10th Maths - Book Back Answers - Chapter 4 Unit Exercise 4.2 - English Medium Guides

 

SSLC / 10th - Maths - Book Back Answers - Chapter 4 Unit Excercise 4.2 - English Medium

Tamil Nadu Board 10th Standard Maths - Chapter 4 Unit Exercise 4.2: Book Back Answers and Solutions

    This post covers the book back answers and solutions for Chapter 4 Unit Exercise 4.2 – 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 4.2 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 4.2

1. In ∆ABC, D and E are points on the sides AB and AC respectively such that DE || BC
(i) If ADDB = 34 and AC = 15 cm find AE.
(ii) If AD = 8x – 7 , DB = 5x – 3 , AE = 4x – 3 and EC = 3x – 1, find the value of x.
Answer  Key:

 

(i) Let AE be x
∴ EC = 15 – x
In ∆ABC we have DE || BC
By Basic proportionality theorem, we have

$\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$

$\frac{3}{4}=\frac{\mathrm{x}}{15-\mathrm{x}}$ 
4x = 3 (15 – x)
4x = 45 – 3x
7x = 45 ⇒ x = $\frac{45}{7}$ = 6.43
The value of x = 6.43

 

(ii) Given AD = 8x – 7; 

BD = 5x – 3; AE = 4x – 3; EC = 3x – 1
In ∆ABC we have DE || BC
By Basic proportionality theorem
$\frac{\mathrm{AD}}{\mathrm{DB}}=\frac{\mathrm{AE}}{\mathrm{EC}}$
$\frac{8\mathrm{x}-7}{5\mathrm{x}-3}=\frac{4\mathrm{x}-3}{3\mathrm{x}-1}$

(8x – 7) (3x – 1) = (4x – 3) (5x – 3)
24x² – 8x – 21x + 7 = 20x² – 12x – 15x + 9
24x² – 20x² – 29x + 27x + 7 – 9 = 0
4x² – 2x – 2 = 0
2x² – x – 1 = 0 (Divided by 2)

2x² – 2x + x – 1 = 0
2x(x -1) + 1 (x – 1) = 0
(x – 1) (2x + 1) = 0
x – 1 = 0 or 2x + 1 = 0
x = 1 or 2x = -1 ⇒ x = $-\frac{1}{2}$ (Negative value will be omitted)
The value of x = 1

 

2. ABCD is a trapezium in which AB || DC and P,Q are points on AD and BC respectively, such that PQ || DC if PD = 18 cm, BQ
Answer Key Key:

Join AC intersecting PQ at S.
Let AP be x
∴ AD = x + 18
In the ∆ABC, QS || AB
By basic proportionality theorem.

$\frac{\mathrm{AS}}{\mathrm{SC}}=\frac{\mathrm{BQ}}{\mathrm{QC}}$
$\frac{\mathrm{AS}}{\mathrm{SC}}=\frac{35}{15}....\left(1\right)$
In the ∆ACD; PS || DC
By basic proportionality theorem.
$\frac{\mathrm{AS}}{\mathrm{SC}}=\frac{\mathrm{AP}}{\mathrm{PD}}$
$\frac{\mathrm{AS}}{\mathrm{SC}}=\frac{\mathrm{x}}{18}....\left(2\right)$ ………..(2)
From (1) and (2) we get
$\frac{35}{15}=\frac{\mathrm{x}}{18}$
15x = 35 × 18 ⇒ x = $\frac{35\times 18}{15}$ = 42
AD = AP + PD
= 42 + 18 = 60
The value of AD = 60 cm

 

3. In ∆ABC, D and E are points on the sides AB and AC respectively. For each of the following cases show that DE || BC.
(i) AB = 12 cm, AD = 8 cm, AE = 12 cm and AC = 18 cm.
(ii) AB = 5.6 cm, AD = 1.4 cm, AC = 7.2 cm and AE = 1.8 cm.
Answer  Key:

(i) Here AB = 12 cm; BD =12 – 8 = 4 cm; AE =12 cm; EC = 18 – 12 = 6 cm

$\therefore \frac{AD}{DB}=\frac{8}{4}$=2

$\frac{AE}{EC}=\frac{12}{6}$=2 

$\frac{AD}{DB}=\frac{AE}{EC}$
By converse of basic proportionality theorem DE || BC

 

