Ncert Solutions For Class 10 Maths Ex 6.2

Ncert Solutions For Class 10 Maths Chapter 6 Ex 6.2

 Question 1:

From the given figures , we get that ST ǁ QR. Find TR in (i) and PS in (ii).

2

Solution:

In the triangle, PQR, ST ǁ QR (given)

Therefore, PS/SQ = PT/TR [Using the basic Proportionality theorem]

=> 1.5/3 = 1/ TR

=> ∑ TR = 3/1.5

TR = 3 x 10/15 = 2cm

Hence, TR = 2cm

(ii) In triangle PQR, ST ǁ QR (given)

Therefore, PS/ SQ = PT / TR [Using the basic proportionality theorem]

= > PS/7.2 = 1.8/5.4

= > PS = 1.8 x 7.2/5.4 = 18/10 x 72/10 x 10/54 = 24/10

= > PS = 2.4

Hence, PS = 2.4 cm

 

Question 2:

A and B are two points on the sides of XY and XZ respectively of a XYZ triangle. State whether AB ǁ YZ for the following given cases:

1: XA = 3.9 cm, AY = 3 cm XB = 3.6 cm and BZ= 2.4 cm

2: XA = 4 cm, YA = 4.5 cm, XB = 8 cm and ZB = 9 cm

3: XY = 1.28 cm, XZ = 2.56 cm, XA = 0.18 cm and XB = 0.63 cm

3

Solution:

In triangle XYZ, A and B are the two points on sides XY and XZ respectively.

1:  XA = 3.9 cm and AY = 3 cm (given)

XB = 3.6 cm, BZ = 2.4 cm (Given)

Therefore, XA/AY = 3.9/3 = 39/30 = 13/10 = 1.3 [Using the basic proportionality theorem]

And, XB/BZ = 3.6/2.4 = 36/24 = 3/2 = 1.5

So, XA/AY ≠ XB/BZ

Hence, AB is not parallel to YZ.

2: XA = 4cm, YA = 4.5 cm, XB = 8cm, ZB= 9cm

Therefore, XA/YA = 4/4.5 = 40/45 = 8/9 [Using the basic proportionality theorem]

And, XB/ZB = 8/9

So, XA/YA = XB/ZB

Hence, AB is parallel to YZ

3: XZ = 1.28 cm, XZ = 2.56 cm, XA= 0.18 cm, XB= 0.36 cm (Given)

Here, AY = XY – XA = 1.28-0.18 = 1.10 cm

And, BZ = XZ – XB = 2.56 – 0.36 = 2.20 cm

So, XA/AY = 0.18/1.10 = 18/110 = 9/55 – – – – – – – – – – – – (1)

And, XA/BZ = 0.36/2.20 = 36/220 = 9/55 – – – – – – – – – – – (2)

Therefore, XA/AY = XB/BZ

Hence, AB is parallel to XY.

 

Question 3:

From the given figure, we see that AB ǁ RQ and AC ǁ RS, prove that PB/BQ = PC/PS

4

Solution:

From the given figure, we get AB ǁ RQ

By using the basic proportionality theorem, we get,

PB/BQ = PA/ PR – – – – – – – – (1)

Similarly, AC ǁ RS

Therefore, PC/PS = PA/PR – – – – (2)

From (1) and (2) we get,

PB/BQ = PC/PS

 

Question 4:

From the figure we get, ST ǁ PR and SU ǁ PT. Prove that QU/UT = QT/TR

5

Solution:

In triangle PQR, ST ǁ PR (Given)

Therefore, QS/SP= QT/TR – – – – – – – (1) [Using the proportionality Theorem]

In triangle PQR, SU ǁ PT (Given)

Therefore, QS/SP = QU/UT – – – – – – (2) [Using the basic proportionality theorem]

From equation (1) and (2) we get,

QT/TR = QU/UT

 

Question 5:

From the following figure we get, XZ ǁ AC and XZ ǁ AD, show that YZ ǁ CD.

