Category: Class 8

Students can Download Maths Chapter 5 Information Processing Ex 5.3 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 5 Information Processing Ex 5.3

MISCELLANEOUS QUESTIONS

Question 1.
Shanthi has 5 chudithar sets and 4 Frocks. In how many possible ways, can she wear either a chudithar or a frock ?
Solution:
Shanthi his 5 chudidhar sets and 4 frocks.
She wear either chudidhar or a frock.
∴ Total possible ways = 5 + 4 = 9 ways

Question 2.
In a Higher Secondary School, the following types of groups are available in XI standard
I. Science Group:
(i) Physics, Chemistry, Biology and Mathematics
(ii) Physics, Chemistry, Mathematics and Computer Science
(iii) Physics, Chemistry, Biology and Home Science
II. Arts Group:
(i) 1. Accountancy, Commerce, Economics and Business Maths
(ii) 2. Accountancy, Commerce, Economics and Computer Science
(iii) 3. History, Geography, Economics and Commerce
III. Vocational Group:
(i) Nursing – Biology, Theory, Practical I and Practical II
(ii) Textiles and Dress Designing – Home Science, Theory, Practical I and Practical II
In how many possible ways, can a student choose the group?
Solution:
The student either select any one of science group in 3 ways or any of the arts group in 3 ways or any of the vocational group in 2 ways.
∴ Total possible ways = 3 + 3 + 2 = 8 ways

Question 3.
An examination paper has 3 sections, each with five instructed to answer one question from each section. In can the questions be answered?
Solution:
The tree diagram for this may be

∴ Number of possible ways to select one questions from each of 3 sections is 3 × 5 = 15 ways.

Question 4.
On a sports day, students must take also part in one of the one track events 100m Running and 4 × 100 m Relay. He must take part of any of the field events Long Jump, High Jump and Javelin Throw. In how many different ways can the student take part in the given events?
Solution:
Number of track events ⇒ (100m running, 4 × 100 m Relay) 2.
Number of field events ⇒ (Long jump, High jump, Javelin Throw) 3.
Students can take part in the given events in 2 × 3 = 6 ways.

Question 5.
The given spinner is spun twice and the two numbers got are used to form a 2 digit number. How many different 2 digits numbers are possible?

Solution:
On the first spin we get any of the five numbers to form ones place then insecond spin the number got will fill 10’s place.
∴ Number of ways = 5 × 5 = 25 ways.
Removing the repetitions (11, 22, 33, 44, 55) once we get 25 – 5 = 20 ways.
20 different two digit numbers are possbile

Question 6.
Colour the following pattern with as few colours as possible but make sure that no two adjacent sections are of the same colour.

Solution:

Students can Download Maths Chapter 5 Information Processing Ex 5.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 5 Information Processing Ex 5.2

Question 1.
Colour the following patterns with as few colours as possible but make sure that no two adjacent sections are of the same colour.
Solution:

Question 2.
Ramya wants to paint a pattern in her living room wall with a minimum budget. Help her to colour the pattern with 2 colours but make sure that no two adjacent boxes are the same colour. The pattern is shown in the picture.

Solution:

Question 3.
Colour the countries in the following maps with as few colours as possible but make
sure that no two adjacent countries are of the same colour.

Solution:

Students can Download Maths Chapter 3 Algebra Additional Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Additional Questions

Additional Questions And Answers

Exercise 3.1

Very Short Answers [2 Marks]

Question 1.
Find the product of the following.
(i) (x, y)
(ii) (10x, 5y)
(iii) (2x², 5y²)
Solution:
(i) x × y = xy
(ii) 10x × 5y = (10 × 5) x × xy
= 50 xy
(iii) 2x² x 5y² = (2 x 5) x (x² + y²)
= 10x²y²

Short Answers [3 Marks]

Question 1.
Find the product of the following.
(i) 3ab² c³ by 5a³b²c
(ii) 4x²yz by \(\frac{3}{2}\) x²yz²
Solution:
(i) (3ab³c³) × (5a³b²c)
= (3 × 5)(a × a³ × b² × b² × c² × c)
= 15a¹⁺³.b²⁺².c³⁺¹ = 15a⁴b⁴c⁴

(ii) 4x²yz by \(\frac{3}{2}\) x²yz²
= (4 × \(\frac{3}{2}\)) × (x² × x² × y × y × z × z²)
= -6x²⁺² y¹⁺¹ x¹⁺² = -6x⁴y²z³

