10th Maths - Book Back Answers - Chapter 4 Unit Exercise 4.5 - English Medium Guides

 

SSLC / 10th - Maths - Book Back Answers - Chapter 4 Unit Excercise 4.5 - English Medium

Tamil Nadu Board 10th Standard Maths - Chapter 4 Unit Exercise 4.5: Book Back Answers and Solutions

    This post covers the book back answers and solutions for Chapter 4 Unit Exercise 4.5 – Maths from the Tamil Nadu State Board 10th Standard textbook. These detailed answers have been carefully prepared by our expert teachers at KalviTips.com.

    We have explained each answer in a simple, easy-to-understand format, highlighting important points step by step under the relevant subtopics. Students are advised to read and memorize these subtopics thoroughly. Once you understand the main concepts, you’ll be able to connect other related points with real-life examples and confidently present them in your tests and exams.

    By going through this material, you’ll gain a strong understanding of Chapter 4 Unit Exercise 4.5 along with the corresponding book back questions and answers (PDF format).

Question Types Covered:

• 1 Mark Questions: Choose the correct answer, Fill in the blanks, Identify the correct statement, Match the following 
• 2 Mark Questions: Answer briefly 
• 3, 4, and 5 Mark Questions: Answer in detail

All answers are presented in a clear and student-friendly manner, focusing on key points to help you score full marks.

All the best, Class 10 students! Prepare well and aim for top scores. Thank you!

Chapter 4 Algebra Unit Ex 4.5

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 Key:
(3) ∠B = ∠D

Hint:

$\frac{AB}{ED}=\frac{BC}{DF}=\frac{AC}{EF}$

$\therefore \angle B=\angle D$

 

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 Key:
(2) 70°
Hint:
Since ∆LMN ~ ∆PQR
∠N = ∠R
∠N = 180 – (60 + 50)
= 180 – 110°
∠N = 70° ∴ ∠R = 70°

 

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 Key:
(4) $\sqrt{2}$ cm
Hint:

AB² = AC² + BC²
AB² = 5² + 5²
(It is an isosceles triangle)
AB = $\sqrt{50}=\sqrt{25\times 2}$
AB = 5$\sqrt{2}$

 

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 …………….

(1) 25 : 4
(2) 25 : 7
(3) 25 : 11
(4) 25 : 13
Answer Key:
(1) 25 : 4

Hint. 

Area of ∆PQR : Area of ∆PST
$\frac{Area of△PQR }{Area of △PST}=\frac{P{Q}^2}{P{S}^2}=\frac{5^2}{2^2}=\frac{25}{4}$
Area of ∆PQR : Area of ∆PST = 25 : 4

 

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 Key:
(4) 15 cm
Hint:

$\frac{Perimeter of △ABC}{Perimeter of△PQR}=\frac{AB}{PQ}$

$\frac{36}{24}=\frac{AB}{10}$

$36\times 10=24AB$

$AB=\frac{36\times 10}{24}$

$AB=15cm$ 

 

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 Key:
(1) 1.4 cm
Hint:

 ∆ABC, ADE

$\frac{AB}{AD}=\frac{AC}{AE}\Rightarrow \frac{3.6}{2.1}=\frac{2.4}{AE}$

$3.6\times AE=2.4\times 2.1$ 

$\mathrm{AE}=\frac{2.4\times 2.1}{3.6}=\frac{24\times 21}{360}$ 

$\mathrm{AE}=1.4\mathrm{cm}$ 

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 Key:
(2) 4 cm
Hint:

AD is the bisector of ∠A

$\frac{BD}{DC}=\frac{AB}{AC}\Rightarrow \frac{6}{3}=\frac{8}{AC}$

$\mathrm{AC}=\frac{3\times 8}{6}=\frac{24}{6}=4\mathrm{cm}$

  

8. In the adjacent figure ∠BAC = 90° and AD ⊥ BC then ………..

(1) BD.CD = BC²
(2) AB.AC = BC²
(3) BD.CD = AD²
(4) AB.AC = AD²
Answer Key:
(3) BD CD = AD²
Hint:

$△\mathrm{BAC}\sim △\mathrm{BDA}$

$\frac{\mathrm{AC}}{\mathrm{AD}}=\frac{\mathrm{AB}}{\mathrm{BD}}....\left(1\right)$

$△\mathrm{CAB}\sim △\mathrm{CDA}$

$\frac{\mathrm{AC}}{\mathrm{CD}}=\frac{\mathrm{AB}}{\mathrm{AD}}....\left(2\right)$

$\mathrm{From} \left(1\right) \mathrm{and} \left(2\right)\mathrm{we} \mathrm{get}$

$\frac{\mathrm{BD}}{\mathrm{AD}}=\frac{\mathrm{AD}}{\mathrm{CD}}$

${\mathrm{AD}}^2=\mathrm{BD}.\mathrm{CD}$ 

  

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 Key:
(1) 13 m
Hint:

AC² (Distance between the two tops)
= AE² + EC²
= 5² + 122
= 25 + 144= 169
AC = $\sqrt{169}$= 13 cm

 

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

(1) 80°
(2) 85°
(3) 75°
(4) 90°
Answer Key:
(4) 90°
Hint.
PR = 26 cm, QR = 24 cm, ∠PAQ = 90°
In the ∆PQR,

$\mathrm{PQ}=\sqrt{{\mathrm{PR}}^2-{\mathrm{QR}}^2}=\sqrt{{26}^2-{24}^2}=\sqrt{675-576}$

$\mathrm{PQ}=\sqrt{100}=10$ 

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°

 

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

 

12. How many tangents can be drawn to the circle from an exterior point?
(1) one
(2) two
(3) infinite
(4) zero
Answer Key:
(2) 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 ……….
(1) 100°
(2) 110°
(3) 120°
(4) 130°
Answer Key:
(2) 110°
Hint.

∠OAP = 90°
∠APO = 35°
∠AOP = 180 – (90 + 35)
= 180 – 125 = 55
∠AOB = 2 × 55 = 110°

 

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 …….

(1) 6 cm
(2) 5 cm
(3) 8 cm
(4) 4 cm
Answer Key:
(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

 

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°
Answer Key:
(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°