6

Solution:

In triangle BCA, XY ǁ AC (Given)

Therefore, BX/XA = BY/YC – – – – – – – (1) [Using the basic proportionality theorem]

In triangle BCA, XY ǁ AC (Given)

Therefore, BX/XA = BZ/ZD – – – – – – – (2) [Using the basic proportionality theorem]

From the equation (1) and (2) we get,

BY/YC = BZ/ZD

In triangle BCA, YZ ǁ CD [By converse of the basic proportionality theorem]

 

  Question 6:

From the figure, three points X, Y and Z are points on AB, AC and AD respectively such that XY ǁ BC and XZ ǁ BD. Show that YZ ǁ CD

7

Solution:

In triangle ABC, XY ǁ BC (Given)

Therefore, AX/XB = AY/YC – – – – – – – (1) [Using the basic proportionality theorem]

In triangle ABC, XZ ǁ BD (Given)

Therefore, AX/XB = AZ/ ZD – – – – – – – (2) [Using the basic proportionality theorem]

From the equations (1) and (2), we get

AY/ YC = AZ/ZD

In triangle ACD, YZ ǁ CD [By the converse of basic proportionality theorem]

 

Question 7:

Using the basic proportionality theorem, prove that a line drawn through the midpoints of one side of a triangle is parallel to the other side that bisects the third side.

Solution:

8

From the given diagram we get,

The triangle PQR in which S is the midpoint of P and Q such that PS=SQ

A line parallel to QR intersects PR at T such that ST ǁ QR

To prove: T is the midpoint of PR

Proof: S is the midpoint of PQ

Therefore, PS=SQ

=>PS/QS = 1 – – – – – – – (1)

In triangle PQR, ST ǁ QR,

Therefore, PS/SQ = PT / TR [Using the basic proportionality theorem]

=>1 = PT/ TR [from equation (1)]

Therefore, PT = TR

Hence, T is the midpoint of PR

 

Question 8:

Prove that the line joining the midpoints of any two sides of a triangle is parallel to the third side by using the converse of basic proportionality theorem.

Solution:

9

Given:

From the triangle PQR in which ST are the midpoints of PQ and PR respectively such that PS = SQ and PT= TR

To prove: ST ǁ QR

Proof: S is the midpoint of PQ (Given)

Therefore, PS = SQ

=>PS/QS = 1 – – – – – – – – – (1)

Also, T is the midpoint of PR (Given)

Therefore, PT= TR

=>PT/TR = 1 [from equation (1)]

From equation (1) and (2) we get,

PS/ QS = PT/ TR

Hence, ST ǁ QR [By the converse of the basic proportionality theorem]

 

Question 9:

PQRS is a trapezium in which PQ ǁ SR and its diagonals intersect each other at a point A. show that PA/QA = RA/SA

Solution:

10

Given:

PQRS is a trapezium in which PQ ǁ RS in which the diagonals PR and QS intersect each other at A.

To prove: PA/QA = RA/ SA

Construction: Through A, draw TA ǁ SR ǁ PQ

Proof: In triangle PSR, we have

AT ǁ SR (By construction)

Therefore, PT/TS = PA/RA – – – – – – – (1) [Using the basic proportionality theorem]

In triangle PQS, we have

AT ǁ PQ (by construction)

Therefore, ST/TP = SA/QA – – – – – – – – – (2)  [Using the basic proportionality theorem]

From the equations (1) and (2) we get,

PA/RA = QA / SA

=>PA / QA = RA/SA

 

 

Question 10:

The diagonals of a quadrilateral PQRS intersect each other at the point A such that PA/QA = RA/SA. Show that PQRS is a trapezium.

11

Solution:

Given:

Quadrilateral PQRS in which diagonals PR and QS intersect each other at A such that PA/ QA = RA/SA

To prove: PQRS is a trapezium

Construction: Through A, draw line TA, where TA ǁ PQ, which meets PS at T

Proof: In triangle SPQ, we have

TA ǁ PQ

Therefore, ST/TP = SA/AQ – – – – – – (1) [Using the basic proportionality theorem]

Also, PA/QA = RA/SA (Given)

=>PA/ RA = QA/ SA

=> RA/PA = QA/SA

=> SA/AQ = RA/PA – – – – – – – – (2)

From the equations (1) and (2) we get,

ST / TP = RA/ PA

Therefore, By using converse of Basic proportionality theorem,

TA ǁ SR also TA ǁ PQ

=>PQ ǁ SR

Hence, quadrilateral PQRS is a trapezium with PQ ǁ RS

 

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