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.