NCERT Solutions Maths Class 10 Exercise 6.2

# NCERT Solutions Maths Class 10 Exercise 6.2

NCERT Solutions Maths Class 10 Exercise 6.2

1. In figure (i) and (ii), DE BC. Find EC in (i) and AD in (ii).

Solution: (i) Since DE BC,

1.5/3 = 1/EC

EC = 3/1.5

EC = 2 cm

(ii) Since DE BC,

EC = 2.4 cm

2. E and F are points on the sides 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 = 0.36 cm

Solution: (i) The figure for the above question is as follows:

Given: PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

Now, PE/EQ = 3.9/3 = 1.3 cm

And PF/FR = 3.6/2.4 = 1.5 cm

Since, PE/EQ ≠ PF/FR

Therefore, EF is not parallel to QR.

(ii) Given: PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

Now, PE/EQ = 4/4.5 = 8/9 cm

And PF/FR = 8/9 cm

Since, PE/EQ = PF/FR

Therefore, EF is parallel to QR.

(iii) Given: PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

EQ = PQ – PE = 1.28 – 0.18 = 1.10 cm

And FR = PR – PF = 2.56 – 0.36 = 2.20 cm

Now, PE/EQ = 0.18/1.10 = 18/110 = 9/55 cm

And PF/FR = 0.36/2.20 = 36/220 = 9/55 cm

Since, PE/EQ = PF/FR

Therefore, EF is parallel to QR.

3. In figure, if LM CB and LN CD, prove that AM/AB = AN/AD.

Solution: In ABC, LM CB

Therefore, AM/AB = AL/AC    [From the Basic Proportionality Theorem] ………. (i)

And in ACD, LN CD

Therefore, AL/AC = AN/AD    [From the Basic Proportionality Theorem] ………. (ii)

From equations (i) and (ii), we have

4. In figure, DE AC and DF AE. Prove that BF/FE = BE/EC.

Solution: In ∆BCA, DE AC

Therefore, BE/EC = BD/DA    [From the Basic Proportionality Theorem] ………. (i)

And in BEA, DF AE

Therefore, BF/FE = BD/DA    [From the Basic Proportionality Theorem] ………. (ii)

From equations (i) and (ii), we have

BF/FE = BE/EC                                 Hence proved.

5. In figure, DE OQ and DF OR. Show that EF QR.

Solution: In ∆PQO, DE  OQ

Therefore, PE/EQ = PD/DO    [From the Basic Proportionality Theorem] ………. (i)

And in POR, DF OR

Therefore, PD/DO = PF/FR    [From the Basic Proportionality Theorem] ………. (ii)

From equations (i) and (ii), we have

PE/EQ = PF/FR

Since in ∆PQR, PE/EQ = PF/FR

Therefore, EF QR [By the converse of Basic Proportionality Theorem]

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.

Solution: Given: O is any point in PQR, in which AB PQ and AC PR.

To prove: BC QR

Construction: Join BC.

Proof: In OPQ, AB PQ

Therefore, OA/AP = OB/BQ    [From the Basic Proportionality Theorem] ………. (i)

And in OPR, AC PR

Therefore, OA/AP = OC/CR    [From the Basic Proportionality Theorem] ………. (ii)

From equations (i) and (ii), we have

OB/BQ = OC/CR

Since, iOQR, B and C are points dividing the sides OQ and OR in the same ratio.

Therefore, AB PQ [By the converse of Basic Proportionality Theorem]

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).

Solution:  Given: A ABC, in which D is the mid-point of side AB and the line DE is drawn parallel to BC, meeting AC at E.

To prove: AE = EC

Proof: Since DE BC

Therefore, AD/DB = AE/EC    [From the Basic Proportionality Theorem] ………. (i)

AE/EC = 1 [From equation (i)]

AE = EC

Hence, E bisects the third side AC.

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).

Solution: Given: A ABC, in which D and E are the mid-points of sides AB and AC respectively.

To Prove: DE BC

Proof: Since D and E are the mid-points of AB and AC respectively.

Therefore, AD = DB and AE = EC

And AE = EC

AE/EC = 1

Therefore, AD/DB = AE/EC = 1

Thus, in ABC, D and E are points dividing the sides AB and AC in the same ratio.

Therefore, DE BC [By the converse of Basic Proportionality Theorem]

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.

Solution: Given: A trapezium ABCD, in which AB DC and its diagonals AC and BD intersect each other at O.

To ProveAO/BO = CO/DO

Construction: Through O, draw OE AB, and hence OE DC.

Proof: In ADC, we have OE DC

Therefore, AE/ED = AO/CO       [From the Basic Proportionality Theorem] ………. (i)

Again, in ABD, we have OE AB     [By construction]

Therefore, ED/AE = DO/BO       [From the Basic Proportionality Theorem] ………. (ii)

AE/ED = BO/DO ………. (ii)

From equations (i) and (ii), we get

AO/CO = BO/DO

AO/BO = CO/DO

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.

Solution: Given: A quadrilateral ABCD, in which its diagonals AC and BD intersect each other at O such that AO/BO = CO/DO or AO/CO = BO/DO

To Prove: Quadrilateral ABCD is a trapezium.

Construction: Through O, draw OE AB meeting AD at E.

Proof: In ADB, we have OE AB [By construction]

Therefore, DE/EA = DO/BO       [From the Basic Proportionality Theorem]

EA/DE = BO/DO

EA/DE = BO/DO = AO/CO      [Since AO/CO = BO/DO]

EA/DE = AO/CO

Thus in ADC, E and O are points dividing the sides AD and AC in the same ratio. Therefore, EO DC         [By the converse of Basic Proportionality Theorem]

But EO AB       [By construction]

Therefore, AB DC

Therefore, the quadrilateral ABCD is a trapezium.