Long Answers [5 Marks]

Question 1.
Simplify (3x – 2) (x – 1) (3x + 5).
Solution:
(3x – 2) (x – 1) (3x + 5)
= {(3x – 2) (x – 1)} × (3x + 5) [∴Multiplication in associative]
= {3x (x – 1) – 2 x – 1)} × (3x + 5)
= (3x² – 3x – 2x + 2) × (3x + 5)
= (3x² – 5x + 2) (3x + 5)
= 3x² × (3x + 5) – 5x (3x + 5) + 2 (3x + 5)
= (9x³ + 15x²) + (-15x² – 25x) + (6x + 10)
= 9x³ + 15x² – 15x² – 25x + 6x + 10
= 9x³ – 19x + 10

Question 2.
Simplify (5 – x) (3 – 2x) (4 – 3x).
Solution:
(5 – x) (3 – 2x) (4 – 3x)
= {(5 – x)(3 – 2x)} × (4 – 3x) [∴ Multiplication in association]
= {5 (3 – 2x) -x (3 – 2x)} × (4 – 3x)
= (15 – 10x – 3x + 2x²) × (4 – 3x)
= (2x² – 13x + 15) (4 – 3x)
= 2x² × (4 – 3x) – 13x (4 – 3x) + 15 (4 – 3x)
= 8x³ – 6³ – 52x + 39x² + 60 – 45x
= -6x³ + 47x² – 97x + 60

Exercise 3.2

Very Short Answers [2 Marks]

Question 1.
Divide.
(i) 12x³y³ by 3x²y
(ii) -15a² bc³ by 3ab
(iii) 25x³y² by – 15x²y
Solution:

Short Answers [3 Marks]

Question 1.
Divide
(i) 15m²n³ by 5m²n²
(ii) 24a³b³ by -8ab
(iii) -21 abc² by 7 abc
Solution:

Question 2.
Divide
(i) 16m³y² by 4m²y
(ii) 32m² n³p² by 4mnp
Solution:

Long Answers [5 Marks]

Question 1.
Divide.
(i) 9m⁵ + 12m⁴ – 6m² by 3m²
(ii) 24x³y + 20x²y² – 4xy by 2xy
Solution:

Exercise 3.3

Question 1.
Evaluate:
(i) (2x + 3y)²
(ii) (2x – 3y)²
Solution:
(i) (2x + 3y)²
= (2x)² + 2 × (2x) × (3y) + (3y)²
[using (a + b)² = a² + 2ab + b²]
= 4x² + 12xy + 9y²

(ii) (2x – 3y)²
= (2x)² – 2(2x) (3y) + (3y)²
[∵ using (a – b)² = a² – 2ab + b²]
= 4x² – 12xy + 9y²

Short Answers [3 Marks]

Question 1.
Evaluate the following
(i) (2x – 3) (2x + 5)
(ii) (y – 7) (y + 3)
(iii) 107 × 103
Solution:
(i) (2x – 3) (2x + 5)
= (2x)² + (-3 + 5) (2x) + (-3) (5)
[∵ (x + a) (x + b) = x² + (a + b)x + ab]
= 2²x² + 2 × 2x + (-15)
= 4x² + 4x – 15

(ii) (y – 7) (y + 3)
= y² + (-7 + 3)y + (-7) (3)
[∵ (x + a)(x + b) = x² + (a + b)x + ab]
= y² – 4y + (-21) = y² – 4y – 21

(iii) 107 × 103
= (100 + 7) × (100+ 3)
= 100² + (7 + 3) × 100 +(7 × 3)
= 10000 + 10 × 100 + 21 = 10000 + 1000 + 21 = 11021

Long Answers [5 Marks]

Question 1.
If x + y = 12 and xy = 14 find x² + y².
Solution:
(x + y)² = x² + y² + 2xy
12² = x² + y² + 2 × 14
144 = x² + y² + 28
x² + y² = 144 – 28
x² + y² = 116

Question 2.
If 3x + 2y = 12 and xy = 6 find the value of 9x² + 4y².
Solution:
(3x + 2y)² = (3x)² + (2y)² + 2 (3x) (2y)
= 9x² + 4y² + 12xy
12² = 9x² + 4y² + 12 × 6
144 = 9x² + 4y² + 72
144 – 72 = 9x² + 4y²
∴ 9x² + 4y² = 72

