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Karnataka State Syllabus Class 10 Maths Chapter 2 Triangles Ex 2.2
Question 1.
In the following fig. (i) and (ii). DE || BC.
Find EC in (i) and AD in (ii).
Answer:
(i) In ∆ABC, DE || BC, EC = ?
1.5 EC = 1 × 3
∴ EC = 2 cm
(ii) In ∆ABC, DE || BC, AD =?
= 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.
Answer:
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
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
Answer:
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 2.19, DE ||AC and DF || AE. Prove that
Answer:
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.2.20, DE || OQ and DF || OR. Show that EF || QR.
Answer:
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 figures 2.21, A, B, and C are points on OP. OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.
Answer:
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).
Answer:
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)
Answer:
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 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
Answer:
Data: ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O.
To Prove:
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.
Question 10.
The diagonals of a quadrilateral ABCD intersect each other at the point O such that . Show that ABCD is a trapezium.
Answer:
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 pairs of lines and equal in area.
∴ AB || DC.