10th Maths - Book Back Answers - Chapter 4 Unit Exercise 4.1 - English Medium Guides

 

SSLC / 10th - Maths - Book Back Answers - Chapter 4 Unit Excercise 4.1 - English Medium

Tamil Nadu Board 10th Standard Maths - Chapter 4 Unit Exercise 4.1: Book Back Answers and Solutions

    This post covers the book back answers and solutions for Chapter 4 Unit Exercise 4.1 – 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 4.1 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 4.1

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

Answer Key:

(i) In ∆ABC and ∆AED
$\frac{\mathbf{A}\mathbf{B}}{\mathbf{A}\mathbf{D}}\mathbf{=}\frac{\mathbf{A}\mathbf{C}}{\mathbf{A}\mathbf{E}}$

$\frac{\mathbf{8}}{\mathbf{3}}\mathbf{=}\frac{\mathbf{11}}{\frac{\mathbf{2}}{\mathbf{2}}}$
$\frac{\mathbf{8}}{\mathbf{3}}\mathbf{=}\frac{\mathbf{11}}{\mathbf{4}}\mathbf{\Rightarrow }\mathbf{32}$
∴ The two triangles are not similar.

 

(ii) In ∆ABC and ∆PQC
∠PQC = 70°
∠ABC = ∠PQC = 70°
∠ACB = ∠PCQ (common)
∆ABC ~ ∆PQC
$\frac{5}{\mathrm{X}}=\frac{6}{3}$
6x = 15
$\mathrm{X}=\frac{15}{6}=\frac{5}{2}$
∴ x = 2.5

∆ ABC and ∆PQC are similar. The value of x = 2.5

 

2. A girl looks the reflection of the top of the lamp post on the mirror which is 66 m away from the foot of the lamppost. The girl whose height is 12.5 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.
Answer Key:
Let the height of the tower ED be “x” m. In ∆ABC and ∆EDC.
∠ABC = ∠CED = 90° (vertical Pole)
∠ACB = ∠ECD (Laws of reflection)
∆ ABC ~ ∆DEC

$\frac{\mathrm{AB}}{\mathrm{DE}}=\frac{\mathrm{BC}}{\mathrm{EC}}$ 
$\frac{1.5}{\mathrm{x}}=\frac{0.4}{\mathrm{87.6 }}$ 

 

$\mathrm{X}=\frac{1.5\times 87.6}{0.4}=\frac{1.5\times 876}{4}$
= 1.5 × 219 = 328.5
The height of the Lamp Post = 328.5 m

 

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.
Answer Key:

In ∆ABC and ∆PQR,
∠ABC = ∠PQR = 90° (Vertical Stick)
∠ACB = ∠PRQ (Same time casts shadow)
∆BCA ~ ∆QRP

 $\frac{AB}{PQ}=\frac{BC}{QR}$

 $\frac{6}{\mathrm{x}}=\frac{4}{28}$
4x = 6 × 28 ⇒ x = $\frac{6\times 28}{4}$ = 42
Length of the lamp post = 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.
Answer Key:
In ∆PQT and ∆STR we have
∠P = ∠S = 90° (Given)
∠PTQ = ∠STR (Vertically opposite angle)

By AA similarity

∆PTQ ~ ∆STR we get
$\frac{\mathrm{PT}}{\mathrm{ST}}=\frac{\mathrm{TQ}}{\mathrm{TR}}$ 

PT × TR = ST × TQ
Hence it is 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. 

Answer Key:

In ∆ABC and ∆ADE
∠ACB = ∠AED = 90°
∠A = ∠A (common)
∴ ∆ABC ~ ∆ADE (By AA similarity)
$\frac{\mathrm{BC}}{\mathrm{DE}}=\frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{AC}}{\mathrm{AE}}$

$\frac{12}{\mathrm{DE}}=\frac{13}{3}=\frac{5}{\mathrm{AE}}$ 

In ∆ABC, AB² = BC² + AC²
= 122 + 52 = 144 + 25 = 169
AB = 169 $\sqrt{169}$ = 13
Consider, $\frac{13}{3}=\frac{5}{\mathrm{AE}}$
∴ AE =$\frac{5\times 3}{13}=\frac{15}{13}$
AE = $\frac{15}{13}$ and DE = $\frac{36}{13}$
Consider, $\frac{12}{\mathrm{DE}}=\frac{13}{3}$
DE =$\frac{12\times 3}{13}=\frac{36}{13}$

 

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. 