Exercise 3.4

Question 1.
Factorize:
(i) 7(2x + 5) + 3 (2x + 5)
(ii) 12x³y⁴ + 16x²y⁵ – 4x⁵y²
Solution:
(i) 7(2x + 5) + 3 (2x + 5)
= (2x + 5) (7 + 3)
(ii) 12x³y⁴ + 16x²y⁵ – 4x⁵y²
= 4x²y² (3xy² + 4y³ – x³)

Short Answers [3 Marks]

1. Factorize
(i) 81a² – 121b²
(ii) x² + 8x + 16
Solution:
(i) 81a² – 121b²
= (9a)² – (11b)²
[∵ using a² – b² = (a + b)²]
= (9a + 11b) (9a – 11b)

(ii) x² + 8x + 16 = x² + 2 × x × 4 + 4²
[∵ using a² + 2ab + b² = (a + b)²]
= (x + 4)² = (x + 4)(x + 4)

Long Answers [5 Marks]

Question 1.
Factorize
(i) x² + 2xy + y² – a² + 2ab – b²
(ii) 9 – a⁶ + 2a³ – b⁶
Solution:
(i) x² + 2xy + y² – a² + 2ab – b².
= (x² + 2xy + y²) – (a² – 2ab + b²)
= (x + y)² – (a – b)²
= {(x + y) + (a – b)} {(x + y) – (a – b)}
= (x + y + a – b) (x + y – a + b)

(ii) a – a⁶ + 2a³b³ – b⁶
= 9 – (a⁶ – 2a³b³ + b⁶)
= 3² -{(a³)² – 2 × a³ × b³ + (b³)²}
= 3² – (a³ – 6³)²
= {3 + (a³ – b³)} {3 – (a³ – b³)}
= (3 + a³ – b³) (3 – a³ + b³)
= (a³ – b³ + 1){-a³ + b³ + 3)

Question 2.
Factorize
(i) 100 (x + y )² – 81 (a + b)²
(ii)(x + 1)² – (x – 2)²
Solution:
(i) 100 (x + y)² – 81 (a + b)²
= {10 (x + y)}² – {(a (a + b)}²
= {10 (x + y) + 9 (a + b)}
{10 (x + y) – 9(a + b)}
= (10x + 10y + 9a – 9b)}
(10x + 10y – 9a – 9b)

(ii) (x – 1)² – (x – 2)²
= {(x – 1 +(x – 2)}
{(x – 1) – (x – 2)}
= (2x – 3) – (x – 1 – x + 2)
= (2x – 3) × 1 = 2x – 3

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Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.5

Miscellaneous Practice Problems

Question 1.
Subtract: -2(xy)² (y³ + 7x²y + 5) from 5y² (x²y³ – 2x⁴y + 10x²)
Solution:
5y² (x²y³ – 2x⁴y + 10x²) – [(-2)(xy)² (y³ + 7x²y + 5)]
= [5y² (x²y³) – 5y² (2x⁴y) + 5y² (10x²)] – [(-2)x²y² (y³ + 7x²y + 5)]
= (5y⁵5x² – 10x⁴y³ + 50x²y²)
= 5x²y⁵ – 10x⁴y³ + 50x²y² – [(-2x²y⁵) – 14x⁴y³ – 10x²y²]
= 5x²y⁵ – 10x⁴y³ + 50x²)y² + 2x² y⁵ + 14x⁴)y³ + 10x²)y²
= (5 + 2)x²y⁵ + (-10 + 14)x⁴y³ + (+50 + 10)x²y²
= 7x²y⁵ + 4x⁴y³ + 60x²y²

Question 2.
Multiply (4x² + 9) and (3x – 2).
Solution:
(4x² + 9) (3x – 2) = 4x²(3x – 2) + 9(3x – 2)
= (4x²)(3x) – (4x²)(2) + 9(3x) – 9(2) = (4 × 3 × x × x²) – (4 × 2 × x²) + (9 × 3 × x) – 18
= 12x³ – 8x² + 27x – 18 (4x³ + 9) (3x – 2) = 12x³ – 8x² + 27x – 18

Question 3.
Find the simple interest on Rs. 5a²b² for 4ab years at 7b% per annum.
Solution:

Question 4.
The cost of a note book is Rs. 10ab. If Babu has Rs. (5a²b + 20ab² + 40ab). Then how many note hooks can he buy?
Solution:
For ₹ 10 ab the number of note books can buy = 1.