(ii) Here AB = 5.6 cm; AD = 1.4 cm;
BD = AB – AD
= 5.6 – 1.4 = 4.2

AC = 7.2 cm; AE = 1.8 cm
EC = AC – AE
= 7.2 – 1.8
EC = 5.4 cm
$\therefore \frac{AD}{DB}=\frac{1.4}{4.2}=\frac{1}{3}$
$\frac{AE}{EC}=\frac{1.8}{5.4}=\frac{1}{3}$
$\frac{AE}{EC}=\frac{AD}{DB}$
By converse of basic proportionality theorem DE || BC

 

4. In fig. if PQ || BC and BC and PR || CD prove that

(i)$\frac{\mathrm{AR}}{\mathrm{AD}}=\frac{\mathrm{AQ}}{\mathrm{AB}}$
(ii)$\frac{QB}{AQ}=\frac{DR}{AR}$
Answer Key:

 
(i) In ∆ABC, We have PQ || BC
By basic proportionality theorem

$\frac{\mathrm{AQ}}{\mathrm{AB}}=\frac{AP}{AC}....\left(1\right)$
In ∆ACD, We have PR || CD
basic proportionality theorem
$\frac{AP}{AC}=\frac{AR}{AD}....\left(2\right)$
From (1) and (2) we get
$\frac{\mathrm{AQ}}{\mathrm{AB}}=\frac{AR}{AD}\left(or\right)\frac{AR}{AD}=\frac{\mathrm{AQ}}{\mathrm{AB}}$

 

(ii) In ∆ABC, PQ || BC (Given)
By basic proportionality theorem
$\frac{AP}{AC}=\frac{AQ}{QB}....\left(1\right)$
In ∆ADC, PR || CD (Given)
By basic proportionality theorem
$\frac{AP}{AC}=\frac{AR}{RD}....\left(2\right)$
From (1) and (2) we get
$\frac{AQ}{QB}=\frac{AP}{RD}\left(or\right)\frac{QB}{AQ}=\frac{RD}{AR}$

 

5. Rhombus PQRB is inscribed in ∆ABC such that ∠B is one of its angle. P, Q and R lie on AB, AC and BC respectively. If AB = 12 cm and BC = 6 cm, find the sides PQ, RB of the rhombus.
Answer Key:

Let the side of the rhombus be “x”. Since PQRB is a Rhombus PQ || BC
By basic proportionality theorem

$\frac{\mathrm{AP}}{\mathrm{AB}}=\frac{\mathrm{PQ}}{\mathrm{BC}}\Rightarrow \frac{12-\mathrm{x}}{\mathrm{BC}}=\frac{\mathrm{x}}{6}$
12x = 6 (12 – x)
12x = 72 – 6x
12x + 6x = 72
18x = 72 ⇒ x = $\frac{72}{18}$ = 4
Side of a rhombus = 4 cm
PQ = RB = 4 cm

 

6. In trapezium ABCD, AB || DC , E and F are points on non-parallel sides AD and BC respectively, such that EF || AB.
Show that = $\frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{BF}}{\mathrm{FC}}$
Answer Key:
Given: ABCD is a trapezium AB || DC
E and F are the points on the side AD and BC
EF || AB
To Prove: $\frac{\mathrm{AE}}{\mathrm{ED}}=\frac{\mathrm{BF}}{\mathrm{FC}}$

Construction: Join AC intersecting AC at P
Proof:
In ∆ABC, PF || AB (Given)
By basic proportionality theorem
$\frac{AP}{PC}=\frac{BF}{FC}....\left(1\right)$
In the ∆ACD, PE || CD (Given)
By basic Proportionality theorem
$\frac{AP}{PC}=\frac{\mathrm{AE}}{\mathrm{ED}}....\left(2\right)$
From (1) and (2) we get
$\frac{\mathrm{AE}}{\mathrm{ED}}=\frac{BF}{FC}$

 

7. In figure DE || BC and CD || EE Prove that AD² = AB × AF.

Answer Key:

Given: In ∆ABC, DE || BC and CD || EF

To Prove: AD² = AB × AF
Proof: In ∆ABC, DE || BC (Given)
By basic proportionality theorem
$\frac{\mathrm{AB}}{\mathrm{AD}}=\frac{\mathrm{AC}}{\mathrm{AE}}....\left(1\right)$
In ∆ADC; FE || DC (Given)
By basic Proportionality theorem
$\frac{AD}{AF}=\frac{AC}{AE}....\left(2\right)$
From (1) and (2) we get
$\frac{\mathrm{AB}}{\mathrm{AD}}=\frac{AD}{AF}$
AD² = AB × AF
Hence it is proved

 

