Samacheer Kalvi 10th Maths Solutions Chapter 4 Exercise 4.2 Pdf
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TN Samacheer Kalvi 10th Maths Solutions Chapter 4 – Geometry
Board | TNSCERT Class 10th Maths |
Class | 10 Class |
Subject | Maths |
Chapter | 4 (Exercise 4.2) |
Chapter Name | Geometry |
TNSCERT Class 10th Maths Pdf | all Exercise Solution
Exercise – 4.2
(1) In ∆ABC, D and E are points on the sides AB and AC respectively such that DE||BC
(i) If AD/DB = 3/4 and AC = 15cm find AE.
(ii) If AD = − 8 7 x, DB = − 5 3 x, AE = − 4 3 x and EC = − 3 1 x , find the value of x.
Solution:
(i) Given that AD/DB = 3/4 and AC = 15cm
By Thales theorem,
AD/DB = AE/EC
AD/DB + 1 = AE/EC + 1 [∵ both side adding 1]
AD/DB +1 = AE+EC/EC
AD/DB + 1 = AC/EC [∵ AC = AE + EC]
3/4 + 1 = 15/EC
7/4 = 15/EC
EC = 15×4/7
EC = 60/7
Now, AC = AE + EC
AE = AC – EC
AE = 15 – 60/7 = 105-60/7 = 45/7 = 6.4 cm
Therefore, the value of AE = 6.4cm
(ii) By Thales theorem,
AD/DB = AE/EC
= > 8x–7/5x-3 = 4x-3/3x-1 [∵ AD = 8x – 7, DB = 5x-3, AE = 4x-3, EC = 3x-1]
= (8x – 7) (3x – 1) = (4x – 3) (5x – 3)
= 24x2 – 8x – 21x + 7 = 20x2 – 12x – 15x + 9
= 24x2 – 20x2 – 29x + 27x + 7 – 9 = 0
= 4x2 – 2x – 2 = 0
= 2x2 – x – 1 = 0
= 2x2 – 2x + x – 1 = 0
= 2x (x – 1) + 1 (x – 1) = 0
= (x – 1) (2x + 1) = 0
= (x – 1) = 0
x – 1 = 0
x = 1
(2x + 1) = 0
2x = -1
x = -1/2
x ≠ -1/2 [∵ Any length value is not negative]
∴ therefore the value of x is 1.
(2) ABCD is a trapezium in which AB||DC and P, Q are Points on AD and BC respectively, such that PQ||DC if PD = 18cm, BQ = 35cm and QC = 15cm, Find AD.
Solution:
Give that PD = 18cm, BQ = 35cm, QC = 15cm
Let, AP = x cm
By Thales theorem,
AP/PD = BQ/QC
x/18 = 35/15
x = 35×18/15
x = 42
Now, AP = 42 cm
Then, AD = AP + PD
= (42 + 18) cm
= 60cm
Thus, the value of AD = 60cm
(3) In ∆ABC, D and E are points on the sides AB and AC respectively. For each of the following cases show that DE||BC
(i) AB = 12cm, AD = 8cm, AE = 12cm and AC = 18cm
(ii) AB = 5.6cm, AD = 1.4cm, AC = 7.2cm and AE = 1.8 cm.
Solution:
(i) Now, given that
AB = 12cm
AD = 8cm
Then DB = 12-8 = 4cm
AC = 18cm
AE = 12cm
EC = (18-12) cm
= 6cm
Now, AD/DB = 8/4 = 2 and AE/EC = 12/6 = 2
Now, AD/DB = AE/EC
∴ So, DE||BC [Proved]
(ii) Given that,
AB = 5.6cm
AD = 1.4cm
Then DB = AB – AD
= (5.6 – 1.4) cm
= 4.2 cm
and AC = 7.2 cm
AE = 1.8 cm
EC = (AC – AE)
= (7.2 – 1.8) cm
= 5.4 cm
Now, AD/DB = 1.4/4.2 = 1/3 and AE/EC = 1.8/5.4 = 1/3
Now, AD/DB = AE/EC
∴ DE||BC [Proved]
(4) In figure. if PQ||BC and PR||CD prove that
(i) AR/AD = AQ/AB
(ii) QB/AQ = DR/AR
Solution:
Now, PQ||BC,
then by Thales theorem,
AQ/QB = AP/PC
QB/AQ = PC/AP
1 + QB/AQ = PC/AP + 1 [Both side adding 1]
AQ+QB/AQ = AP+PC/AP
AB/AQ = AC/AP [∵ AQ + QB = AB, AP + DC = AC]
AB/AQ = AC/AP —- (i)
Now, PR||CD, then by Thales theorem,
AR/RD = AP/PC
RD/AR = PC/AP
1 + RD/AR = DC/AP + 1 [Both side adding 1]
AR + RD/AR = AP + PC/AD
AD/AR = AC/AP [∵ AR + RD = AD, AP + PC = AC]
AD/AR = AC/AP
AD/AR = AB/AQ [By equation (i)]
AR/AD = AQ/AB [Proved]
(ii) PQ||BC then, by Thales theorem,
AQ/QB = AD/PC —- (i) and PR||CD, then, by Thales theorem.
