**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,

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

Here sides are dividing in ratio.

∴ EF || QR

Question 3.

In the figure, if LM || CB and LN || CD, prove that \(\frac{A M}{A B}=\frac{A N}{A D}\)

Solution:

In ∆ABC, LM || CB [given]

∴ Using the Basic proportionality theorem, we have

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

Solution: In ∆ABC, DE || AC.

Similarly, In ∆ABE, DF ||AE.

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

Question 5.

In the figure, DE || OQ and DF || OR. Show that EF || QR.

Solution:

In ∆PQO,

∵ DE || OQ [given]

∴ Using the Basic proportionality theorem, we have

\(\frac{P E}{E Q}=\frac{P D}{D O}\) …………… (1)

Again, in ∆POR, DF || OR [given]

∴ Using the Basic proportionality theorem, we have

Now, in ∆PQR,

∵ E and F are two distinct points on PQ and PR respectively and \(\frac{P E}{E Q}=\frac{P F}{F R}\),

i.e., E and F divide the two sides PQ and PR of ∆PQR in the same ratio.

∴ By converse of Basic proportionality theorem, EF || QR.

Question 6.

In the following 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:

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.

Similarly, In ∆OPR. AC|| PR.

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

∴ BC || QR (∵ Theorem 2)

Question 7.

Using Basic proportionality theorem, prove that a line drawn through the mid-point of one side of a triangle parallel to another side bisects the third side. A

Solution:

We have ∆ABC, in which D is the midpoint of AB and E is a point on AC such that DE || BC.

∵ DE || BC [given]

∴ Using the Basic proportionality theorem, we get

⇒ E is the mid point of AC. Hence, it is proved that a line through the midpoint of one side of a triangle parallel to another side bisects the third side.

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

As per S.S.S. Postulate.

∆ADE ~ ∆ABC

∴ They are equiangular triangles.

∴ ∠A is common.

∠ADE = ∠ABC

∠ADE = ∠ACE

These are a 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 \(\frac{A O}{B O}=\frac{C O}{D O}\). Show that ABCD is a trapezium.

Solution:

It is given that \(\frac{A O}{B O}=\frac{C O}{D O}\)

From \(\frac{A O}{B O}=\frac{C O}{D O}\), we have \(\frac{A O}{C O}=\frac{B O}{D O}\)

Let us draw OE such that OE || BA

In ∆ADB, OE || AB [By construction]

i.e., the points O and E divide the sides AC and AD of ∆ADC respectively in the same ratio.

∴ Using the converse of Basic proportionality theorem, we get OE || DC and OE || AB

⇒ AB || DC

⇒ ABCD is a trapezium.

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.