Answer Key:

Given ∆ACB ~ ∆APQ
$\frac{\mathrm{AC}}{\mathrm{AP}}=\frac{\mathrm{BC}}{\mathrm{PQ}}=\frac{\mathrm{AB}}{\mathrm{AQ}}$

$\frac{\mathrm{AC}}{2.8}=\frac{8}{4}=\frac{6.5}{\mathrm{AQ}}$ 

Consider $\frac{\mathrm{AC}}{2.8}=\frac{8}{4}$
4 AC = 8 × 2.8
AC = $\frac{8\times 2.8}{4}$ = 5.6 cm
Consider $\frac{8}{4}=\frac{6.5}{\mathrm{AQ}}$
8 AQ = 4 × 6.5
AQ = $\frac{4\times 6.5}{8}$= 3.25 cm
Length of AC = 5.6 cm; Length of 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.
Answer Key:

(i) In ∆LOP and ∆QMO
∠OLP = ∠OQM = 90°
∠LOP = ∠OMQ (Since OQRP is a square OP || MN)
∴ ∆LOP~ ∆QMO (By AA similarity)

 

(ii) In ∆LOP and ∆RPN
∠OLP = ∠PRN = 90°
∠LPO = ∠PNR (OP || MN) .
∴ ∆LOP ~ ∆RPN (By AA similarity)

 

(iii) In ∆QMO and ∆RPN
∠MQO = ∠NRP = 90°
∠RPN = ∠QOM (OP || MN)
∴ ∆QMO ~ ∆RPN (By AA similarity)

 

(iv) We have ∆QMO ~ ∆RPN
$\frac{\mathrm{MQ}}{\mathrm{PR}}=\frac{\mathrm{QO}}{\mathrm{RN}}$

$\frac{\mathrm{MQ}}{\mathrm{QR}}=\frac{\mathrm{QR}}{\mathrm{RN}}$ 
QR² = MQ × RN
Hence it is proved.

 

8. If ∆ABC ~ ∆DEF such that area of ∆ABC is 9cm² and the area of ∆DEF is 16 cm² and BC = 2.1 cm. Find the length of EF.
Answer Key:

Given ∆ABC ~ ∆DEF

$\therefore \frac{\mathrm{Area} \mathrm{of} △\mathrm{ABC}}{\mathrm{Area} \mathrm{of} △\mathrm{DEF}}=\frac{{\mathrm{BC}}^2}{{\mathrm{EF}}^2}$ (square of their corresponding sides)

$\frac{9}{16}=\frac{{\left(\mathrm{}\right)}^{}}{}$($\frac{{\left(\mathrm{2.1)}\right)}^2}{{\mathrm{EF}}^2}$
($\frac{3}{4}$)² = ($\frac{2.1}{\mathrm{EF}}$)²
$\frac{3}{4}=\frac{2.1}{\mathrm{EF}}$
EF = $\frac{4\times 2.1}{3}$ = 2.8 cm
Legth of EF = 2.8 cm

 

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

Answer Key:
In the ∆PAC and ∆BQC
∠PAC = ∠QBC = 90°
∠C is common
∆PAC ~ QBC
$\frac{\mathrm{AP}}{\mathrm{BQ}}=\frac{\mathrm{AC}}{\mathrm{BC}}$
$\frac{6}{\mathrm{y}}=\frac{\mathrm{AC}}{\mathrm{BC}}$
$\therefore \frac{BC}{AC}=\frac{y}{6}....\left(1\right)$
In the ∆ACR and ∆QBC
∠ACR = ∠QBC = 90°
∠A is common
∆ACR ~ ABQ
$\frac{\mathrm{RC}}{\mathrm{QB}}=\frac{\mathrm{AC}}{\mathrm{AB}}$
$\frac{3}{y}=\frac{\mathrm{AC}}{\mathrm{AB}}$
$\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{y}}{3}....\left(2\right)$
By adding (1) and (2)
$\frac{\mathrm{BC}}{\mathrm{AC}}+\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{y}{3}+\frac{\mathrm{y}}{3}$
$1=\frac{3\mathrm{y}+6\mathrm{y}}{18}$
9y = 18 ⇒ y = $\frac{18}{9}$ = 2
The Value of y = 2m

