Samacheer Kalvi 10th Maths Solutions Chapter 4 Exercise 4.5 Pdf

Samacheer Kalvi 10th Maths Solutions Chapter 4 Exercise 4.5 Pdf

TN Samacheer Kalvi 10th Maths Solutions Chapter 4 Exercise 4.5 : Hello students, Are You looking for Samacheer Kalvi 10th Maths Solutions Chapter 4 Exercise 4.5 – Geometry on internet. If yes, here we have given TNSCERT Class 10 Maths Solution Chapter 4 Exercise 4.5 Pdf Guide for Samacheer Kalvi 10th Mathematics Solution for Exercise 4.5

TN Samacheer Kalvi 10th Maths Solutions Chapter 4 – Geometry

Board	TNSCERT Class 10th Maths
Class	10 Class
Subject	Maths
Chapter	4 (Exercise 4.5)
Chapter Name	Geometry

TNSCERT Class 10th Maths Pdf | all Exercise Solution

 

Exercise – 4.5

 

(1) If in triangles ABC and EDF, AB/DE = BC/FD then they will be similar, when

(A) ∠B = ∠E

(B) ∠A = ∠D

(C) ∠B = ∠D

(D) ∠A = ∠F

Solution:

Now, ∠B = ∠D

Then, ∆ABC and ∆EDF similar

∴ (c) ∠B = ∠D

 

(2) In ∆LMN, ∠L = 60°, ∠M = 50°. If ∆LMN ~ ∆PQR then the value of ∠R is –

(A) 40°

(B) 70°

(C) 30°

(D) 110°

Solution:

Now, ∆LMN ~ ∆PQR

∠L = 60°, ∠M = 50°

∴ ∠N = 188° – (60+50)

∠N = 70°

Now, ∠N = ∠R

∴ ∠R = 70° (B)

 

(3) If ∆ABC is an isosceles triangle with ∠C = 90° and AC = 5cm, then AB is

(A) 2.5cm

(B) 5 cm

(C) 10 cm

(D) 5√2 cm

Solution:

Now, AC = BC = 5 (Given)

∆ABC, ∠C = 90°

By Pythagoras theorem,

(AB)² = (AC)² + (BC)²

(AB)² = (5)² + (5)

(AB)² = 25 + 25

(AB)² = (5√2)²

AB = 5√2 (D)

 

(4) In a given figure ST||QR, PS = 2 cm and SQ = 3 cm. Then the ratio of the area of ∆PQR to the area of ∆PST is

(A) 25:4

(B) 25:7

(C) 25:11

(D) 25:13

Solution:

Now, ∆PQR ~ ∆PST then,

Area of ∆PQR/area of ∆PST = (PQ)²/(PS)² = (5)²/(2)² = 25/4

∴ required ratio is 25:4 (A)

 

(5) The perimeters of two similar triangles ∆ABC and ∆PQR are 36cm and 24cm respectively. If PQ = 10 cm, then length of AB is

(A) 6 2/3 cm

(B) 10√6/3 cm

(C) 66 2/3 cm

(D) 15cm

Solution:

Now, ∆ABC ~ ∆PQR

∴ AB/PQ = AB+BC+CA/PQ+QR+PR

AB/PQ = 36/24

AB = 36×10/24

AB = 15cm (D)

 

(6) If in ∆ABC, DE||BC.AB = 3.6 cm, AC = 2.4 cm and AD = 2.1 cm then the length of AE is

(A) 1.4 cm

(B) 1.8 cm

(C) 1.2 cm

(D) 1.05 cm

Solution:

Now, ∠ADE = ∠ABC and ∠A is common.

