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

 

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

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

    This post covers the book back answers and solutions for Chapter 4 Unit Exercise 4 – 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 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. In the figure, if BD ⊥ AC and CE ⊥ AB , prove that
(i) ∆ AEC ~ ∆ADB
(ii) $\frac{\mathbf{C}\mathbf{A}}{\mathbf{A}\mathbf{B}}\mathbf{=}\frac{\mathbf{C}\mathbf{E}}{\mathbf{D}\mathbf{B}}$

Answer Key:

(i) ∠AEC = ∠ADB = 90° ∠A is common By AA – Similarity.
∴ ∆AEC ~ ∆ADB
Since the two triangles are similar

 

(ii) $\frac{\mathrm{AE}}{\mathrm{AD}}=\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{\mathrm{EC}}{\mathrm{DB}}$
$\frac{\mathrm{AC}}{\mathrm{AB}}=\frac{\mathrm{CE}}{\mathrm{DB}}$

 

2. In the given figure AB || CD || EF. If AB = 6 cm, CD = x cm, EF = 4 cm, BD = 5 cm and DE = y cm. Find x and y.

Answer Key:
In the given diagram ∆AEF and ∆ACD
∠AEF = ∠ACD = 90°
∠A is common
By AA – Similarity.
∴ ∆AEF ~ ∆ACD
$\frac{\mathrm{AE}}{\mathrm{AC}}=\frac{\mathrm{AF}}{\mathrm{AD}}=\frac{\mathrm{EF}}{\mathrm{CD}}$
$\frac{\mathrm{AE}}{\mathrm{AC}}=\frac{\mathrm{EF}}{\mathrm{CD}}$
$\frac{\mathrm{AE}}{\mathrm{AC}}=\frac{4}{x}$
$AC=\frac{AE\times x}{4}....\left(1\right)$
In ∆EAB and ∆ECD,
∠EAB = ∠ECD = 90°
∠E is common
∆ ECD ~ ∆EAB
$\frac{\mathrm{EC}}{\mathrm{EA}}=\frac{\mathrm{ED}}{\mathrm{EB}}=\frac{\mathrm{CD}}{\mathrm{AB}}$

$\frac{\mathrm{EC}}{\mathrm{EA}}=\frac{\mathrm{x}}{6}$

$\mathrm{EC}=\frac{\mathrm{EA}\times \mathrm{x}}{6}....\left(2\right)$ 
In ∆AEB; CD || AB
By Basic Proportionality Theorem
$\frac{\mathrm{AB}}{\mathrm{CD}}=\frac{\mathrm{EB}}{\mathrm{ED}}$

$\frac{6}{\mathrm{x}}=\frac{5+\mathrm{y}}{\mathrm{y}}$

$\mathrm{x}=\frac{6\mathrm{y}}{\mathrm{y}+5}(\mathrm{EC}=\mathrm{x})...\left(3\right)$ 
Add (1) and (2) we get
AC+AE=$\frac{AE\times x}{4}+\frac{x\times EA}{6}$

$\therefore 1=\frac{5x}{12}\Rightarrow 5x=12$

$\mathrm{x}=\frac{12}{5}=2.4\mathrm{cm}$

Substitute the value of x = 2.4 in (3)
2.4 = $\frac{6y}{y+5}$
6y = 2.4y + 12
6y – 2.4y = 12 ⇒ 3.6 y = 12
$\mathrm{y}=\frac{12}{3.6}=\frac{120}{6}=\frac{10}{3}=3.3\mathrm{cm}$
The value of x = $\frac{12}{5}$(or) 2.4 cm and y = $\frac{10}{3}$ (or) 3.3 cm

 

3. O is any point inside a triangle ABC. The bisector of ∠AOB, ∠BOC and ∠COA meet the sides AB, BC and CA in point D, E and F respectively. Show that AD × BE × CF = DB × EC × FA.
Answer Key:

In ∆ABC the bisector meets AB at D, BC at E and AC at F.

The angle bisector AO, BO and CO intersect at “O”.
By Cevas Theorem
$\frac{\mathrm{AD}}{\mathrm{DB}}\times \frac{\mathrm{BE}}{\mathrm{EC}}\times \frac{\mathrm{CF}}{\mathrm{AF}}=1$
AD × BE × CF = DB × EC × AF
Hence it is proved

 

4. In the figure, ABC is a triangle in which AB = AC. Points D and E are points on the side AB and AC respectively such that AD = AE. Show that the points B, C, E and D lie on a same circle.

Answer Key:
∠B = ∠C (Given AB = AC)
AD + DB = AE + EC
BD = EC (Given AD = AE)
DE parallel BC Since AEC is a straight line.
∠AED + ∠CED = 180°
∠CBD + ∠CED = 180°
Similarly of the opposite angles = 180°
∴ BCED is a cyclic quadrilateral

 

5. Two trains leave a railway station at the same time. The first train travels due west and the second train due north. The first train travels at a speed of 20 km/hr and the second train travels at 30 km/hr. After 2 hours, what is the distance between them?
Answer Key:
A is the position of the 1^st train.
B is the position of the 2^nd train.

