Question 2 iiiE and F are points on the sides PQ and PR respectively of a - PQR- 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

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Standard X

Mathematics

Criteria for Similarity of Triangles

Question 2 ii...

Question

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

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Solution

In ΔPQR, E and F are two points on the sides PQ and PR respectively.
(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;
FR = PR - PF = 2.56 - 0.36 = 2.20 cm
So, $\frac{P E}{E Q} = \frac{0.18}{1.10} = \frac{18}{110} = \frac{9}{55} . . . (i)$
And, $\frac{P E}{F R} = \frac{0.36}{2.20}$ = $\frac{36}{220} = \frac{9}{55} . . . (i i)$
$∴ \frac{P E}{E Q} = \frac{P F}{F R}$ .
Hence, EF is parallel to QR.

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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

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E and F are points on the sides PQ and PR respectively of a ΔPQR. For each of the following cases, state whether EF || QR.

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(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

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Q. In

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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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Criteria for Similarity of Triangles

Standard X Mathematics

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Question 2 (iii) E and F are points on the sides PQ and PR respectively of a ΔPQR. 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