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# Ex 6.2 NCERT Solutions- Triangles Class 10 Notes | EduRev

## Class 10 : Ex 6.2 NCERT Solutions- Triangles Class 10 Notes | EduRev

The document Ex 6.2 NCERT Solutions- Triangles Class 10 Notes | EduRev is a part of the Class 10 Course Mathematics (Maths) Class 10.
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Exercise 6.2
Q.1. In figures (i) and (ii), DE y BC. Find EC in (i) and AD in (ii). Sol. [By basic proportional theorem] (ii) In ΔABC, DE || BC [By basic proportional theorem] Q.2. 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:
(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 = 0.36 cm
Sol. (i)   ⇒ EF is not parallel to QR
[By converse of B.P.T.]
(ii)   ⇒ EF || QR [By converse of B.P.T.]
(iii)   ⇒ EF || QR [By converse of B.P.T.]

Q.3. In the figure, if LM || CB and LN || CD, prove that . Sol. From equation (i) and (ii) Q.4. In the figure, DE || AC and DF || AE. Prove that  Sol. In ΔABC, From equation (i) and (ii) Q.5. In the figure, DE || OQ and DF || OR. Show that EF || QR. Sol. In Δ PQO, In ΔPOR, DF || OR ...(ii)
From equation (i) and (ii), we get ∴  EF || QR [By converse of B.P.T.]

Q.6. In the figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR. Sol. In Δ PQR,
AB || PQ In ΔPOR, AC || PR
∴  OP/PA = OR/RC    .......(ii)

From equation (i) and (ii), we get BC || QR [By converse of B.P.T.]

Q.7. Using Theorem 6.1, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side.
Sol. Given: In Δ ABC, D is the mid-point of AB and DE || BC To Prove: AE = EC
Proof: In ΔABC,
DE || BC [By B.P.T.]
⇒ 1 = AE/EC ⇒ AE = EC
Hence, DE bisects AC.

Q.8. Using Theorem 6.2, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.
Sol. Given. A ΔABC in which D and E are mid-points of sides AB and AC respectively.
To Prove: DE || BC
Proof: In ΔABC, AD = DB and AE = EC  ∴ DE || BC   [By converse of B.P.T.]

Q.9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that .
Sol.
Given: ABCD is a trapezium in which AB || DC  Construction: Draw EO || DC
Proof: In ΔABD, EO || DC     [By construction]
DC || AB     [Given]
⇒  EO || AB From equation (i) and (ii) Q.10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that . Show that ABCD is a trapezium.
Sol. Given: A quadrilateral ABCD, whose diagonals intersect at O. To Prove: ABCD is a trapezium.
Construction: Draw EO || AB Proof: In ΔABC, OE || AB  From equation (i) and (ii) ⇒ OE || DC [By converse of B.P.T.]
OE || AB and OE || DC ⇒ AB || DC
∴ ABCD is a trapezium.

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## Mathematics (Maths) Class 10

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