Distance Covered in 2 hours
OA = 2 × 20 = 40 km
OB = 2 × 30 = 60 km
Distance between the train after 2 hours

$\mathrm{AB}=\sqrt{{\mathrm{OA}}^2+{\mathrm{OB}}^2}$

$\mathrm{AB}=\sqrt{{40}^2+{60}^2}$

$\mathrm{AB}=\sqrt{1600+3600}$

$\mathrm{AB}=\sqrt{5200}\left(or\right)\sqrt{52\times 100}$

$10\sqrt{4\times 13}=20\sqrt{13}$

$=72.11\mathrm{km}$ 

Distance between

the two train = 72.11 km (or) $20\sqrt{13}$km

 

6. D is the mid point of side BC and AE ⊥ BC. If BC = a, AC = b, AB = c, ED = x, AD = p and AE = h, prove that

(i) b² = p² + ax + $\frac{{\mathrm{a}}^2}{4}$
(ii) c² = p² – ax +$\frac{{\mathrm{a}}^2}{4}$
(iii) a² + c² = 2 p² + $\frac{{\mathrm{a}}^2}{2}$
Answer Key:
(i) Given ∠AED = 90°

ED = x; DC = $\frac{\mathrm{a}}{2}$
(D is the mid point of BC)
∴ EC = x + $\frac{\mathrm{a}}{2}$, BE = $\frac{\mathrm{a}}{2}$ – x
∴ In the right ∆ AED
AD² = AE² + ED²
p² = h² + x²
In the right ∆ AEC,
AC² = AE² + EC²

 ${\mathrm{h}}^{2+}{\mathrm{x}}^2+\frac{{\mathrm{a}}^2}{4}+2\times \mathrm{x}\times \frac{\mathrm{a}}{2}$

${\mathrm{b}}^2={\mathrm{p}}^2+\frac{{\mathrm{a}}^2}{4}+ax(from 1)$

${\mathrm{b}}^2={\mathrm{p}}^2+ax+\frac{1}{4}{a}^2....\left(2\right)$ 

 

(ii) In the right triangle ABE,
AB² = AE² + BE²
c² = h² + ($\frac{\mathrm{a}}{2}$ – x)²
c² = h² + $\frac{a^2}{4}$ + x² – ax
c² = h² + x² + $\frac{1}{4}$a² – ax
c² = p² – ax + $\frac{a^2}{4}$ (from 1)

 

(iii) By adding (2) and (3)
b² + c² = p² + ax + $\frac{a^2}{4}$ + p² – ax + $\frac{a^2}{4}$
= 2p² + $\frac{2a^2}{4}$
= 2p² + $\frac{a^2}{4}$

 

7.A man whose eye-level is 2 m above the ground wishes to find the height of a tree. He places a mirror horizontally on the ground 20 m from the tree and finds that if he stands at a point C which is 4 m from the mirror B, he can see the reflection of the top of the tree. How height is the tree?
Answer Key:
Let the height of the tree AD be “h”.

In ∆ ACD and ∆ BCF,
∠A = ∠B = 90°
∠C is common
∆ ACD ~ ∆ BCF by AA similarity
$\frac{AD}{BF}=\frac{AC}{BC}$
$\frac{h}{x}=\frac{24}{2}=6$
h = 6x ………(1)
In ∆ ACE and ∆ ABF,
∠C = ∠B = 90°
∠A is common
∴ ∆ ACE ~ ∆ ABF
$\frac{\mathrm{CE}}{\mathrm{BF}}=\frac{\mathrm{AC}}{\mathrm{AB}}$
$\frac{2}{\mathrm{x}}=\frac{24}{20}$
24x = 20 × 2
$\mathrm{x}=\frac{20\times 2}{24}=\frac{5\times 2}{6}=\frac{10}{6}$
$\mathrm{x}=\frac{5}{3}$
Substitute the value of x in (1)
h = 6 ×$\frac{5}{3}$ = 10 m
∴ Height of the tree is 10 m

 

8.An emu which is 8 ft tail is standing at the foot of a pillar which is 30 ft high. It walks away from the pillar. The shadow of the emu falls beyond emu. What is the relation between the length of the shadow and the distance from the emu to the pillar?
Answer Key:
Let the shadow of the emu AE be “x” and BE be “y” ED || BC

By basic proportionality theorem
$\frac{AE}{AB}=\frac{ED}{BC}$
$\frac{x}{x+y}=\frac{8}{30}$
30x = 8x + 8y
22x – 8y = 0
(÷ by 2) 11x – 4y = 0
11x = 4y
x = $\frac{4}{11}$ × y
x = $\frac{4}{11}$ × distance from the pillar to emu
Length of = $\frac{4}{11}$ × distance from the shadow the pillar to emu

 

9. Two circles intersect at A and B. From a point P on one of the circles lines PAC and PBD are drawn intersecting the second circle at C and D. Prove that CD is parallel to the tangent at P.
Answer Key:
Proof:

A and B are the points intersecting the circles. Join AB.
∠P’PB = ∠PAB (alternate segment theorem)
∠PAB + ∠BAC = 180° …(1)
(PAC is a straight line)
∠BAC + ∠BDC = 180° …(2)
ABDC is a cyclic quadrilateral.
From (1) and (2) we get
∠P’PB = ∠PAB = ∠BDC
P’P and DC are straight lines.
PD is a transversal alternate angles are equal.
∴ P’P || DC.

 

10. Let ABC be a triangle and D, E, F are points on the respective sides AB, BC, AC (or their extensions).
Let AD : DB = 5 : 3, BE : EC = 3 : 2 and AC = 21. Find the length of the line segment CF.
Answer Key:

$\frac{AD}{DB}=\frac{5}{3};\frac{BE}{EC}=\frac{3}{2};AC=21$
By Ceva’s theorem

$\frac{BE}{EC}\times \frac{CF}{FA}\times \frac{AD}{DB}=1$

$\frac{3}{2}\times \frac{CF}{21-CF}\times \frac{5}{3}=1$

$\frac{5CF}{2(21-CF)}=1$

$\frac{CF}{\left(21-CF\right)}=\frac{2}{5}$

$5CF=42-2CF$

$7CF=42$

$CF=\frac{42}{7}=6$ 

Length of the line segment CF = 6 units