8. In ∆ABC, AD is the bisector of ∠A meeting side BC at D, if AB = 10 cm, AC = 14 cm and BC = 6 cm, find BD and DC.

Answer Key:

In ∆AABC AD is the internal bisector of ∠A
Given BC = 6 cm
Let BD = x ∴ DC = 6 – x cm
By Angle bisector theorem
$\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\mathrm{AB}}{\mathrm{AC}}$ 

$\frac{\mathrm{x}}{6-\mathrm{x}}=\frac{10}{14}$

14x = 60 – 10x
24x = 60
x = $\frac{60}{24}=\frac{10}{4}$ = 2.5
BD = 2.5 cm;
DC = 6 – x ⇒ 2.5 = 3.5 cm

9. Check whether AD is bisector of ∠A of ∆ABC in each of the following,
(i) AB = 5 cm, AC = 10 cm, BD = 1.5 cm and CD = 3.5 cm.
(ii) AB = 4 cm, AC 6 cm, BD = 1.6 cm and CD = 2.4 cm.
Answer Key:

(i) In ∆ABC, AB = 5 cm, AC = 10 cm, BD = 1.5 cm, CD = 3.5 cm
$\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{1.5}{3.5}=\frac{15}{35}=\frac{3}{7}$

$\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{5}{10}=\frac{1}{2}$

$\frac{\mathrm{BD}}{\mathrm{DC}}\ne \frac{\mathrm{AB}}{\mathrm{AC}}$
∴ AD is not a bisector of ∠A.

 

(ii) In ∆ABC, AB = 4 cm, AC = 6 cm, BD = 1.6 cm, CD = 2.4 cm

$\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{1.6}{2.4}=\frac{16}{24}=\frac{2}{3}$
$\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{4}{6}=\frac{2}{3}$
∴ $\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{\mathrm{AB}}{\mathrm{AC}}$
By angle bisector theorem; AD is the internal bisector of ∠A

 

10. In figure ∠QPR = 90°, PS is its bisector.
If ST ⊥ PR, prove that ST × (PQ + PR) = PQ × PR.
Answer Key:

Given: ∠QPR = 90°; PS is the bisector of ∠P. ST ⊥ ∠PR

To prove: ST × (PQ + PR) = PQ × PR
Proof: In ∆ PQR, PS is the bisector of ∠P.
$\therefore \frac{\mathrm{PQ}}{\mathrm{QR}}=\frac{\mathrm{QS}}{\mathrm{SR}}$
Adding (1) on both side
$1+\frac{\mathrm{PQ}}{\mathrm{QR}}=1+\frac{\mathrm{QS}}{\mathrm{SR}}$
$\frac{\mathrm{PR}+\mathrm{PQ}}{\mathrm{PR}}=\frac{SR+QS}{\mathrm{SR}}$
$\frac{\mathrm{PR}+\mathrm{PQ}}{\mathrm{PR}}=\frac{QR}{\mathrm{SR}}.....\left(1\right)$
In ∆ RST And ∆ RQP
∠SRT = ∠QRP = ∠R (Common)
∴ ∠QRP = ∠STR = 90°
(By AA similarity) ∆ RST ~ RQP
$\frac{\mathrm{SR}}{\mathrm{QR}}=\frac{\mathrm{ST}}{\mathrm{PQ}}$
$\frac{QR}{\mathrm{SR}}=\frac{\mathrm{PQ}}{\mathrm{ST}}....\left(2\right)$
From (1) and (2) we get
$\frac{\mathrm{PR}+\mathrm{PQ}}{\mathrm{PR}}=\frac{\mathrm{PQ}}{\mathrm{ST}}$
ST (PQ + PR) = PQ × PR

11. ABCD is a quadrilateral in which AB = AD, the bisector of ∠BAC and ∠CAD intersect the sides BC and CD at the points E and F respectively. Prove that EF || BD.
Answer Key:

ABCD is a quadrilateral. AB = AD.
AE and AF are the internal bisector of ∠BAC and ∠DAC.

