RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2

Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 6 Triangles Ex 6.2 Textbook Exercise Questions and Answers.

RBSE Class 10 Maths Solutions Chapter 6 Triangles Exercise 6.2

Question 1.
In Fig. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

Solution:
(i) In △ABC
DE || BC ….(Given)
∴ \(\frac{\mathrm{AD}}{\mathrm{BD}}\) = \(\frac{\mathrm{AE}}{\mathrm{EC}}\)
(By Basic Proportionality Theorem)
\(\frac{1.5}{3}\) = \(\frac{1}{\mathrm{EC}}\)
EC = \(\frac{3}{1.5}\) = \(\frac{3 \times 10}{15}\) = 2
∴ EC = 2 cm

(ii) In △ABC,
DE || BC (Given)
∴ \(\frac{\mathrm{AD}}{\mathrm{BD}}\) = \(\frac{\mathrm{AE}}{\mathrm{EC}}\)
(By 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 = 9 cm
(iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm.
Solution:
According to the question in △PQR two points E and F are on the sides PQ and PR respectively.

(i) PE = 3.9 cm., EQ = 3 cm.
PF = 3.6 cm., FR = 2.4 cm.
\(\frac{\mathrm{PE}}{\mathrm{EQ}}\) = \(\frac{3.9}{3}\) = \(\frac{39}{30}\) = \(\frac{13}{10}\) = 1.3 cm ….(i)
\(\frac{\mathrm{PF}}{\mathrm{FR}}\) = \(\frac{3.6}{2.4}\) = \(\frac{36}{24}\) = \(\frac{3}{2}\) = 1.5 cm ……..(ii)
From (i) and (ii)
\(\frac{\mathrm{PE}}{\mathrm{EQ}}\) ≠ \(\frac{\mathrm{PF}}{\mathrm{FR}}\)
∴ EF is not parallel to QR.

(ii) PE = 4 cm., QE = 4.5 cm.,
PF = 8 cm. and RF = 9 cm.
\(\frac{\mathrm{PE}}{\mathrm{QE}}\) = \(\frac{4}{4.5}\) = \(\frac{40}{45}\) = \(\frac{8}{9}\) ….(i)
\(\frac{\mathrm{PF}}{\mathrm{RF}}\) = \(\frac{8}{9}\) ……….(ii)
From (i) and (ii), \(\frac{\mathrm{PE}}{\mathrm{QE}}\) = \(\frac{\mathrm{PF}}{\mathrm{RF}}\)
∴ By converse of Basic Proportionality Theorem,
EF || QR

(iii) PQ = 1.28 cm., PR = 2.56 cm.,
PE = 0.18 cm., PF = 0.36 cm.
EQ = PQ – PE = 1.28 – 0.18 = 1.10 cm.
FR = PR – PF = 2.56 – 0.36 = 2.20 cm.
Here \(\frac{\mathrm{PE}}{\mathrm{EQ}}\) = \(\frac{0.18}{1.10}\) = \(\frac{18}{110}\) = \(\frac{9}{55}\) ….(i)
and \(\frac{\mathrm{PF}}{\mathrm{FR}}\) = \(\frac{0.36}{2.20}\) = \(\frac{36}{220}\) = \(\frac{9}{55}\) ….(ii)
From (i) and (ii), \(\frac{\mathrm{PE}}{\mathrm{EQ}}\) = \(\frac{\mathrm{PF}}{\mathrm{FR}}\)
∴ By converse of Basic Proportionality Theorem,
EF || QR

Question 3.
In Fig., if LM || CB and LN || CD, prove \(\frac{\mathrm{AM}}{\mathrm{AB}}\) = \(\frac{\mathrm{AN}}{\mathrm{AD}}\)

Solution:
According to the question in △ABC,
ML || BC (Given)
∴ \(\frac{\mathrm{AM}}{\mathrm{MB}}\) = \(\frac{\mathrm{AL}}{\mathrm{LC}}\) ….(i)
By converse of Basic Proportionality Theorem,
Again in △ADC
LN || DC
∴ \(\frac{\mathrm{AN}}{\mathrm{ND}}\) = \(\frac{\mathrm{AL}}{\mathrm{LC}}\) …(ii)
(By Basic Proportionality Theorem)
From (i) and (ii),

Question 4.
In Fig., DE || AC and DF || AE. Prove that \(\frac{\mathrm{BF}}{\mathrm{FE}}\) = \(\frac{\mathrm{BE}}{\mathrm{EC}}\).

Solution:
According to the question in △ABC,
DE || AC (Given)
∴ \(\frac{\mathrm{BD}}{\mathrm{DA}}\) = \(\frac{\mathrm{BE}}{\mathrm{EC}}\) ….(i)
(By Basic Proportionality Theorem)
Again in △ABE, DF || AE (Given)
∴ \(\frac{\mathrm{BD}}{\mathrm{DA}}\) = \(\frac{\mathrm{BF}}{\mathrm{FE}}\) ….(ii)
(By Basic Proportionality Theorem)
From (i) and (ii), \(\frac{\mathrm{BE}}{\mathrm{EC}}\) = \(\frac{\mathrm{BF}}{\mathrm{FE}}\) (Hence Proved)

Question 5.
In Fig., DE || OQ and DF || OR. Show that EF || QR.

