In this post, we will share NCERT Class 10th Maths Book Solutions Chapter 6 Triangles Ex 6.2. These solutions are based on new NCERT Syllabus.

## NCERT Class 10th Maths Solutions Chapter 6 Triangles Ex 6.2

Question 1.

In figures (i) and (ii), DE || SC. Find EC in (i) and AD in (ii).

Solution:

(i) Since DE || BC [Given]

∴ Using the Basic proportionality theorem,

We have \(\frac{A D}{D B}=\frac{A E}{E C}\)

Since, AD = 1.5 cm, DB = 3 cm and AE = 1 cm,

∴ \(\frac{1.5 \mathrm{cm}}{3 \mathrm{cm}}=\frac{1 \mathrm{cm}}{E C}\)

By cross-multiplication, we have

EC × 1.5 = 1 × 3

⇒ EC = \(\frac{1 \times 3}{1.5}=\frac{1 \times 3 \times 10}{15}\)

EC = 2 cm

(ii) In ∆ABC, DE || BC

Using the Basic proportionality theorem,

∴ AD = 2.4 cm.

Question 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 = 9cm

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

Solution:

(i) We have, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

⇒ EF is not parallel to QR.

(ii) We have, PE = 4 cm, QE = 4.5 cm PF = 8 cm and RF = 9 cm

⇒ EF is parallel to QR.

Question 3.

In the figure, if LM || CB and LN || CD, prove that \(\frac{A M}{A B}=\frac{A N}{A D}\)

Solution:

In ∆ABC, LM || CB [given]

∴ Using the Basic proportionality theorem, we have

Question 4.

In the figure, DE || ACand DF || AE.

Prove that \(\frac{B F}{F E}=\frac{B E}{E C}\)

Solution:

In ∆ABC

∵ DE || AC [given]

Using the basic proportionality theorem, we have

Question 5.

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

Solution:

In ∆PQO,

∵ DE || OQ [given]

∴ Using the Basic proportionality theorem, we have

\(\frac{P E}{E Q}=\frac{P D}{D O}\) …………… (1)

Again, in ∆POR, DF || OR [given]

∴ Using the Basic proportionality theorem, we have

Now, in ∆PQR,

∵ E and F are two distinct points on PQ and PR respectively and \(\frac{P E}{E Q}=\frac{P F}{F R}\),

i.e., E and F divide the two sides PQ and PR of ∆PQR in the same ratio.

∴ By converse of Basic proportionality theorem, EF || QR.

Question 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.

Solution:

In ∆PQR, O is a point and OP, OQ and OR are joined. We have points A, B and C on OP, OQ and OR respectively such that AB || PQ and AC || PR.

Now, in ∆OPQ,

∵ AB || PQ [Given]

Using the Basic proportionality theorem, we have

i.e., B and C divide the sides OQ and OR of ∆OQR in the same ratio.

By converse of Basic proportionality theorem, BC || QR.

Question 7.

Using Basic proportionality theorem, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. A

Solution:

We have ∆ABC, in which D is the midpoint of AB and E is a point on AC such that DE || BC.

∵ DE || BC [given]

∴ Using the Basic proportionality theorem, we get

⇒ E is the mid point of AC. Hence, it is proved that a line through the midpoint of one side of a triangle parallel to another side bisects the third side.

Question 8.

Using converse of basic proportionality theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.

Solution:

We have ∆ABC, in which D and E are the mid-points of sides AB and AC respectively.

∴ AD = DB ……………. (1)

and AE = EC ………….. (2)

From (1) and (2), we have

\(\frac{A D}{D B}\) = 1 and \(\frac{A E}{E C}\) = 1

⇒ \(\frac{A D}{D B}=\frac{A E}{E C}\)

⇒ DE || BC (By converse proportionality theorem).

Question 9.

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that \(\frac{A O}{B O}=\frac{C O}{D O}\).

Solution:

We have, a trapezium ABCD such that AB || DC. The diagonals AC and BD intersect each other at O.

Let us draw OE parallel to either AB or DC.

In ∆ADC

OE || DC [By construction]

∴ Using the Basic proportionality theorem, we get

\(\frac{A E}{E D}=\frac{A O}{C O}\) …………. (1)

In ∆ABD

OE || AB [By construction]

∴ Using the Basic proportionality theorem, we get

Question 10.

The diagonals of a quadrilateral ABCD intersect each other at the point O such that \(\frac{A O}{B O}=\frac{C O}{D O}\). Show that ABCD is a trapezium.

Solution:

It is given that \(\frac{A O}{B O}=\frac{C O}{D O}\)

From \(\frac{A O}{B O}=\frac{C O}{D O}\), we have \(\frac{A O}{C O}=\frac{B O}{D O}\)

Let us draw OE such that OE || BA

In ∆ADB, OE || AB [By construction]

i.e., the points O and E divide the sides AC and AD of ∆ADC respectively in the same ratio.

∴ Using the converse of Basic proportionality theorem, we get OE || DC and OE || AB

⇒ AB || DC

⇒ ABCD is a trapezium.