∴ ∆ABC ~ ∆ADE

Then, by Thales theorem,

AB/AD = AC/AE

3.6/2.1 = 2.4/AE

AE = 2.4×2.1/3.6

AE = 1.4 cm (A)

 

(7) In a ∆ABC, AD is the bisector of ∠BAC. If AB = 8cm, BD = 6 cm and DC = 3cm. The length of the side AC is

(A) 6cm

(B) 4cm

(C) 3cm

(D) 8cm

Solution:

Thus, AB/BC = BD/DC = AC = AB×DC/BD = 8×3/6 = 4cm (B)

 

(8) In the adjacent figure ∠BAC = 90° and AD⊥BC then

(A) BD.CD = BC²

(B) AB.AC = BC²

(C) BD.CD = AD²

(D) AB.AC = AD²

Solution:

∠ADB = ∠ADC = 90°

Then, ∠BAD = ∠DAC = ∠BAC/2 = 45°

Now, ∆ABD ~ ∆ADC, then,

By Thales theorem,

AD/BD = CD/AD

AD² = BD.CD (C)

 

(9) Two poles of heights 6 m and 11 m stand vertically on a plane ground. If the distance between their feet is 12 m, what is the distance between their tops?

(A) 13m 

(B) 14m

(C) 15m

(D) 12.8m

Solution:

Now, AE = 11cm,

CD = BE then CD = 6cm

AB = AE – BE

= (11 – 6) = 5 cm

CB = DE = 12cm

Now, ∆ABC, ∠B = 90°

By Pythagoras theorem,

(AC)² = (AB)² + (BC)²

(AC)²  (5)² + (12)²

(AC)² = 25 + 144 = 169

(AC)² = (13)²

∴ AC = 13 (A)

 

(10) In the given figure, PR = 26cm, QR = 24cm, ∠PAQ = 90°, PA = 6 cm and QA = 8cm. Find ∠PQR

(A) 80°

(B) 85°

(C) 75°

(D) 90°

Solution:

Now, ∆APQ, ∠A = 90°

Then by Pythagoras theorem,

(PQ)² = (AP)² + (AQ)²

(PQ)² = (6)² + (8)² [∵ AP = 6, AQ = 8]

(PQ)² = 36 + 64 = 100 = (10)²

PQ = 10

Now, ∆PQR, PQ = 10cm PR = 26cm and RQ = 24cm

∴ ∆PQR, right angle triangle.

∴ Hypotenuse is RP = 26cm

Then, ∠PQR = 90° (D)

 

(11) A tangent is perpendicular to the radius at the

(A) Centre

(B) Point of contact

(C) Infinity

(D) Chord

Solution:

(B) Point of contact.

 

(12) How many tangents can be drawn to the circle from an exterior point?

(A) One

(B) Two

(C) infinite

(D) Zero

Solution:

(B) Two.

 

(13) The two tangents from an external points P to a circle with centre at O are PA and PB. If ∠APB = 70° then the value of ∠AOB is

(A) 100°

(B) 110°

(C) 120°

(D) 130°

Solution:

Now, ∠APB = 70°

Then, ∠APO = ∠BPO = ∠APO/2 = 35°

∆APO,

∠AOP = 180° – (∠OAP + ∠APO)

= 180° – (90 + 35)

= 180° – 125

= 55°

∆BPO then,

∠BOP = 180° – (∠OPB + ∠OBP)

= 180° – (35 + 90)

= 180° – 125 = 55°

∴ Now, ∠AOB = ∠AOP + ∠BOP

= 55° + 55° = 110° (B)

 

(14) In figure CP and CQ are tangents to a circle with centre at O. ARB is another tangent touching the circle at R. If CP = 11 cm and BC = 7 cm, then the length of BR is

(A) 6 cm

(B) 5 cm

(C) 8 cm

(D) 4 cm

Solution:

Now, CP = CQ and CA = CB

Then, AP = CP – CA

= (11 – 7) cm

= 4cm

Now, AB = AR + RB

Now, Point A to AP and AR same tangent then, AP = AR

We know that AR = RB

∴ RB = AR = AP = 4cm (D)

 

(15) In figure if PR is tangent to the circle at P and O is the centre of the circle, then ∠POQ is

(A) 120°

(B) 100°

(C) 110°

(D)  90°

Solution:

Now, ∠OPQ = 90-60 = 30° [∵ ∠OPR = 90°]

Now, ∠OPQ = ∠OQP [∵ OP = OQ = radius]

∆POQ, ∴ ∠POQ = 180° – (∠OPQ + ∠OQP) = 180° – 30+30 = 180 – 60 = 120°

 

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