Well-designed AP 10th Class Maths Textbook Solutions Chapter 6 Triangles Exercise 6.2 offers step-by-step explanations to help students understand problem-solving strategies.

## Triangles Class 10 Exercise 6.2 Solutions – 10th Class Maths 6.2 Exercise Solutions

Question 1.

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

(i)

(ii)

Solution:

i) Given in ∆ABC, DE || BC.

By using basic proportionality the-orem (Thales theorem)

where, AD = 1.5 cm, DB = 3 cm, AE = 1 cm, EC = ?

⇒

Therefore, EC = 2 cm

ii) Given in ∆ABC, DE || BC.

By using basic proportionality the-orem (Thales theorem)

where, AD = ?, BD = 7.2 cm, AE = 1.8 cm, EC = 5.4 cm

AD = \(\frac{7.2}{3}\) = 2.4 cm

Therefore, 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

Solution:

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

So, in ∆PQR, EF is not parallel to QR according to basic proportionality theorem.

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

Solution:

Given, PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm

So, in ∆PQR according to basic propor-tionality theorem EF // QR.

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

Solution:

Given PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

So, in ∆PQR according to basic propor-tionality theorem EF // QR.

Question 3.

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

Solution:

Given in ∆ABC, LM || CB.

According to basic proportionality theorem,

\(\frac{A M}{A B}\) = \(\frac{\mathrm{AL}}{\mathrm{AC}}\) → (1)

In ∆ACD, LN || CD

According to basic proportionality theorem,

\(\frac{A L}{A C}\) = \(\frac{\mathrm{AN}}{\mathrm{AD}}\) → (2)

From (1) and (2) \(\frac{\mathrm{AM}}{\mathrm{AB}}\) = \(\frac{\mathrm{AN}}{\mathrm{AD}}\).

Question 4.

In given figure, DE || AC and DF || AE. Prove that \(\frac{\mathrm{BF}}{\mathrm{FE}}\) = \(\frac{\mathrm{BE}}{\mathrm{EC}}\).

Solution:

Given, DE || AC and DF || AE.

In ∆BC, DE||AC

According to basic proportionality theorem,

\(\frac{\mathrm{BD}}{\mathrm{DA}}\) = \(\frac{\mathrm{BE}}{\mathrm{EC}}\) → (1)

In ∆ABC, DF || AE

According to basic proportionality theorem,

\(\frac{\mathrm{BD}}{\mathrm{DA}}\) = \(\frac{\mathrm{BF}}{\mathrm{FE}}\) → (2)

From (1) and (2) \(\frac{\mathrm{BF}}{\mathrm{FE}}\) = \(\frac{\mathrm{BE}}{\mathrm{EC}}\).

Question 5.

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

Solution:

Given DE || OQ and DF || OR.

In ∆PQO, DE || OQ.

According to basic proportionality theorem,

\(\frac{\mathrm{PE}}{\mathrm{EQ}}\) = \(\frac{\mathrm{PD}}{\mathrm{DO}}\) → (1)

In ∆POR, DF || OR

According to basic proportionality theorem,

\(\frac{\mathrm{PD}}{\mathrm{DO}}\) = \(\frac{\mathrm{PF}}{\mathrm{FR}}\) → (2)

From (1) and (2) \(\frac{\mathrm{PE}}{\mathrm{EQ}}\) = \(\frac{\mathrm{PF}}{\mathrm{FR}}\)

In ∆PQR, \(\frac{\mathrm{PE}}{\mathrm{EQ}}\) = \(\frac{\mathrm{PF}}{\mathrm{FR}}\)

According to converse of basic proportionality theorem, EF || QR.

Question 6.

In given 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 A, B and C are points on OP, OQ and OR respectively.

AB || PQ and AC || PR.

In ∆OPQ, AB || PQ

According to basic proportionality the-orem,

\(\frac{\mathrm{OA}}{\mathrm{AP}}\) = \(\frac{\mathrm{OB}}{\mathrm{BQ}}\) → (1)

According to basic proportionality the-orem,

\(\frac{\mathrm{OA}}{\mathrm{AP}}\) = \(\frac{\mathrm{OC}}{\mathrm{CR}}\) → (2)

From (1) and (2) \(\frac{\mathrm{OB}}{\mathrm{BQ}}\) = \(\frac{\mathrm{OC}}{\mathrm{CR}}\)

In ∆OQR, \(\frac{\mathrm{OB}}{\mathrm{BQ}}\) = \(\frac{\mathrm{OC}}{\mathrm{CR}}\)

According to converse of basic proportionality theorem, BC || QR

Question 7.

Using Theorem 6.1, prove that a line drawn through the midpoint 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 in ∆ABC, DE || BC and AD = DB.

According to basic proportionality theorem,

Therefore, AE = EC

Hence DE can bisect the third side 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).

Solution:

Given in ∆PQR, M and N are mid points of PQ and PR respectively. That is

PM = MQ ⇒ \(\frac{\mathrm{PM}}{\mathrm{MQ}}\) = 1

and PN = NR ⇒ \(\frac{\mathrm{PN}}{\mathrm{NR}}\) = 1

In ∆PQR, \(\frac{\mathrm{PM}}{\mathrm{MQ}}\) = \(\frac{\mathrm{PN}}{\mathrm{NR}}\)

According to converse of basic proportionality theorem, MN || QR.

Therefore, line joining the mid-points of any two sides of a triangle is parallel to the third side.

Question 9.

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

.

Solution:

Given in trapezium ABCD, AB || DC and AC, BD can intersect at O.

Let OP || AB and CD.

In ∆ABD, PO || AB.

According to basic proportionality theorem,

In ∆ACD, PO || DC

According to basic proportionality theorem,

From (1) and (2) \(\frac{\mathrm{OA}}{\mathrm{OC}}\) = \(\frac{\mathrm{BO}}{\mathrm{DO}}\)

Therefore, \(\frac{\mathrm{AO}}{\mathrm{BO}}\) = \(\frac{\mathrm{CO}}{\mathrm{DO}}\)

Question 10.

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

Solution:

Given in quadrilateral ABCD, AC and BD intersect at O and \(\frac{\mathbf{A O}}{\mathbf{B O}}\) = \(\frac{\mathbf{C O}}{\mathbf{D O}}\).

\(\frac{\mathbf{A O}}{\mathbf{C O}}\) = \(\frac{\mathbf{B O}}{\mathbf{D O}}\) → (1)

Let PO || AB in ∆ABD.

According to basic proportionality theorem,

\(\frac{\mathbf{A P}}{\mathbf{P D}}\) = \(\frac{\mathbf{B O}}{\mathbf{D O}}\) → (2)

From (1) and (2) \(\frac{\mathbf{A P}}{\mathbf{P D}}\) = \(\frac{\mathbf{A O}}{\mathbf{C O}}\)

In ∆ACD, \(\frac{\mathbf{A P}}{\mathbf{P D}}\) = \(\frac{\mathbf{A O}}{\mathbf{C O}}\)

According to converse of basic proportionality theorem, PO || DC

So, PO || AB and PO || DC.

Therefore, AB || DC.

In quadrilateral ABCD, AB || DC.

If one pair of opposite sides are parallel, then it is a trapezium. Hence, ABCD is a trapezium.