Question 5.
Factorise : (7y² – 19y – 6)
Solution:
7y² – 19y – 6 is of the form ax² + bx + c where a = 7; b = -19; c = – 6

The product a × c = 7 × -6 = -42
sum b = – 19

The middle term – 19y can be written as – 21y + 2y
7y² – 19y – 6 = 7y² – 21y + 2y – 6
= 7y(y – 3) + 2(y – 3) = (y – 3)(7y + 2)
7y² – 19y – 6 = (y – 3)(7y + 2)

Challenging Problems

Question 6.
A contractor uses the expression 4x² + 11x + 6 to determine the amount of wire to order when wiring a house. If the expression comes from multiplying the number of rooms times the number of outlets and he knows the number of rooms to be (x + 2), find the number of outlets in terms of ‘x’ [Hint: factorise 4x² + 11x + 6]
Solution:
Given Number of rooms = x + 2
Number of rooms × Number of outlets = amount of wire.

Now factorising 4x² + 11x + 6 which is of the form ax² + bx + c with a = 4 b = 11 c = 6.
The product a × c = 4 × 6 = 24
sum b = 11


The middle term 11x can be written as 8x + 3x
∴ 4x² + 11x + 6 = 4x² + 8x + 3x + 6 = 4x (x + 2) + 3 (x + 2)
4x² + 11x + 6 = (x + 2) (4x + 3)
Now from (1) the number of outlets

∴ Number of outlets = 4x + 3

Question 7.
A mason uses the expression x² + 6x + 8 to represent the area of the floor of a room. If the decides that the length of the room will be represented by (x + 4), what will the width of the room be in terms of x ?
Solution:
Given length of the room = x + 4
Area of the room = x² + 6x + 8
Length × breadth = x² + 6x + 8

Question 8.
Find the missing term: y² + (-) x + 56 = (y + 7)(y + -)
Solution:
We have (x + a) (x + b) = x² + (a + b)x + ab
56 = 7 × 8 .
∴ y² + (7 + 8)x + 56 = (y + 7)(y + 8)

Question 9.
Factorise: 16p⁴ – 1
Solution:
16p⁴ – 1 = 2⁴p⁴ – 1 = (2²)²(p²)² – 1²
Comparing with a² – b² = (a + b)(a – b) where a = 2²p² and b = 1
∴ (2²p²)² – 1² = (2²p² + 1) (2²p² – 1) = (4p² + 1) (4p² – 1)
∴ 16p⁴ – 1 = (4p² + 1)(4p² – 1)(4p² + 1)(2²p² – 1²)
= (4p² + 1) [(2p)²– 1²] = (4p² + 1) (2p + 1)(2p – 1)
[∵ using a² – b² = (a + b) (a – b)]
∴ 16p⁴ – 1 = (4psup>2 + 1)(2p + 1)(2p – 1)

Question 10.
Factorise : x⁶ – 64y³
Solution:
x⁶ – 64y³ = (x²)³ – 4³y³ = (x²)³ – (4y)³
This is of the form a³ – b³ with a = x², b = 4y
a³ – b³ = (a – b)(a² + ab + b²)
(x²)³ – (4y)³ = (x² – 4y) [(x²)² + (x²)(4y) + (4y)²]
= (x² – 4y) [x⁴ + 4x²y + 16y²]
∴ x⁶ – 64y³ = (x² – 4y) [x⁴ + 4x²y + 16y²]

Students can Download Maths Chapter 4 Geometry Additional Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 4 Geometry Additional Questions

Additional Questions And Answers

Very Short Answers [2 Marks]

Question 1.
In the given figure if ∠A = ∠C then prove that ∆AOB ~ ∆COD.

Solution:
In triangles ∆AOB and ∆COD
∠A = ∠C (Y given)
∠AOB = ∠COD [∵ Vertically opposite angles]
∠ABO = ∠CDO [Remaining angles of ∆AOB and ∆COD]
∴ ∆AOB ~ ∆COD [∵ AAA similarity]
∵ ∆AOB ~ ∆COD [∵ AAA similarity]

Question 2.
In the figure AB ⊥ BC and DE ⊥ AC prove that ∆ABC ~ ∆AED.

Solution:
In triangles ∆ABC and ∆AED
∠ABC = ∠AED = 90°
∠BAC = ∠EAD [Each equal to A]
∠ADE = ∠ACB [∵ Remaining angles]
∴ By AAA criteria of similarity ∆ABC ~ ∆AED

Short Answers [3 Marks]

Question 1.
In the figure with respect to ABEP and CPD prove that BP × PD = EP × PC.