 

10. Construct a triangle similar to a given triangle PQR with its sides equal to $\frac{2}{3}$ of the corresponding sides of the triangle PQR (scale factor $\frac{2}{3}$ ).
Answer Key:
Given ∆PQR, we are required to construct another triangle whose sides are $\frac{2}{3}$ of the corresponding sides of the ∆PQR

Steps of construction:

(i) Construct a ∆PQR with any measurement.
(ii) Draw a ray QX making an acute angle with QR on the side opposite to the vertex P.
(iii) Locate 3 points Q₁, Q₂ and Q₃ on QX.
So that QQ₁ = Q₁Q₂ = Q₂Q₃
(iv) Join Q₃ R and draw a line through Q₂ parallel to Q₃ R to intersect QR at R’.
(v) Draw a line through R’ parallel to the line RP to intersect QP at P’. Then ∆ P’QR’ is the required triangle.

 

11. Construct a triangle similar to a given triangle LMN with its sides equal to $\frac{4}{5}$ of the corresponding sides of the triangle LMN (scale factor $\frac{4}{5}$ ).
Answer Key:

Given a triangle LMN, we are required to construct another ∆ whose sides are $\frac{4}{5}$ of the corresponding sides of the ∆LMN. 

Steps of Construction:

1. Construct a ∆LMN with any measurement.
2. Draw a ray MX making an acute angle with MN on the side opposite to the vertex L.
3. Locate 5 Points Q₁, Q₂, Q₃, Q₄, Q₅ on MX.
   So that MQ₁ = Q₁Q₂ = Q₂Q₃ = Q₃Q₄ = Q₄Q₅
4. Join Q₅ N and draw a line through Q₄. Parallel to Q₅N to intersect MN at N’.
5. Draw a line through N’ parallel to the line LN to intersect ML at L’.

12. Construct a triangle similar to a given triangle ABC with its sides equal to $\frac{6}{5}$ of the corresponding sides of the triangle ABC (scale factor $\frac{6}{4}$).
Answer Key:

Given triangle ∆ABC, we are required to construct another triangle whose sides are $\frac{6}{5}$ of the corresponding sides of the ∆ABC.
Steps of construction
(i) Construct an ∆ABC with any measurement.
(ii) Draw a ray BX making an acute angle with BC.
(iii) Locate 6 points Q₁, Q₂, Q₃, Q₄, Q₅, Q₆ on BX such that
BQ₁ = Q₁Q₂ = Q₂Q₃ = Q₃Q₄ = Q₅Q₆

(iv) Join Q₅ to C and draw a line through Q₆ parallel to Q₅ C intersecting the extended line BC at C’.
(v) Draw a line through C’ parallel to AC intersecting the extended line segment AB at A’.
∆A’BC’ is the required triangle.

 

13. Construct a triangle similar to a given triangle PQR with its sides equal to $\frac{7}{3}$ of the corresponding sides of the triangle PQR (scale factor $\frac{7}{3}$).
Answer Key:

Given triangle ABC, we are required to construct another triangle whose sides are $\frac{7}{3}$ of the corresponding sides of the ∆ABC.

Steps of construction:

(i) Construct a ∆PQR with any measurement.
(ii) Draw a ray QX making an acute angle with QR on the side opposite to the vertex P.
(iii) Locate 7 points Q₁, Q₂, Q₃, Q₄, Q₅, Q₆, Q₇ on QX.
So that
QQ₁ = Q₁Q₂ = Q₂Q₃ = Q₃Q₄ = Q₅Q₆ = Q₆Q₇

(iv) Join Q₃ to R and draw a line through Q₇ parallel to Q₃R intersecting the extended line segment QR at R’.

(v) Draw a line through parallel to RP.
Intersecting the extended line segment QP at P’.
∴ ∆P’QR’ is the required triangle.