**KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangle Ex 2.2** are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangle Exercise 2.2.

## Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangle Exercise 2.2

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 =?

\(\quad \frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)

\(\frac{1.5}{3}=\frac{1}{\mathrm{EC}}\)

1.5 EC = 1 × 3

\(\mathrm{EC}=\frac{3}{1.5}\)

\(E C=\frac{30}{15}\)

∴ EC = 2 cm

(i) In ∆ABC, DE || BC, AD =?

\(\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{AE}}{\mathrm{EC}}\)

\(\frac{\mathrm{AD}}{7.2}=\frac{1.8}{5.4}\)

\(\quad A D=\frac{1.8}{5.4} \times \frac{7.2}{1}\)

\(=\frac{1.8}{54} \times \frac{72}{1}\)

= 0.6 × 4

∴ 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 = 3.9cm. 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,

\(\frac{\mathrm{PE}}{\mathrm{QR}}=\frac{\mathrm{PF}}{\mathrm{FR}}\)

\(\frac{3.9}{3}=\frac{3.6}{2.4}\)

1.3 ≠ 1.5

Sides are not dividing In ratio.

∴ EF is not parallel to QR.

(ii)

In ∆PQR, if EF || QR then

\(\frac{\mathrm{PE}}{\mathrm{QR}}=\frac{\mathrm{PF}}{\mathrm{FR}}\)

\(\frac{4}{4.5}=\frac{8}{9}\)

\(\frac{8}{9}=\frac{8}{9}\)

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

\(\frac{\mathrm{PE}}{\mathrm{QE}}=\frac{\mathrm{PF}}{\mathrm{FR}}\)

\(\frac{0.18}{1.1}=\frac{0.36}{2.2}\)

\(\frac{1.8}{11}=\frac{3.6}{22}\)

\(\frac{1.8}{11}=\frac{11.8}{11}\)

Here sides are dividing in ratio.

∴ EF || QR

Question 3.

In the following figure. If LM || CB and LN || CD, prove that \(\frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}\)

Solution:

Data: LM || CB and LN || CD then prove that \(\frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}\)

Solution: In ∆ACB, LM || CB

\(\quad \frac{\mathrm{AL}}{\mathrm{AC}}=\frac{\mathrm{AM}}{\mathrm{AB}} \quad \ldots \ldots(1)(\text { Theorem } 1)\)

Similarly in ∆ADC, LN || CD

\(\quad \frac{\mathrm{AL}}{\mathrm{AC}}=\frac{\mathrm{AN}}{\mathrm{AD}} \quad \ldots \ldots(2) \quad(\ldots \text { Theorem } 1)\)

From equation (1) and (2) we have

\(\frac{\mathrm{AL}}{\mathrm{AC}}=\frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}\)

\(\quad \frac{\mathrm{AM}}{\mathrm{AB}}=\frac{\mathrm{AN}}{\mathrm{AD}}\)

Question 4.

In the following figure, DE ||AC and DF || AE. Prove that \(\frac{\mathrm{BF}}{\mathrm{FE}}=\frac{\mathrm{BE}}{\mathrm{EC}}\)

Solution:

Data: In this figure, AE || AC and DF || AE, then we have to prove that \(\frac{\mathrm{BF}}{\mathrm{FE}}=\frac{\mathrm{BE}}{\mathrm{EC}}\)

Solution: In ∆ABC, DE || AC.

\(\quad \frac{\mathrm{BD}}{\mathrm{AD}}=\frac{\mathrm{BE}}{\mathrm{EC}} \ldots \ldots(1) \quad(\text { Theorem } 1)\)

Similarly, In ∆ABE, DF ||AE.

\(\quad \frac{\mathrm{BD}}{\mathrm{AD}}=\frac{\mathrm{BF}}{\mathrm{FE}}\)

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

\(\frac{\mathrm{BD}}{\mathrm{AD}}=\frac{\mathrm{BE}}{\mathrm{EC}}=\frac{\mathrm{BF}}{\mathrm{FE}}\)

\(\quad \frac{\mathrm{BE}}{\mathrm{ED}}=\frac{\mathrm{BF}}{\mathrm{FE}}\)

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.

\(\quad \frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{\mathrm{PD}}{\mathrm{DO}} \quad \ldots \ldots \text { (i) }(\text { Theroem } 1)\)

Similarly. In ∆POR. DF || OR.

\(\quad \frac{\mathrm{PD}}{\mathrm{DO}}=\frac{\mathrm{PF}}{\mathrm{FR}} \quad \ldots \ldots \text { (ii) }(\text { Theorem } 1)\)

from equation (1) and (11), we have

\(\frac{P F}{E Q}=\frac{P D}{D O}=\frac{P F}{F R}\)

\(\frac{\mathrm{PE}}{\mathrm{EQ}}=\frac{\mathrm{PF}}{\mathrm{FR}}\)

In ∆PQR. if \(\frac{P E}{E Q}=\frac{P F}{F R}\) then EF || QK. (∵ Theorem 1).

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.

\(\quad \frac{\mathrm{OA}}{\mathrm{AP}}=\frac{\mathrm{OB}}{\mathrm{BQ}} \quad \ldots \ldots \text { (i) }(\text { Theorem } 1)\)

Similarly, In ∆OPR. AC|| PR.

\(\quad \frac{\mathrm{OA}}{\mathrm{AP}}=\frac{\mathrm{OC}}{\mathrm{CR}} \quad \ldots \ldots \text { (i) }(\text { Theorem } 1)\)

from equation (i) and (ii), we have

\(\frac{\mathrm{OA}}{\mathrm{AP}}=\frac{\mathrm{OB}}{\mathrm{BQ}}=\frac{\mathrm{OC}}{\mathrm{CR}}\)

\(\quad \frac{\mathrm{OB}}{\mathrm{BQ}}=\frac{\mathrm{OC}}{\mathrm{CR}}\)

∴ BC || QR (∵ Theorem 2)

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

\(\quad \frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}\)

\(\frac{1}{2}=\frac{\mathrm{AE}}{\mathrm{AC}}\)

AC = 2AD

∴ AC = AE + EC

∴ E is the mid-point of AC

∴ DE bisects AC at E.

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

\(\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}=\frac{1}{2}\)

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

Question 9.

ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that \(\frac{\mathrm{AO}}{\mathrm{CO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\)

Solution:

Data : ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O.

To Prove: \(\frac{\mathrm{AO}}{\mathrm{CO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\)

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.

\(\frac{\mathrm{AO}}{\mathrm{CO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\)

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:

Data: In the quadrilateral ABCD. the diagonals intersect at 0’ such that \(\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\)

To Prove: ABCD is a trapezium.

Solution: In the qudrilateral ABCD, \(\frac{\mathrm{AO}}{\mathrm{BO}}=\frac{\mathrm{CO}}{\mathrm{DO}}\)

\(\quad \frac{\mathrm{AO}}{\mathrm{CO}}=\frac{\mathrm{OB}}{\mathrm{OD}}\)

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.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangle Ex 2.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangle Exercise 2.2, drop a comment below and we will get back to you at the earliest.