Solution:
Proof:
In ∆EPB and ∆DPC
∠PEB = ∠PDC = 90° [given]
∠EPB = ∠DPC [Vertically opposite angles]
∠EPB = ∠PCD [∵ Remaining angles]
Thus,

Long Answers [5 Marks]

Question 1.
P and Q are points on sides AB and AC respectively of ∆ABC. If AP = 3 cm PB = 6cm, AQ = 5 cm and QC = 10 cm, show that BC = 3 PQ.
Solution:

AB = AP + PB
= 3 + 6 cm = 9 cm
AC = AQ + QC = 510 cm = 15

Students can Download Maths Chapter 4 Geometry Ex 4.3 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 4 Geometry Additional Questions

Construct the quadrilaterals with the following measurements and also find their area.

Question 1.
ABCD, AB = 5 cm, BC 4.5 cm, CD = 3.8 cm, DA = 4.4 cm and AC = 6.2 cm.
Solution:
Given AB = 5 cm,
BC = 4.5 cm,
CD = 3.8 cm,
DA = 4.4 cm,
AC = 6.2 cm

Steps:
1. Draw a line segment AB = 5 cm
2. With A and B as centers drawn arcs of radii 6.2 cm and 4.5cm respectively and let them cut at C.
3. Joined AC and BC.
4. With A and C as centrers drawn arcs of radii 4.4cm and 3.8 cm respectively and let them at D.
5. Joined AD and CD.
6. ABCD is the required quadrilateral.
Calculation of Area:

Question 2.
KITE, KI = 5.4 cm, IT = 4.6 cm, TE= 4.5 cm, KE = 4.8 cm and IE = 6 cm.
Solution:
Given, KI = 5.4 cm,
IT = 4.6 cm,
TE= 4.5 cm,
KE = 4.8 cm,
IE = 6 cm.

Steps:
1. Draw a line segment KI = 5.4 cm
2. With K and I as centers drawn arcs of radii 4.8 cm and 6 cm respectively and let them cut at E.
3. Joined KE and IE.
4. With E and I as centers, drawn arcs of radius 4.5cm and 4.6 cm respectively and let them cut at T.
5. Joined ET and IT.
6. KITE is the required quadrilateral.
Calculation of Area:

Question 3.
PLAY, PL = 7 cm, LA = 6 cm, AY= 6 cm, PA = 8 cm and LY = 7 cm.
Solution:
Given PL = 7 cm,
LA = 6 cm,
AY= 6 cm,
PA = 8 cm,
LY = 7 cm

Steps:
1. Draw a line segment PL = 7 cm
2. With P and L as centers, drawn arcs of radii 8 cm and 6 cm respectively, let them cut at A.
3. Joined PA and LA.
4. With L and A as centers, drawn arcs of radii 7 cm and 6 cm respectively and let them cut at Y.
5. Joined LY, PY and AY.
6. PLAY is the required quadrilateral.
Calculation of Area:

Question 4.
LIKE, LI = 4.2 cm, IK = 7 cm, KE = 5 cm, LK = 6 cm and IE = 8 cm.
Solution:
LI = 4.2 cm,
IK = 7 cm,
KE = 5 cm,
LK = 6 cm,
IE = 8 cm

Steps:
1. Draw a line segment LI = 4.2 cm
2. With L and I as centers, drawn arcs of radii 6 cm and 7 cm respectively and let them cut at K.
3. Joined LK and IK.
4. With I and K as centers, drawn arcs of radius 8 cm and 5 cm respectively and let them cut at E.
5. Joined LE, IE and KE.
6. LIKE is the required quadrilateral.
Calculation of Area:

Question 5.
PQRS, PQ = QR = 3.5 cm, RS = 5.2 cm, SP = 5.3 cm and ∠Q =120° .
Solution:
PQ = QR = 3.5 cm,
RS = 5.2 cm,
SP = 5.3 cm ,
∠Q =120°

Steps:
1. Draw a line segment PQ = 3.5 cm
2. Made ∠Q = 120°. Drawn the ray QX.
3. With Q as centre drawn an arc of radius 3.5 cm. Let it cut the ray QX at R.
4. With R and P as centres drawn arcs of radii 5.2cm and 5.5 cm respectively and let them cut at S.
5. Joined PS and RS.
6. PQRS is the required quadrilateral.
Calculation of Area:
Area of the quadrilateral PQRS = 18 cm²

