Samacheer Kalvi 10th Maths Solutions Chapter 4 Exercise 4.2 Pdf

Samacheer Kalvi 10th Maths Solutions Chapter 4 Exercise 4.2 Pdf

TN Samacheer Kalvi 10th Maths Solutions Chapter 4 Exercise 4.2 : Hello students, Are You looking for Samacheer Kalvi 10th Maths Solutions Chapter 4 Exercise 4.2 – Geometry on internet. If yes, here we have given TNSCERT Class 10 Maths Solution Chapter 4 Exercise 4.2 Pdf Guide for Samacheer Kalvi 10th Mathematics Solution for Exercise 4.2

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 STPR, 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]

 

Here we posted the solution of TN Tamilnadu Board Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry. TNSCERT Samacheer Kalvi 10th Maths Chapter 4 Geometry Solved by Expert Teacher.

Updated: January 7, 2022 — 2:52 pm

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