In this chapter, we provide KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.2 for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.2 pdf, free KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.2 book pdf download. Now you will get step by step solution to each question.

**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 =?

ADBD=AEEC

1.53=1EC

1.5 EC = 1 × 3

EC=31.5

EC=3015

∴ EC = 2 cm

(i) In ∆ABC, DE || BC, AD =?

ADBD=AEEC

AD7.2=1.85.4

AD=1.85.4×7.21

=1.854×721

= 0.6 × 4

∴ AD = 2.4 cm

**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

ALAC=ANAD……(2)(… Theorem 1)

From equation (1) and (2) we have

ALAC=AMAB=ANAD

AMAB=ANAD

**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.

BDAD=BEEC……(1)( Theorem 1)

Similarly, In ∆ABE, DF ||AE.

BDAD=BFFE

from equation (1) and (2). we have

BDAD=BEEC=BFFE

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.

Solution: In ∆ADE and ∆ABC,

∠D = ∠B (corresponding angles)

∠E = ∠C (corresponding angles)

∴∆ADE || ∆ABC

ADAB=AEAC

12=AEAC

AC = 2AD

∴ 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.

∴ AD = DB

AE = EC

AB = 2AD

AC = 2AE

ADAB=AEAC=12

As per S.S.S. Postulate.

∆ADE ~ ∆ABC

∴ They are equiangular triangles.

∴ ∠A is common.

∠ADE = ∠ABC

∠ADE = ∠ACE

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**

—————————————————————————–**All Subject KSEEB Solutions For Class 10**

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