Question 6.
EASY, EA = 6 cm, AS = 4 cm, SY = 5 cm, EY = 4.5 cm and ∠E = 90°.
Solution:
EA = 6 cm,
AS = 4 cm,
SY = 5 cm,
EY = 4.5 cm,
∠E = 90°

1. Drawn a line segment EA = 6 cm
2. Made ∠E = 90°. From E drawn the ray EX.
3. With E as center drawn an arc of 4.5 cm radius. Let of cut the ray EX at Y.
4. With A and Y as centres drawn arcs of radii 4 cm and 5 cm respectively and let them cut at S.
5. Joined AS and YS.
6. EASY is the required quadrilateral.
Calculation of Area:

∴ Area of the quadrilateral = 22.87 cm²

Question 7.
MIND, MI = 3.6 cm, ND = 4 cm, MD = 4 cm, ∠M = 50° and ∠D = 100°.
Solution:
MI = 3.6 cm,
ND = 4 cm,
MD= 4 cm,
∠M = 50°,
∠D = 100°

1. Draw a line segment MI = 3.6 cm
2. At M on MI made an angle ∠IMX = 50°
3. Drawn an arc with center M and radius 4 cm let it cut MX it D
4. At D on DM made an angle ∠MDY = 100°
5. With I as center drawn an arc of radius 4 cm, let it cut DY at N.
6. Joined DN and IN.
7. MIND is the required quadrilateral.
Calculation of Area:

Area of the quadrilateral = 9.6 cm²

Question 8.
WORK, WO = 9 cm, OR = 6 cm, RK = 5 cm, ∠O = 100° and ∠R = 60°.
Solution:
WO = 9 cm,
OR = 6 cm,
RK = 5 cm,
∠O = 100°,
∠R = 60°

Steps:
1. Drawn a line segment WO = 9 cm
2. At O on WO made an angle ∠WOR = 100° and drawn the ray OX.
3. Drawn an arc of radius 6 cm with center O. Let it intersect OX at R.
4. At R on OR, made ∠ORY = 60°, and drawn the ray RY.
5. With center R drawn an arc of radius 5 cm, let it intersect RY at K.
6. Joined WK.
7. WORK is the required quadrilateral.
Calculation of Area:

Area of the quadrilateral = 31.59 cm²

Question 9.
AGRI, AG = 4.5 cm, GR = 3.8 cm, ∠A = 90°, ∠G = 110° and ∠R = 90°.
Solution:
AG = 4.5 cm,
GR = 3.8 cm,
∠A = 90°,
∠G = 110°,
∠R = 90°

1. Draw a line segment AG = 4.5 cm
2. At G on AG made ∠AGX =110°
3. With G as centre drawn an arc of radius 3.8 cm let it cut GX at R.
4. At R on GR made ∠GRZ = 90°
5. At A on AG made ∠GAY = 90°
6. AY and RZ meet at I.
7. AGRI is the required quadrilateral.
Calculation of Area:

Question 10.
YOGA, YO = 6 cm, OG = 6 cm, ∠O = 55°, ∠G = 55° and ∠A = 55°.
Solution:
YO = 6 cm,
OG = 6 cm,
∠O = 55°,
∠G = 55°,
∠A = 55°

Steps:
1. Draw a line segment OG = 6 cm
2. At G on DG made an angle ∠OGY = 55°
3. AT G on GO made ∠GOX = 55°.
4. GY and OX meet cut A.
5. At A on OA made ∠OAZ = 55°
6. Drawn an arc of radius 6 cm with center O. It cut AZ at Y. Joined OY.
7. YOGA is the required quadrilateral.
Calculation of Area:

Area of the quadrilateral YOGA = 28.08 cm²

Students can Download Maths Chapter 4 Geometry Ex 4.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 4 Geometry Ex 4.2

Miscellaneous Practice Problems

Question 1.
In the given figure, find PT given that l₁ || l₂

Solution:
Given that l₁ || l₂
∴ In ∆PQS and ∆PRT
∠P is common
∠Q = ∠R [∵ PR is the transversal for l₁ and l₂ corresponding angles]
∠S = ∠T [∵ corresponding angles]
∴ ∆PQS ~ ∆PRT [∵ By AAA congruency]
In similar triangles, corresponding angles are proportional.

