# AP 10th Class Maths 6th Chapter Triangles Exercise 6.2 Solutions

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

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.