# NCERT Solutions for Class 10 Maths Exercise 6.2

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NCERT solutions for Maths Triangles

## NCERT Solutions for Class 10 Maths Triangles

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

Ans. (i) Since DE BC,

EC =

EC = 2 cm

(ii)Since DE BC,

EC = 2.4 cm

NCERT Solutions for Class 10 Maths Exercise 6.2

###### 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 = 4 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

Ans. (i)Given: PE = 3.9 cm, EQ = 4 cm, PF = 3.6 cm and FR = 2.4 cm

Now, = 0.97 cm

And = 1.2 cm

Therefore, EF does not divide the sides PQ and PR of PQR in the same ratio.

###### EF is not parallel to QR.

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

Now, = cm

And cm

Therefore, EF divides the sides PQ and PR of PQR in the same ratio.

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 ER = PR – PF = 2.56 – 0.36 = 2.20 cm

Now, = cm

And = cm

Therefore, EF divides the sides PQ and PR of PQR in the same ratio.

EF is parallel to QR.

NCERT Solutions for Class 10 Maths Exercise 6.2

###### 3. In figure, if LM CB and LN CD, prove that

Ans. In ABC, LM CB

[Basic Proportionality theorem] ……….(i)

And in ACD, LN CD

[Basic Proportionality theorem] ……….(ii)

From eq. (i) and (ii), we have

NCERT Solutions for Class 10 Maths Exercise 6.2

###### 4. In figure, DE AC and DF AE. Prove that

Ans. In BCA, DE AC

[Basic Proportionality theorem] ……….(i)

And in BEA, DF AE

[Basic Proportionality theorem] ……….(ii)

From eq. (i) and (ii), we have

NCERT Solutions for Class 10 Maths Exercise 6.2

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

Ans. In PQO, DE OQ

[Basic Proportionality theorem] ……….(i)

And in POR, DF OR

[Basic Proportionality theorem] ……….(ii)

From eq. (i) and (ii), we have

EF QR [By the converse of BPT]

NCERT Solutions for Class 10 Maths Exercise 6.2

###### 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.

Ans. 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

[Basic Proportionality theorem] ……….(i)

And in OPR, AC PR

[Basic Proportionality theorem] ……….(ii)

From eq. (i) and (ii), we have

In OQR, B and C are points dividing the sides OQ and OR in the same ratio.

By the converse of Basic Proportionality theorem,

BC QR

NCERT Solutions for Class 10 Maths Exercise 6.2

###### 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).

Ans. Given: A triangle 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

[Basic Proportionality theorem] ……….(i)

[From eq. (i)]

AE = EC

Hence, E is the mid-point of the third side AC.

NCERT Solutions for Class 10 Maths Exercise 6.2

###### 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).

Ans. Given: A triangle 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.

AD = DB and AE = EC

and AE = EC

= 1

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

Therefore, by the converse of Basic Proportionality theorem, we have

DE BC

NCERT Solutions for Class 10 Maths Exercise 6.2

###### 9. ABCD is a trapezium in which AB DC and its diagonals intersect each other at the point O. Show that

Ans. Given: A trapezium ABCD, in which AB DC and its diagonals

AC and BD intersect each other at O.

To Prove:

###### Construction: Through O, draw OE AB, i.e. OE DC.

Proof: In ADC, we have OE DC

[By Basic Proportionality theorem]……….(i)

Again, in ABD, we have OE AB[Construction]

[By Basic Proportionality theorem]

……….(ii)

From eq. (i) and (ii), we get

NCERT Solutions for Class 10 Maths Exercise 6.2

###### 10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that Such that ABCD is a trapezium.

Ans. Given: A quadrilateral ABCD, in which its diagonals AC and

BD intersect each other at O such that , i.e.

.

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]

[By Basic Proportionality theorem]

=

=

Thus in ADC, E and O are points dividing the sides AD and AC in the same ratio. Therefore by the converse of Basic Proportionality theorem, we have

EO DC

But EO AB[By construction]

AB DC

## NCERT Solutions for Class 10 Maths Exercise 6.2

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### 14 thoughts on “NCERT Solutions for Class 10 Maths Exercise 6.2”

1. it is easy to understand

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4. Please send gujarat board book chapter 6 similarity of triangles

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