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

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 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 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 $\frac{\mathrm{BF}}{\mathrm{FE}}=\frac{\mathrm{BE}}{\mathrm{EC}}$
Solution: In ∆ABC, DE || AC. $\therefore \quad \frac{\mathrm{BD}}{\mathrm{AD}}=\frac{\mathrm{BE}}{\mathrm{EC}} \ldots \ldots(1) \quad(\because \text { Theorem } 1)$
Similarly, In ∆ABE, DF ||AE. $\therefore \quad \frac{\mathrm{BD}}{\mathrm{AD}}=\frac{\mathrm{BF}}{\mathrm{FE}}$
from equation (1) and (2). we have $\frac{\mathrm{BD}}{\mathrm{AD}}=\frac{\mathrm{BE}}{\mathrm{EC}}=\frac{\mathrm{BF}}{\mathrm{FE}}$ $\therefore \quad \frac{\mathrm{BE}}{\mathrm{ED}}=\frac{\mathrm{BF}}{\mathrm{FE}}$

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. $\therefore \quad \frac{\mathrm{OA}}{\mathrm{AP}}=\frac{\mathrm{OB}}{\mathrm{BQ}} \quad \ldots \ldots \text { (i) }(\because \text { Theorem } 1)$
Similarly, In ∆OPR. AC|| PR. $\therefore \quad \frac{\mathrm{OA}}{\mathrm{AP}}=\frac{\mathrm{OC}}{\mathrm{CR}} \quad \ldots \ldots \text { (i) }(\because \text { Theorem } 1)$
from equation (i) and (ii), we have $\frac{\mathrm{OA}}{\mathrm{AP}}=\frac{\mathrm{OB}}{\mathrm{BQ}}=\frac{\mathrm{OC}}{\mathrm{CR}}$ $\therefore \quad \frac{\mathrm{OB}}{\mathrm{BQ}}=\frac{\mathrm{OC}}{\mathrm{CR}}$
∴ 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.
AE = EC
AC = 2AE $\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}=\frac{1}{2}$
As per S.S.S. Postulate.
∴ They are equiangular triangles.
∴ ∠A is common.
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.

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