In this post, we will share NCERT Class 10th Maths Book Solutions Chapter 6 Triangles Ex 6.2. These solutions are based on new NCERT Syllabus.

## NCERT Class 10th Maths Solutions Chapter 6 Triangles Ex 6.2

Question 1.

In figures (i) and (ii), DE || SC. Find EC in (i) and AD in (ii).

Solution:

(i) Since DE || BC [Given]

∴ Using the Basic proportionality theorem,

We have \(\frac{A D}{D B}=\frac{A E}{E C}\)

Since, AD = 1.5 cm, DB = 3 cm and AE = 1 cm,

∴ \(\frac{1.5 \mathrm{cm}}{3 \mathrm{cm}}=\frac{1 \mathrm{cm}}{E C}\)

By cross-multiplication, we have

EC × 1.5 = 1 × 3

⇒ EC = \(\frac{1 \times 3}{1.5}=\frac{1 \times 3 \times 10}{15}\)

EC = 2 cm

(ii) In ∆ABC, DE || BC

Using the Basic proportionality theorem,

∴ AD = 2.4 cm.

Question 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 = 9cm

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

Solution:

(i) We have, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

⇒ EF is not parallel to QR.

(ii) We have, PE = 4 cm, QE = 4.5 cm PF = 8 cm and RF = 9 cm

⇒ EF is parallel to QR.

Question 3.

In the figure, if LM || CB and LN || CD, prove that \(\frac{A M}{A B}=\frac{A N}{A D}\)

Solution:

In ∆ABC, LM || CB [given]

∴ Using the Basic proportionality theorem, we have

Question 4.

In the figure, DE || ACand DF || AE.

Prove that \(\frac{B F}{F E}=\frac{B E}{E C}\)

Solution:

In ∆ABC

∵ DE || AC [given]

Using the basic proportionality theorem, we have

Question 5.

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

Solution:

In ∆PQO,

∵ DE || OQ [given]

∴ Using the Basic proportionality theorem, we have

\(\frac{P E}{E Q}=\frac{P D}{D O}\) …………… (1)

Again, in ∆POR, DF || OR [given]

∴ Using the Basic proportionality theorem, we have

Now, in ∆PQR,

∵ E and F are two distinct points on PQ and PR respectively and \(\frac{P E}{E Q}=\frac{P F}{F R}\),

i.e., E and F divide the two sides PQ and PR of ∆PQR in the same ratio.

∴ By converse of Basic proportionality theorem, EF || QR.

Question 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.

Solution:

In ∆PQR, O is a point and OP, OQ and OR are joined. We have points A, B and C on OP, OQ and OR respectively such that AB || PQ and AC || PR.

Now, in ∆OPQ,

∵ AB || PQ [Given]

Using the Basic proportionality theorem, we have

i.e., B and C divide the sides OQ and OR of ∆OQR in the same ratio.

By converse of Basic proportionality theorem, BC || QR.

Question 7.

Using Basic proportionality theorem, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. A

Solution:

We have ∆ABC, in which D is the midpoint of AB and E is a point on AC such that DE || BC.

∵ DE || BC [given]

∴ Using the Basic proportionality theorem, we get

⇒ E is the mid point of AC. Hence, it is proved that a line through the midpoint of one side of a triangle parallel to another side bisects the third side.

Question 8.

Using converse of basic proportionality theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side.

Solution:

We have ∆ABC, in which D and E are the mid-points of sides AB and AC respectively.

∴ AD = DB ……………. (1)

and AE = EC ………….. (2)

From (1) and (2), we have

\(\frac{A D}{D B}\) = 1 and \(\frac{A E}{E C}\) = 1

⇒ \(\frac{A D}{D B}=\frac{A E}{E C}\)

⇒ DE || BC (By converse proportionality theorem).

Question 9.

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that \(\frac{A O}{B O}=\frac{C O}{D O}\).

Solution:

We have, a trapezium ABCD such that AB || DC. The diagonals AC and BD intersect each other at O.

Let us draw OE parallel to either AB or DC.

In ∆ADC

OE || DC [By construction]

∴ Using the Basic proportionality theorem, we get

\(\frac{A E}{E D}=\frac{A O}{C O}\) …………. (1)

In ∆ABD

OE || AB [By construction]

∴ Using the Basic proportionality theorem, we get

Question 10.

The diagonals of a quadrilateral ABCD intersect each other at the point O such that \(\frac{A O}{B O}=\frac{C O}{D O}\). Show that ABCD is a trapezium.

Solution:

It is given that \(\frac{A O}{B O}=\frac{C O}{D O}\)

From \(\frac{A O}{B O}=\frac{C O}{D O}\), we have \(\frac{A O}{C O}=\frac{B O}{D O}\)

Let us draw OE such that OE || BA

In ∆ADB, OE || AB [By construction]

i.e., the points O and E divide the sides AC and AD of ∆ADC respectively in the same ratio.

∴ Using the converse of Basic proportionality theorem, we get OE || DC and OE || AB

⇒ AB || DC

⇒ ABCD is a trapezium.