# KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.2

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### Karnataka State Syllabus Class 10 Maths Chapter 2 Triangles Ex 2.2

Exercise 2.2 Class 10 Maths Triangles Question 1.
In the following fig. (i) and (ii). DE || BC.
Find EC in (i) and AD in (ii).

Solution:
(i) In ∆ABC, DE || BC, EC =?
1.53=1EC
1.5 EC = 1 × 3
EC=31.5
EC=3015
∴ EC = 2 cm

(i) In ∆ABC, DE || BC, AD =?
=1.854×721
= 0.6 × 4

KSEEB Solutions For Class 10 Maths Triangles 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 = 3.9cm. EQ = 3 cm
PF=3.6cm. FR = 2.4 cm.
(ii) PE = 4 cm. QE = 4.5 cm.
PF = 8 cm. RF = 9 cm.
(iii) PQ = 1.28 cm. PR = 2.56 cm.
PE = 0.18 cm. PF = 0.36 cm.
Solution:

In ∆PQR, if EF || QR then,
PEQR=PFFR
3.93=3.62.4
1.3 ≠ 1.5
Sides are not dividing In ratio.
∴ EF is not parallel to QR.

(ii)

In ∆PQR, if EF || QR then
PEQR=PFFR
44.5=89
89=89
Here sides are dividing in ratio.
∴ EF ||QR

(iii)

PE + EQ = PQ
0.18 + EQ = 1.28
∴ EQ=1.28 – 0.18
EQ = 1.1 cm.
Similarly. PF + FR = PR
0.36 + FR = 2.56
FR = 2.56 – 0.36
FR = 2.2cm.
In ∆PQR, if EF || QR then
PEQE=PFFR
0.181.1=0.362.2
1.811=3.622
1.811=11.811
Here sides are dividing in ratio.
∴ EF || QR

10th Maths Triangles Exercise 2.2 Question 3.
In the following figure. If LM || CB and LN || CD, prove that AMAB=ANAD

Solution:
Data: LM || CB and LN || CD then prove that AMAB=ANAD
Solution: In ∆ACB, LM || CB
ALAC=AMAB……(1)( Theorem 1)
Similarly in ∆ADC, LN || CD
From equation (1) and (2) we have

Triangles Class 10 Exercise 2.2 Question 4.
In the following figure, DE ||AC and DF || AE. Prove that BFFE=BEEC

Solution:
Data: In this figure, AE || AC and DF || AE, then we have to prove that BFFE=BEEC
Solution: In ∆ABC, DE || AC.
Similarly, In ∆ABE, DF ||AE.
from equation (1) and (2). we have
BEED=BFFE

Triangle Lesson Exercise 2.2 Question 5.
In the following figure. DE || OQ and DF || OR. Show that EF || QR.

Solution:
Data: In this figure. DE || OQ and DF || OR.
To Prove: EF || QR
Solution: In ∆POQ, DE || OQ.
PEEQ=PDDO…… (i) ( Theroem 1)
Similarly. In ∆POR. DF || OR.
PDDO=PFFR…… (ii) ( Theorem 1)
from equation (1) and (11), we have
PFEQ=PDDO=PFFR
PEEQ=PFFR
In ∆PQR. if PEEQ=PFFR then EF || QK. (∵ Theorem 1).

Triangles Class 10 Solutions KSEEB Question 6.
In the following figure, A, B an C are points on OP. OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

Solution:
Data: In ∆QPR, AB || PQ and AC || PR. A, B, and C are the points on OP. OQ and OR.
To Prove: BC || QR
Solution: In ∆OPQ, AB || PQ.
OAAP=OBBQ…… (i) ( Theorem 1)
Similarly, In ∆OPR. AC|| PR.
OAAP=OCCR…… (i) ( Theorem 1)
from equation (i) and (ii), we have
OAAP=OBBQ=OCCR
OBBQ=OCCR
∴ BC || QR (∵ Theorem 2)

KSEEB Solutions For Class 10 Maths Question 7.
Using Theorem 2.1, 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:
Data: In ∆ABC. D is the mid-point of AB.
DE Is drawn parallel to BC from D.
To Prove: DE bisects AC side at E.
∠D = ∠B (corresponding angles)
∠E = ∠C (corresponding angles)
12=AEAC
∴ AC = AE + EC
∴ E is the mid-point of AC
∴ DE bisects AC at E.

Triangles Class 10 KSEEB Solutions Question 8.
Using Theorem 2.2, 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:
Solution: In ∆ABC, D and E are mid-points of AB and AC.
AE = EC
AC = 2AE
As per S.S.S. Postulate.
∴ They are equiangular triangles.
∴ ∠A is common.
These are pair of corresponding angles
∴ DE || BC

Triangles Class 10 Exercise 2.2 Solutions Question 9.
ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AOCO=CODO
Solution:
Data : ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O.
To Prove: AOCO=CODO
Proof: In Trapezium ABCD. AB || DC.
∴ In ∆AOB and ∆DOC,
∠OCO = ∠OAB (alternate angles)

∠ODC = ∠OBA (alternate angles)
∠DOC = ∠AOB (vertically opposite angles)
∴ ∆AOB and ∆DOC are equiangular triangles.
∴ ∆AOB ||| ∆DOC
Similar triangles divides sides in ratio.
AOCO=CODO

10th Maths Triangles Exercise 2.2 Solutions Question 10.
The diagonals of a quadrilateral ABCD intersect each other at the point O such that AOBO=CODO . Show that ABCD is a trapezium.

Solution:
Data: In the quadrilateral ABCD. the diagonals intersect at 0’ such that AOBO=CODO
To Prove: ABCD is a trapezium.
Solution: In the qudrilateral ABCD, AOBO=CODO
AOCO=OBOD
It means sides of ∆AOB and ∆DOC divides proportionately.
∴ ∆AOB ||| ∆DOC.
Similarly. ∆AOD ||| ∆BOC.
Now, ∆AOB + ∆AOD = ∆BOC + ∆DOC
∆ABD = ∆ABC.
Both triangles are on the same base AB and between two pair of lines and equal in area.
∴ AB || DC.

All Chapter KSEEB Solutions For Class 10 Maths

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All Subject KSEEB Solutions For Class 10

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