Question 2.
From the diagram, prove that ∆SUN ~ ∆RAY

Proof:
From the ∆SUN and ∆RAY
SU = 10;
UN = 12;
SN = 14;
RA = 5,
AY = 6;
RY = 7
We have

Question 3.
The height of a tower is measured by a mirror on the ground by which the top of the tower’s reflection is seen. Find the height of the tower.
Solution:
The image and its reflection make similar shapes

Question 4.
In the figure, given that ∠1 = ∠2 and ∠3 ≡ ∠4 Prove that ∆MUG ≡ ∆ TUB.

Proof:

Question 5.
If ∆WAR ≡ ∆MOB, name the additional pair of corresponding parts. Name the criterion used by you.
Solution:

Given ∆WAR ≡ ∆MOB
∠RWA ≡ ∠BMO [∵ sum of three angles of a triangle are 180°]
∴ Criteria used here is angle sum property of triangles.

Question 6.
In the figure, ∠TMA ≡ ∠IAM and ∠TAM ≡ ∠IMA. P is the midpoint of MI and N is the midpoint of AI. Prove that ∆ PIN ~ ∆ ATM.

Solution:

Question 7.
In the figure, if ∠FEG = ∠1 then, prove that DG² = DE.DF.

Proof:

Question 8.
In the figure, ∠TEN ≡ ∠TON = 90° and TO = TE. Prove that ∠ORN ≡ ∠ERN

Solution:

Question 9.
In the figure, PQ ≡ TS, Q is the midpoint of PR, S is the midpoint TR and ∠POU ≡ ∠TSU. Prove that QU ≡ SU.

Proof:

Question 10.
In the figure ∆TOP ≡ ∆ARM . Explain why?
Solution:
In ∆TOP and ∆ARM
OP = RM given
∠TOP = ∠ARM = 90°
given ∠OTP = ∠RAM
given ∠OPT = ∠RMA Remaining angle, by angle sum property.
∴ By ASA criteria we can say that ∆TOP ≡ ∆ARM

Students can Download Maths Chapter 2 Measurements Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 2 Measurements Intext Questions

Try this Page No. 35

Question 1.
\(\frac{22}{7}\) and 3.14 are rational numbers. Is ‘π’ a rational number? Why?
Solution:
\(\frac{22}{7}\) and 3.14 are rational numbers n has non-terminating and non -repeating decimal expansion. So it is not a rational number. It is an irrational number.

Try this Page No. 38

Question 1.
The given circular figure is divided into six equal parts. Can we call the parts as sectors? Why?

Solution:
No, the equal parts are not sectors. Because a sector is a plane surface that is enclosed between two radii and the circular arc of the circle.
Here the boundaries are not radii.

Try these Page No. 38

Question 1.
Fill the central angle of the shaded sector (each circle is divided into equal sectors)

Try this Page No. 44

Question 1.
If the radius of a circle is doubled, what will the area of the new circle so formed?
Solution:
If r = 2r₁ ⇒ Area of the circle = πr² = π(2r₁)² = π4r₁² = 4πr₁²
Area = 4 × old area.

Try this Page No. 49

Question 1.
All the sides of a rhombus are equal. Is it a regular polygon?
Solution:
For a regular polygon all sides and all the angles must be equal. But in a rhombus all the
sides are equal. But all the angles are not equal
∴ It is not a regular polygon.

Try this Page No. 53

Question 1.
In the above example split the given mat as into two trapeziums and verify your answer.
Solution:
Area of the mat = Area of I trapezium + Area of II trapezium

∴ Cost per sq.feet = ₹ 20
Cost for 28 sq. feet = ₹ 20 × 28 = ₹ 560
∴ Total cost for the entire mat = ₹ 560
Both the answers are the same.

Try these Page No. 54

Question 1.
Show that the area of the unshaded regions in each of the squares of side ‘a’ units are the same in all the cases given below.

Solution:

Question 2.
If π = \(\frac{22}{7}\), show that the area of the unshaded part of a square of side ‘a’ units is approximately \(\frac{3}{7}\) a² sq. units and that of the shaded part is approximately \(\frac{4}{7}\) a² sq. units for the given figure.

Solution:

Try this Page No. 57

Question 1.
List out atleast three objects in each category which are in the shape of cube, cuboid,
cylinder, cone and sphere.
Solution:
(i) Cube – dice, building blocks, jewel box.
(ii) Cuboid – books, bricks, containers.
(iii) Cylinder – candles, electric tube, water pipe.
(iv) Cone – Funnel, cap, ice cream cone
(v) Sphere – ball, beads, lemon.

