR, E and F are points on the sides PQ and PR respectively. Then verify that EF || QR if PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm.

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Solution

The basic proportionality theorem states that if a line is drawn parallel to one side of a triangle and it intersects the other two sides at two distinct points then it divides the two sides in the same ratio.

It is given that $P Q = 1.28$ cm, $P R = 2.56$ cm, $P E = 0.18$ cm and $P F = 0.36$ cm.

We find the following:

$E Q = P Q - P E = 1.28 - 0.18 = 1.10$ cm
$F R = P R - P F = 2.56 - 0.36 = 2.20$ cm

Using the basic proportionality theorem, we have

$\frac{P E}{E Q} = \frac{0.18}{1.10} = 0.1636 \frac{P F}{F R} = \frac{0.36}{2.20} = 0.1636$

Since $\frac{P E}{E Q} = \frac{P F}{F R}$

Hence, $E F$ is parallel to $Q R$ .

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Question 2 (iii)
E and F are points on the sides PQ and PR respectively of a $Δ P Q R$ . For the following case, state whether EF || QR.
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.63 cm

Question 2 (iii)
E and F are points on the sides PQ and PR respectively of a $Δ P Q R$ . For the following case, state whether EF || QR.
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.63 cm