Samacheer Kalvi 10th Maths Solutions Chapter 4 Exercise 4.1 Pdf

Samacheer Kalvi 10th Maths Solutions Chapter 4 Exercise 4.1 Pdf

TN Samacheer Kalvi 10th Maths Solutions Chapter 4 Exercise 4.1 : Hello students, Are You looking for Samacheer Kalvi 10th Maths Solutions Chapter 4 Exercise 4.1 – Geometry on internet. If yes, here we have given TNSCERT Class 10 Maths Solution Chapter 4 Exercise 4.1 Pdf Guide for Samacheer Kalvi 10th Mathematics Solution for Exercise 4.1

TN Samacheer Kalvi 10th Maths Solutions Chapter 4 – Geometry

Board TNSCERT Class 10th Maths
Class 10 Class
Subject Maths
Chapter 4 (Exercise 4.1)
Chapter Name Numbers and Sequences

TNSCERT Class 10th Maths Pdf | all Exercise Solution

 

Exercise – 4.1

 

(1) Check whether the which triangles are similar and find the value of x.

(i)  

(ii) 

Solution:

(i)

Now, ∆ABC and ∆ADE

AB/AD = AC/AE

8/3 = (11/2)/2

8/3 = 11/4

33 = 32

33 ≠ 32

So, ∆ABC and ∆ADE are not similar triangles.

 

(ii)

Now ∠PQC = 180 – 110 = 70°

∴ ∠ABC = 70°

∴ ∠ABC = ∠PQC and

So, ∆ABC and ∆PQC are similar triangles.

∴ AB/PQ = BC/QC

5/x = 6/3

x = 155/62

∴ x = 5/2 = 2.5

Thus, ∆ABC ~ ∆PQC and value of x is 2.5.

 

(2) A girl looks the reflection of the top of the lamp post on the mirror which is 6.6 m away from the foot of the lamppost. The girl whose height is 1.25 m is standing 2.5 m away from the mirror. Assuming the mirror is placed on the ground facing the sky and the girl, mirror and the lamppost are in a same line, find the height of the lamp post.

Solution:

Now, BE = 2.5m, AB = 6.6m and DE = 1.25m.

Let, AC = h = Lamppost height.

Now, ∆ABC and ∆BED are similar

So, AC/DE = AB/BE

h/1.25 = 6.6/2.5

h = 6.6×1.25×10/2.5×10×100

h = 3.3m

Therefore, the height of the lamppost is 3.3m.

 

(3) A vertical stick of length 6 m casts a shadow 400 cm long on the ground and at the same time a tower casts a shadow 28 m long. Using similarity, find the height of the tower.

Solution:

Now, AB = 6m

BC = 400cm = 4m

PHOTO

PR = h (Let)

PQ = 28m

Now, ∆ABC and ∆PQR similar

So, AB/PR = BC/PQ

6/h = 4/28

h = 6×28/4

h = 42m

Therefore, The height of the tower is 42m.

 

(4) Two triangles QPR and QSR, right angled at P and S respectively are drawn on the same base QR and on the same side of QR. If PR and SQ intersect at T, prove that PT × TR = ST × TQ.

Solution:

Now, ∠QPR = ∠QSR = 90° and ∠PTQ = ∠STR

So, ∆PQT and ∆TSR are similar triangles.

∴ PT/TS = TQ/TR

PT.TR = TS.TQ

PT × TR = ST × TQ (Proved)

 

(5) In the adjacent figure, ABC is right angled at C and DE AB. Prove that ABC ~ ADE and hence find the lengths of AE and DE.

Solution:

Now, ∠AED = ∠ACB = 90° and ∠CAB = ∠DAE

∴ ∆ABC and ∆ADE are similar.

