TOPICS

Exercise - 6.2

Triangles

**Question-1 :-** In Figure, (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

(i) Given that DE || BC. Let EC = x cm By using basic proportionality theorem, we get AD/DB = AE/EC 1.5/3 = 1/x 1.5x = 3 x = 3/1.5 x = 30/15 x = 2 Hence, EC = x cm = 2 cm

(ii) Given that DE || BC. Let AD = x cm By using basic proportionality theorem, we get AD/DB = AE/EC x/7.2 = 1.8/5.4 5.4x = 1.8 x 7.2 5.4x = 12.96 x = 12.96/5.4 x = 2.4 Hence, AD = x cm = 2.4 cm

**Question-2 :-** E and F are points on the sides PQ and PR respectively of a Δ 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 = 0.36 cm

(i) Given that PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm PE/EQ = 3.9/3 = 1.3 PF/FR = 3.6/2.4 = 1.5 Hence, PE/EQ ≠ PF/FR Therefore, EF is not parallel to QR.

(ii) Given that PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm PE/EQ = 4/4.5 = 8/9 PF/FR = 8/9 Hence, PE/EQ = PF/FR Therefore, EF || QR.

(iii) Given that PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm PE/PQ = 0.18/1.28 = 18/128 = 9/64 PF/PR = 0.36/2.56 = 9/64 Hence, PE/PQ = PF/PR Therefore, EF || QR.

**Question-3 :-** In Figure, if LM || CB and LN || CD, prove that
AM/AB = AN/AD

Given that : LM || CB and LN || CD Prove that : AM/AB = AN/AD Proof : LM || CB (Given) By using basic proportionality theorem, we get AM/AB = AL/AC .........(i) Similarly, LN || CD AN/AD = AL/AC .........(ii) From (i) & (ii), We get AM/AB = AN/AD

**Question-4 :-** In Figure, DE || AC and DF || AE. Prove that
BF/FE = BE/EC

Given that : DE || AC and DF || AE Prove that : BF/FE = BE/EC Proof : In ΔABC, DE || AC (Given) By using basic proportionality theorem, we get BD/DA = BE/EC .........(i) Similarly, In ΔBAE, DF || AE BD/DA = BF/FE .........(ii) From (i) & (ii), We get BE/EC = BF/FE

**Question-5 :-** In Figure, DE || OQ and DF || OR. Show that EF || QR.

Given that : DE || OQ and DF || OR Prove that : EF || QR Proof : In Δ POQ, DE || OQ (Given) By using basic proportionality theorem, we get PE/EQ = PD/DO .........(i) Similarly, In Δ POR, DF || OR PF/FR = PD/DO .........(ii) From (i) & (ii), We get PE/EQ = PF/FR By using converse of basic proportionality theorem Therefore, EF || QR

**Question-6 :-** In Figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

Given that : AB || PQ and AC || PR Prove that : BC || QR Proof : In Δ POQ, AB || PQ (Given) By using basic proportionality theorem, we get OA/AP = OB/BQ .........(i) Similarly, In Δ POR, AC || PR OA/AP = OC/CR .........(ii) From (i) & (ii), We get OB/BQ = OC/CR By using converse of basic proportionality theorem Therefore, BC || QR

**Question-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. (Recall that you have proved it in Class IX).

Given that : Consider the given figure in which PQ is a line segment drawn through the mid-point P of line AB, such that PQ || BC Prove that : Q is the mid-point of AC. Proof : In Δ ABC, PQ || BC By using basic proportionality theorem, we get AQ/QC = AP/PB AQ/QC = 1/1 (P is the mid-point of AB. So, AB = PB) AQ = QC or Q is the mid point of AC.

**Question-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. (Recall that you have done it in Class IX).

Given that : Consider the given figure in which PQ is a line segment joining the mid-points P and Q of line AB and AC respectively. So, we have AP = PB and AQ = QC. Prove that : PQ || BC Proof : In Δ ABC AP/PB = 1/1 (Given AP = PB) AQ/QC = 1/1 (Given AQ = QC) Therefore, AQ/QC = AP/PB By using basic proportionality theorem, we get PQ || BC

**Question-9 :-** ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO.

Given that : ABCD is a trapezium in which AB || DC. Prove that : AO/BO = CO/DO Construction : Draw a line EF through point O, such that EF || CD Proof : In Δ ADC, EO || CD By using basic proportionality theorem, we get AE/ED = OA/OC ........(i) In Δ ABD, OE || AB By using basic proportionality theorem, we get ED/AE = OD/OB AE/ED = OB/OD ........(ii) From equations (i) & (ii), we get OA/OC = OB/OD Therefore, AO/BO = OC/OD

**Question-10 :-** The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO. Show that ABCD is a trapezium.

Given that : The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO. Prove that : ABCD is a trapezium. Construction : Draw a line OE || AB. Proof : In Δ ABD, OE || AB By using basic proportionality theorem, we get AE/ED = OB/OD ........(i) AO/BO = CO/DO ........(ii) (Given) From equations (i) & (ii), we get AE/ED = AO/OC EO || DC (By the converse of basic proportionality theorem) AB || OE || DC AB || CD Therefore, ABCD is a trapezium.

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