Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.5

Students can download Maths Chapter 4 Geometry Ex 4.5 Questions and Answers, Notes, Samacheer Kalvi 10th Maths Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus, helps students complete homework assignments and to score high marks in board exams.

Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Ex 4.5

Question 1.
If in triangles ABC and EDF, \(\frac { AB }{ DE } \) = \(\frac { BC }{ FD } \) then they will be similar, when ……….
(1) ∠B = ∠E
(2) ∠A = ∠D
(3) ∠B = ∠D
(4) ∠A = ∠F
Answer:
(3) ∠B = ∠D
Hint:
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.5 1

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.5

Question 2.
In ∆LMN, ∠L = 60°, ∠M = 50°. If ∆LMN ~ ∆PQR then the value of ∠R is ……………
(1) 40°
(2) 70°
(3) 30°
(4) 110°
Answer:
(2) 70°
Hint:
Since ∆LMN ~ ∆PQR
∠N = ∠R
∠N = 180 – (60 + 50)
= 180 – 110°
∠N = 70° ∴ ∠R = 70°
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.5 2

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.5

Question 3.
If ∆ABC is an isosceles triangle with ∠C = 90° and AC = 5 cm, then AB is ………
(1) 2.5 cm
(2) 5 cm
(3) 10 cm
(4) 5 \(\sqrt { 2 }\) cm
Answer:
(4) 5 \(\sqrt { 2 }\) cm
Hint:
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.5 3
AB2 = AC2 + BC2
AB2 = 52 + 52
(It is an isosceles triangle)
AB = \(\sqrt { 50 }\) = \(\sqrt{25 \times 2}\)
AB = 5 \(\sqrt { 2 }\)

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.5

Question 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 …………….
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.5 4
(1) 25 : 4
(2) 25 : 7
(3) 25 : 11
(4) 25 : 13
Answer:
(1) 25 : 4
Hint. Area of ∆PQR : Area of ∆PST
\(\frac{\text { Area of } \Delta \mathrm{PQR}}{\text { Area of } \Delta \mathrm{PST}}=\frac{\mathrm{PQ}^{2}}{\mathrm{PS}^{2}}=\frac{5^{2}}{2^{2}}=\frac{25}{4}\)
Area of ∆PQR : Area of ∆PST = 25 : 4

Question 5.
The perimeters of two similar triangles ∆ABC and ∆PQR are 36 cm and 24 cm respectively. If PQ =10 cm, then the length of AB is ………….
(1) 6 \(\frac { 2 }{ 3 } \) cm
(2) \(\frac{10 \sqrt{6}}{3}\)
(3) 66 \(\frac { 2 }{ 3 } \) cm
(4) 15 cm
Answer:
(4) 15 cm
Hint:
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.5 5

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.5

Question 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 ………..
(1) 1.4 cm
(2) 1.8 cm
(3) 1.2 cm
(4) 1.05 cm
Answer:
(1) 1.4 cm
Hint:
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.5 6

Question 7.
In a ∆ABC, AD is the bisector of ∠BAC. If AB = 8 cm, BD = 6 cm and DC = 3 cm.
The length of the side AC is ………….
(1) 6 cm
(2) 4 cm
(3) 3 cm
(4) 8 cm
Answer:
(2) 4 cm
Hint:
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.5 7

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.5

Question 8.
In the adjacent figure ∠BAC = 90° and AD ⊥ BC then ………..
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.5 8
(1) BD.CD = BC2
(2) AB.AC = BC2
(3) BD.CD = AD2
(4) AB.AC = AD2
Answer:
(3) BD CD = AD2
Hint:
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.5 9

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.5

Question 9.
Two poles of heights 6 m and 11m stand vertically on a plane ground. If the distance between their feet is 12 m, what is the distance between their tops?
(1) 13 m
(2) 14 m
(3) 15 m
(4) 12.8 m
Answer:
(1) 13 m
Hint:
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.5 10
AC2 (Distance between the two tops)
= AE2 + EC2
= 52 + 122
= 25 + 144= 169
AC = \(\sqrt { 169 }\) = 13 cm

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.5

Question 10.
In the given figure, PR = 26 cm, QR = 24 cm, ∠PAQ = 90°, PA = 6 cm and QA = 8 cm. Find ∠PQR.
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.5 11
(1) 80°
(2) 85°
(3) 75°
(4) 90°
Answer:
(4) 90°
Hint.
PR = 26 cm, QR = 24 cm, ∠PAQ = 90°
In the ∆PQR,
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.5 12
In the right ∆ APQ
PQ2 = PA2 + AQ2
= 62 + 82
= 36 + 64 = 100
PQ = \(\sqrt { 100 }\) = 10
∆ PQR is a right angled triangle at Q. Since
PR2 = PQ2 + QR2
∠PQR = 90°

Question 11.
A tangent is perpendicular to the radius at the
(1) centre
(2) point of contact
(3) infinity
(4) chord
Solution:
(2) point of contact

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.5

Question 12.
How many tangents can be drawn to the circle from an exterior point?
(1) one
(2) two
(3) infinite
(4) zero
Answer:
(2) two

Question 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 ……….
(1) 100°
(2) 110°
(3) 120°
(4) 130°
Answer:
(2) 110°
Hint.
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.5 13
∠OAP = 90°
∠APO = 35°
∠AOP = 180 – (90 + 35)
= 180 – 125 = 55
∠AOB = 2 × 55 = 110°

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.5

Question 14.
In figure CP and CQ are tangents to a circle with centre at 0. ARB is another tangent touching the circle at R. If CP = 11 cm and BC = 7 cm, then the length of BR is …….
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.5 14
(1) 6 cm
(2) 5 cm
(3) 8 cm
(4) 4 cm
Answer:
(4) 4 cm
Hint.
BQ = BR = 4 cm (tangent of the circle)
PC = QC = 11 cm (tangent of the circle)
QC = 11 cm
QB + BC = 11
QB + 7 = 11
QB = 11 – 7 = 4 cm
BR = BQ = 4 cm

Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.5

Question 15.
In figure if PR is tangent to the circle at P and O is the centre of the circle, then ∠POQ is ………….
Samacheer Kalvi 10th Maths Guide Chapter 4 Geometry Ex 4.5 15
(1) 120°
(2) 100°
(3) 110°
(4) 90°
Answer:
(1) 120°
Hint.
Since PR is tangent of the circle.
∠QPR = 90°
∠OPQ = 90° – 60° = 30°
∠OQB = 30°
In ∆OPQ
∠P + ∠Q + ∠O = 180°
30 + 30° + ∠O = 180°
(OP and OQ are equal radius)
∠O = 180° – 60° = 120°

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