∴ ∆ABC ~ ∆ADE (Proved)

Now, ∆ABC

AC2 + BC2 = AB2

AB2 = (5)2 + (12)2

AB2 = 25 + 144 = 169

AB = 13

Now, ∆ADC ~ ∆ADE

AB/AD = AC/AE

13/3 = 5/AE

AE = 5×3/13 = 15/13

and, AB/AD = BC/DE

13/3 = 12/DE

DE = 12×3/13

DE = 36/13

∴ Therefore, AE = 15/13 and DE = 36/13

 

(6) In the adjacent figure, ACB ~ APQ. If BC = 8 cm, PQ = 4 cm, BA = 6.5 cm and AP = 2.8 cm, find CA and AQ.

Solution:-

Given that,

∆ACB ~ ∆ADQ

and, BC = 8 cm

PQ = 4 cm

BA = 6.5 cm

AP = 2.8 cm

BC/PQ = CA/AP

8/4 = CA/2.8

CA = 8×28/4×10

CA = 5.6

And, BC/PQ = BA/AQ

8/4 = 6.5/AQ

∴ AQ = 65×4/28×10 = 13/4

AQ = 3.25

Therefore, the values of CA = 5.6 cm and AQ = 3.25 cm

 

(7) If figure OPRQ is a square and ∠MLN = 90°. Prove that –

(i) ∆LOP ~ ∆QMO

(ii) ∆LOP ~ ∆RPN 

(iii) ∆QMO ~ ∆RPN

(iv) QR2 = MQ×RN

Solution:

(i) Now, ∠OQM = 90°

∴ ∠OLP = ∠OQM

Now, OP||MN

∴ ∠LOP = ∠OMQ

∴ ∆LOP and ∆OMQ are similar [∵ by AA properties]

∴ ∆LOP ~ ∆QMO [Proved]

 

(ii) Now, ∠PRN = 90°

∴ ∠OLP = ∠PRN

Now, OP||MN

∴ ∠LPO = ∠PNR

So, ∆LOP ~ ∆RPN [By AA] [Proved]

(iii) Now, ∠OQM = ∠PRN = 90°

∴ ∠OQM = ∠PRN

and, OP||MN

Now, ∠QOM = ∠RPN

∴ ∆QMO and ∆PNR are similar [By AA]

∴ ∆QMO ~ ∆RPN [Proved]

 

(iv) ∆MOQ and ∆PRN are similar.

MQ/PR = OQ/RN

MQ/QR = QR/RN [∵ square OPRB, ∴ OP = PR = QR = OQ]

QR2 = MQ × RN [Proved]

 

(8) If ABC ~ DEF such that area of ABC is 9cm2 and the area of DEF is 16cm2 and BC = 2.1 cm. Find the length of EF.

Solution:

We know that, the ratio of area of two similar triangles is equal to the ratio of the squares of any two corresponding sides.

Now, ∴ area ∆ABC/area ∆DEF = BC2/EF2

9/16 = (2.1)2/EF2

EF2 = (2.1)2 × (4)2/(3)2

EF = 21×4/3×10

EF = 28/10 = 2.8

Thus, the length of EF = 2.8 cm

 

(9) Two vertical poles of heights 6m and 3m are erected above a horizontal ground AC. Find the value of y.

Solution:

Now, ∠PAC = ∠QBC = 90°

∴ ∠PAC = ∠QBC and ∠C is common

∴ ∆APC ~ ∆BQC

Now, AP/OQ = AC/BC

6/y = AC/BC

BC/AC = y/6 —– (i)

Now, ∠ACR = ∠ABQ = 90°

∴ ∠ACR = ∠ABQ

and ∠A is common

∴ ∆ACR ~ ∆ABQ

∴ RC/BQ = AC/AB

3/y = AC/AB

AB/AC = y/3 —- (ii)

Now, (i) + (ii), we get

BC/AC + AB/AC = y/6 + y/3

BC+AB/AC = 3y+6y/18

AC/AC = 9y/18 [∵ AC = AB+BC]

9y/18 = 1

9y = 18

y = 18/9

y = 2

∴ Therefore, the value of y = 2 cm

 

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 — 2:31 pm

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