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

Multiple choice questions.
Question 1.
If in triangles ABC and EDF, \(\frac{\mathbf{A B}}{\mathbf{D E}}=\frac{\mathbf{B C}}{\mathbf{F D}}\) then they will be similar, when
(1) ∠B = ∠E
(2) ∠A = ∠D
(3) ∠B = ∠D
(4) ∠A = ∠F
Solution:
(1) ∠B = ∠E
Hint:
Samacheer Kalvi 10th Maths Chapter 4 Geometry Ex 4.5 1

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°
Solution:
(2) 70°
Samacheer Kalvi 10th Maths Chapter 4 Geometry Ex 4.5 2
∆LMN ~ ∆PQR, ∠R = 70°.

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
Solution:
(4) \(5 \sqrt{2}\)cm
Hint:
Samacheer Kalvi 10th Maths Chapter 4 Geometry Ex 4.5 3

Samacheer Kalvi 10th Maths Solutions 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 Chapter 4 Geometry Ex 4.5 4
(1) 25 : 4
(2) 25 : 7
(3) 25 : 11
(4) 25 : 13
Solution:
(1) 25 : 4
Hint:
Ratio of the area of similar triangles is equal to the ratio of the square of their corresponding sides.
∴ 52 : 22 = 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
Samacheer Kalvi 10th Maths Chapter 4 Geometry Ex 4.5 5
Solution:
(4) 15 cm
Hint:
Samacheer Kalvi 10th Maths Chapter 4 Geometry Ex 4.5 6

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
Solution:
(1) 1.4 cm
Samacheer Kalvi 10th Maths Chapter 4 Geometry Ex 4.5 7
\(\frac{3 \cdot 6}{2 \cdot 1}=\frac{2 \cdot 4}{\mathrm{A} \cdot \mathrm{E}}\)
(3.6) (AE) = 2.1 × 2.4
AE = 1.4 cm

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
Solution:
(2) 4 cm
Hint:
Samacheer Kalvi 10th Maths Chapter 4 Geometry Ex 4.5 8

Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Ex 4.5

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

Question 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?
(1) 13 cm
(2) 14 m
(3.) 15 m
(4) 12.8 m
Solution:
(1) 13 cm
Hint:
Samacheer Kalvi 10th Maths Chapter 4 Geometry Ex 4.5 11

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 Chapter 4 Geometry Ex 4.5 12
(1) 80°
(2) 85°
(3) 75°
(4) 90°
Solution:
(4) 90°
Hint:
PR = 26
QR = 24
∠PAQ = 90°
PQ = 10
PQ = \(\sqrt{26^{2}-24^{2}}=\sqrt{100}\) = 10
∴ ∠PAQ = 90°

Samacheer Kalvi 10th Maths Solutions Chapter 4 Geometry Ex 4.5

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

Question 12.
How many tangents can be drawn to the circle from an exterior point?
(1) one
(2) two
(3) infinite
(4) zero
Solution:
(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°
Solution:
(2) 110°
Hint:
Samacheer Kalvi 10th Maths Chapter 4 Geometry Ex 4.5 13

Question 14.
In figure CP and CQ are tangents to a circle T 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
(1) 6 cm
(2) 5 cm
(3) 8 cm
(4) 4 cm
Solution:
(4) 4 cm
Samacheer Kalvi 10th Maths Chapter 4 Geometry Ex 4.5 14
BQ = BR
CP = CQ = 11
BC = 7, ∴ BQ = CQ – BC
= 11 – 7 = 4
BR = BQ = 4cm

Samacheer Kalvi 10th Maths Solutions 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
(1) 120°
(2) 100°
(3) 110°
(4) 90°
Solution:
(1) 120°
∠POQ = 180° -(30° + 30°)
= 180° – 60°
= 120°
Samacheer Kalvi 10th Maths Chapter 4 Geometry Ex 4.5 15

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