AR/RD = AP/DC —- (ii)
Now, AD/PC = AR/RD
AQ/QB = AR/RD (By equation (i)]
QB/A = DR/AR (Proved)
(5) Rhombus PQRB is inscribed in ∆ABC such that ∠B is one of its angle. P, Q and R lie on AB, AC and BC respectively. If AB=12 cm and BC = 6 cm, find the sides PQ, RB of the rhombus.
Solution:
Rhombus PQRB, let PQ = QR = BR = PB = x
Now, PQ||BC, then by Thales theorem,
AP/PB = AQ/QC
1 + AP/PB = 1 + AQ/QC [Both side adding 1]
AP + PB/PB = AQ+QC/QC
AB/PB = AC/QC [∵ AQ + QC = AC, AP + PB = AB]
∴ AB/PB = AC/QC —- (i)
Now, QR||AB, then Thales theorem,
CR/RB = CQ/QA
RB/CR = QA/CQ
CR+RB/CR = CQ+QA/CQ [Both side adding 1]
BC/CR = AC/QC [∵ AQ+QC = AC, CR+RB = BC]
BC/CR = AB/PB [by equation (i)]
6/6-x = 12/x [∵ BC = 6cm, CR = BC – BR, AB = 12, PB = x]
6x = 72 – 12x
18x = 72
x = 72/18 = 4
∴ Therefore rhombus sides PQ = RB = 4cm
(6) In trapezium ABCD, AB||DC, E and F are points on non-parallel sides AD and BC respectively, such that EF||AB. Show that AE/ED = BF/FC.
Solution:
Now, Join BD then 0 point is intersection EF
Now, ∆BDC, then OF||DC
By Thales theorem,
BF/FC = BO/OD —- (i)
Now, ∆ADB, then OE||AB
by Thales theorem,
DE/EA = DO/OB
AE/ED = BO/OD
AE/ED = BF/FC [by equation (i)]
∴ AE/ED = BF/FC [Proved]
(7) In figure DE||BC and CD||EF. Prove that AD2 = AB×AF
Solution:
Now, ∆ABC, DE||BC
By Thales theorem,
BD/AD = CE/AE
BD+AD/AD = CE+AE/AE [Both side adding 1]
AB/AD = AC/AE [∴ AB = AD+DB, AC = AE+EC]
AB/AD = AC/AE —- (i)
Now, ∆ADC, EF||DC
By Thales theorem,
DF/AF = CE/AE
AF+FD/AF = AE+EC/AE [∵ both side adding 1]
AD/AF = AC/AE [∵ AD = AF+FD, AC = AE+EC]
AD/AF = AB/AD [By equation (i)]
AD2 = AB×AF [Proved]
(8) Check whether AD is bisector of ∠A of ∆ABC in each of the following
(i) AB = 5 cm, AC = 10 cm, BD. = 1 5 cm and CD. = 3 5 cm.
(ii) AB= 4 cm, AC = 6 cm, BD = 1.6 cm and CD = 2 4 cm.
Solution:
(i) Now, AB/AC = 5/10 = 1/2
and, BD/DC = 1.5/3.5 = 3/7
Now, AB/AC ≠ BD/DC
So, AD is not bisector of ∠A.
(ii) Now, AB/AC = 4/6 = 2/3
and BD/DC = 1.6/3.4 = 2/3
Now, AB/AC = BD/DC
By converse of angle bisector theorem
Thus, AD is bisector of ∠A.
(9) In figure ∠QPR = 90°, PS is its bisector. If ST⊥PR, prove that ST×(PQ+PR)=PQ×PR.
Solution:
Now, ∠QPR = ∠STR = 90°
and ∠R is common.
∴ ∆PQR and ∆RST are similar.
∴ SR/QR = ST/PQ
QR/SR = PQ/ST —– (i)
Now, ∆PQR; PS is bisector of ∠P
By angle bisector theorem,
QS/SR = PQ/PR
RS + SQ/RS = PQ+RP/RP [Both side adding 1]
QR/RS = PQ+RP/RP [∵ QS + SR = QR]
PQ/ST = PQ+RP/RP [By equation (i)]
ST × (PQ+RP) = PQ × RP
∴ ST × (PQ + PR) = PQ × PR [Proved]
(10) ABCD is a quadrilateral in which AB=AD, the bisector of ∠BAC and ∠CAD intersect the sides BC and CD at the points E and F respectively. Prove that EF||BD.
Solution:
Now, ∆ADC, AF is bisector of ∠DAC,
By angle bisector theorem,
AD/AC = DF/FC —– (i)
Now, ∆ACB, AE is bisector of ∠CAB.
By angle bisector theorem,
AB/AC = BE/EC
AD/AC = BE/EC [∵ Given that AB = AD]
DF/FC = BE/EC [By equation (i)]
CF/FD = CE/EB
∴ By Thales theorem,
∴ EF||BD. [Proved]
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