To prove: EF || BD.
Construction: Join EF and BD
Proof: In ∆ ABC, AE is the internal bisector of ∠BAC.
By Angle bisector theorem, we have,
$\therefore \frac{AB}{AC}=\frac{BE}{EC}....\left(1\right)$
In ∆ ADC, AF is the internal bisector of ∠DAC
By Angle bisector theorem, we have,
$\frac{\mathrm{AD}}{\mathrm{AC}}=\frac{\mathrm{DF}}{\mathrm{FC}}$
$\therefore \frac{AB}{AC}=\frac{\mathrm{DF}}{\mathrm{FC}}(AB=AD given)....\left(2\right)$
From (1) and (2), we get,
$\frac{BE}{EC}=\frac{\mathrm{DF}}{\mathrm{FC}}$
Hence in ∆ BCD,
BD || EF (by converse of BPT)

 

12. Construct a ∆PQR which the base PQ = 4.5 cm, R = 35° and the median from R to RG is 6 cm.
Answer Key:

Steps of construction:

1. Draw a line segment PQ = 4.5 cm
2. At P, draw PE such that ∠QPE = 60°
3. At P, draw PF such that ∠EPF = 90°
4. Draw the perpendicular bisect to PQ, which intersects PF at O and PQ at G.
5. With O as centre and OP as radius draw a circle.
6. From G mark arcs of radius 5.8 cm on the circle. Mark them at R and S
7. Join PR and RQ. PQR is the required triangle.

 

13. Construct a ∆PQR in which QR = 5 cm, ∠P = 40° and the median PG from P to QR is 4.4 cm. Find the length of the altitude from P to QR.
Answer Key:

Steps of construction

1. Draw a line segment RQ = 5 cm.
2. At R draw RE such that ∠QRE = 40°
3. At R, draw RF such that ∠ERF = 90°
4. Draw the perpendicular bisector to RQ, which intersects RF at O and RQ at G.
5. With O as centre and OP as radius draw a circle.
6. From G mark arcs of radius 4.4 cm on the circle. Mark them as P and S.
7. Join PR and PQ. Then ∆PQR is the required triangle.
8. From P draw a line PN which is perpendicular to RQ it meets at N.
9. Measure the altitude PN.
   PN = 2.2 cm.

14. Construct a ∆PQR such that QR = 6.5 cm, ∠P = 60° and the altitude from P to QR is of length 4.5 cm.
Answer Key:

Steps of construction

1. Draw a line segment QR = 6.5 cm.
2. At Q draw QE such that ∠RQE = 60°.
3. At Q, draw QF such that ∠EQF = 90°.
4. Draw the perpendicular of QR which intersects QF at O and QR at G.
5. With O as centre and OQ as radius draw a circle.
6. X Y intersects QR at G. On X Y, from G mark an arc at M. Such that GM = 4.5 cm.
7. Draw AB through M which is parallel to QR.
8. AB Meets the circle at P and S.
9. join QP and RP.
   PQR is the required triangle.

 

15.Construct a ∆ABC such that AB = 5.5 cm, ∠C = 25° and the altitude from C to AB is
Answer Key:

Steps of construction

1. Draw a line segment AB = 5.5 cm.
2. At A draw AE such that ∠BAE = 25°.
3. At A draw AF such that ∠EAF = 90°.
4. Draw the perpendicular bisector of AB which intersects AF at O and AB at G.
5. With O as centre and OB as radius draw a circle.
6. X Y intersects AB at G. On X Y, from G mark an arc at M. Such that GM = 4 cm.
7. Through M draw a line parallel to AB intersect the circle at C and D.
8. Join AC and BC.
   ABC is the required triangle.

 

16. Draw a triangle ABC of base BC = 5.6 cm, ∠A = 40° and the bisector of ∠A meets BC at D such that CD = 4 cm.
Answer Key:

Steps of construction

1. Draw a line segment BC = 5.6 cm.
2. At B draw BE such that ∠CBE = 40°.
3. At B draw BF such that ∠EBF = 90°.
4. Draw the perpendicular bisector to BC which intersects BF at O and BC at G.
5. With O as centre and OB as radius draw a circle.
6. From C mark an arc of 4 cm on CB at D.
7. The perpendicular bisector intersects the circle at I. Joint ID.
8. ID produced meets the circle at A. Now Join AB and AC.
   This ABC is the required triangle.

17. Draw ∆PQR such that PQ = 6.8 cm, vertical angle is 50° and the bisector of the vertical angle meets the base at D where PD = 5.2 cm
Answer Key:

Steps of construction

1. Draw a line segment PQ = 6.8 cm.
2. At P draw PE such that ∠QPE = 50°.
3. At P draw PF such that ∠EPF = 90°.
4. Draw the perpendicular bisector to PQ which intersects PF at O and PQ at G.
5. With O as centre and OP as radius draw a circle.
6. From P mark an arc of 5.2 cm on PQ at D.
7. The perpendicular bisector intersects the circle at I. Join ID.
8. ID produced meets the circle at A. Now Joint PR and QR. This PQR is the required triangle.