Solution:
Given : In △PQR
DE || OQ, DF || OR
To prove: EF || QR
Proof : In △PQO, ED || QO (Given)
∴ \(\frac{\mathrm{PD}}{\mathrm{DO}}\) = \(\frac{\mathrm{PE}}{\mathrm{EQ}}\) ….(i)
(By Basic Proportionality Theorem)
Again in △POR,
DF || OR (Given)
∴ \(\frac{\mathrm{PD}}{\mathrm{DO}}\) = \(\frac{\mathrm{PF}}{\mathrm{FR}}\) ….(ii)
(By Basic Proportionality Theorem)
From (i) and (ii),
\(\frac{\mathrm{PE}}{\mathrm{EQ}}\) = \(\frac{\mathrm{PF}}{\mathrm{FR}}\)
In △PQR, \(\frac{\mathrm{PE}}{\mathrm{EQ}}\) = \(\frac{\mathrm{PF}}{\mathrm{FR}}\)
∴ By converse of Basic Proportionality Theorem,
EF || QR (Hence Proved)

Question 6.
In Fig., 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 : In △PQR points A, B and C are on OP, OQ and OR respectively such that
AB || PQ, AC || PR
To prove : BC || QR
Proof: In △OPQ,
AB || PQ (Given)
∴ \(\frac{\mathrm{OA}}{\mathrm{AP}}\) = \(\frac{\mathrm{OB}}{\mathrm{BQ}}\) …….(i)
(By Basic Proportionality Theorem)
Again in △OPR,
AC || PR (Given)
∴ \(\frac{\mathrm{OA}}{\mathrm{AP}}\) = \(\frac{\mathrm{OC}}{\mathrm{CR}}\) …(ii)
(By Basic Proportionality Theorem)
From (i) and (ii),
\(\frac{\mathrm{OB}}{\mathrm{BQ}}\) = \(\frac{\mathrm{OC}}{\mathrm{CR}}\)
∴ By converse of Basic Proportionality Theorem in △OQR
BC || QR (Hence Proved)

Question 7.
Using Theorem 6.1, prove that a line drawn through the mid-point 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, D is the mid-point of AB, i.e., AD = DB. A line parallel to BC intersects AC at E, i.e., DE || BC

To prove : E is the mid-point of AC.
Proof: D is the mid-point of AB
i.e., AD = DB (Given)
⇒ \(\frac{\mathrm{AD}}{\mathrm{DB}}\) = 1 …..(i)
Again in △ABC,
DE || BC
∴ \(\frac{\mathrm{AD}}{\mathrm{DB}}\) = \(\frac{\mathrm{AE}}{\mathrm{EC}}\)
(By Basic Proportionality Theorem)
∴ 1 = \(\frac{\mathrm{AE}}{\mathrm{EC}}\) [From (i)]
∴ AE = EC
∴ E is the mid-point of AC. (Hence Proved)

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 △ABC, D and E are respectively the mid-points of AB and AC so that AD = BD and AE = EC. D and E have been joined.

To prove : DE || BC
Proof: D is the mid-point of AB. (Given)
i.e., AD = BD
⇒ \(\frac{\mathrm{AD}}{\mathrm{BD}}\) = 1 ….(i)
Again E is the mid-point of AC. (Given)
∴ AE = EC
⇒ \(\frac{\mathrm{AE}}{\mathrm{EC}}\) = 1 ….(ii)
From (i) and (ii),
\(\frac{\mathrm{AD}}{\mathrm{BD}}\) = \(\frac{\mathrm{AE}}{\mathrm{EC}}\)
∴ By converse of Basic Proportionality Theorem,
DE || BC (Hence Proved)

Question 9.
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that \(\frac{\mathrm{AO}}{\mathrm{BO}}\) = \(\frac{\mathrm{CO}}{\mathrm{DO}}\).
Solution:
Given : ABCD is a trapezium in which AB || DC. The diagonals AC and BD intersect each other at point O.

To prove : \(\frac{\mathrm{AO}}{\mathrm{BO}}\) = \(\frac{\mathrm{CO}}{\mathrm{DO}}\)
Construction : Through O draw FO || DC || AB
Proof: Now in △DAB.
FO || AB (By Construction)
∴ \(\frac{\mathrm{DF}}{\mathrm{FA}}\) = \(\frac{\mathrm{DO}}{\mathrm{OB}}\) ….(i)
(By Basic Proportionality Theorem)
Again in △DCA,
FO || DC (By Construction)
∴ \(\frac{\mathrm{DF}}{\mathrm{FA}}\) = \(\frac{\mathrm{CO}}{\mathrm{OA}}\) ….(ii)
(By Basic Proportionality Theorem)
Fron (i) and (ii),
\(\frac{\mathrm{DO}}{\mathrm{OB}}\) = \(\frac{\mathrm{CO}}{\mathrm{OA}}\)
or \(\frac{\mathrm{OA}}{\mathrm{OB}}\) = \(\frac{\mathrm{CO}}{\mathrm{DO}}\) (Hence Proved)

Question 10.
The diagonals of a quadrilateral ABCD intersect each other at the point O such that \(\frac{\mathrm{AO}}{\mathrm{BO}}\) = \(\frac{\mathrm{CO}}{\mathrm{DO}}\). Show that ABCD is a trapezium.
Solution:
Given : In quadrilateral ABCD the diagonals AC and BD interesect each other at the point O such that \(\frac{\mathrm{AO}}{\mathrm{BO}}\) = \(\frac{\mathrm{CO}}{\mathrm{DO}}\).

To prove : The quadrilateral ABCD is a trapezium.
Construction : Through ‘O’ draw a line EO || AB, which meets AD at E.
Proof: In △DAB.
EO || AB (By Construction)

∴ By converse of Basic Proportionality Theorem,
EO || DC
Also, EO || AB
⇒ AB || DC
∴ The quadrilateral ABCD is a trapezium. (Hence Proved)