Try this Page No. 58

Question 1.
Tabulate the number of faces(F), vertices(V) and edges(E) for the following polyhedron. Also find F + V – E
Solution:

From the table F + V – E = 2 for all the solid shapes.

Try this Page No. 58

Question 1.
Find the area of the given nets.

Solution:
Area = 6 × Area of a square of side 6 cm
= 6 × (6 × 6) cm²
= 216 cm²
(ii) Area = Area of 2 rectangles of side (8 × 6) cm² + Area of 2 rectangles of side (8 × 4) cm² + Area of 2 rectangles of side (6 × 4) cm²
= (8 × 6) + (8 × 4) + (6 × 4)cm²
= 48 + 32 + 24 cm²
= 104 cm²

Students can Download Maths Chapter 3 Algebra Intext Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Intext Questions

Exercise 3.1

Recap Page No. 66 and 67

Question 1.
Write the numbers of terms in the following expressions.
(i) x + y + z – xyz
Solution:
4 terms

(ii) m²n²c
Solution:
1 term

(iii) a²b²c – ab²c² + a²bc² + 3abc
Solution:
4 terms

(iv) 8x² – 4xy + 7xy²
Solution:
3 terms

Question 2.
Identify the numerical co-efficient of each term in the following expressions.
Question 1.
2x² – 5xy + 6y² + 7x – 10y + 9
Solution:
Numerical co efficient in 2x² is 2
Numerical co efficient in -5xy is -5
Numerical co efficient in 6y² is 6
Numerical co efficient in 7x is 7
Numerical co efficient in -10y is – 10
Numerical co-efficient in 9 is 9

Question 2.
\(\frac{x}{3}+\frac{2 y}{5}-x y+7\)
Solution:

Numerical co efficient in -xy is -1
Numerical co efficient in 7 is 7

Question 3.
Pick out the like terms from the following.

Solution:

Question 4.
Add : 2x, 6y, 9x – 2y
Solution:
2x + 6y + 9x – 2y = 2x + 9x + 6y – 2y = (2 + 9)x + (6 – 2)y = 11x + 4y

Question 5.
Simplify : (5x³ y³ – 3x² y² + xy + 7) + (2xy + x³y³ – 5 + 2x²y²)
Solution:
(5x³y³ – 3x²y² + xy + 7) + (2xy + x³y³ – 5 + 2x²y²)
= 5x³y³ + x³y³ – 3x²y² + 2x²y² + xy + 2xy + 7 – 5
= (5 + 1)x³y³ + (-3 + 2)x²y² +(1 +2)xy + 2
= 6x³y³ – x²y² + 3xy + 2

Question 6.
The sides of a triangle are 2x – 5y + 9, 3y + 6x – 7 and -4x + y +10 . Find perimeter of the triangle.
Solution:
Perimeter of the triangle = Sum of three sides
= (2x – 5y + 9) + (3y + 6x – 7) + (-4x + y + 10)
= 2x – 5y + 9 + 3y + 6x – 7 – 4x + y + 10
= 2x + 6x – 4x – 5y + 3y + y + 9 – 7 + 10
= (2 + 6 – 4)x + (-5 + 3 + 1)y + (9 – 7 + 10)
= 4x – y + 12
∴ Perimeter of the triangle = 4x – y + 12 units.

Question 7.
Subtract -2mn from 6mn.
Solution:
6 mn – (-2mn) = 6mn + (+2mn) = (6 + 2) mn = 8mn

Question 8.
Subtract 6a² – 5ab + 3b² from 4a² – 3ab + b².
Solution:
(4a² – 3ab+ b²) – (6a²– 5ab + 3b²)
= (4a² – 6a²) + (- 3ab -(-5 ab)] + (b²– 3b²)
= (4 – 6) a² + [-3ab + (+ 5ab)] + (1 – 3) b²
= [4 + (- 6)] a² + (-3 + 5) ab + [1+ (-3)]b²
= -2a² + 2ab – 2b²

Question 9.
The length of a log is 3a + 4b – 2 and a piece (2a – b) is remove from it. What is the length of the remaining log?
Solution:
Length of the log = 3a + 4b – 2

Length of the piece removed = 2a – b
Remaining length of the log = (3a + 4b – 2) – (2a – b)
= (3a – 2a) + [4b – (-b)] – 2
= (3 – 2)a + (4 + 1)b – 2
= a + 5b – 2

Question 10.
A tin had ‘x’ litre oil. Another tin had (3x² + 6x – 5) litre of oil. The shopkeeper added (x + 7) litre more to the second tin. Later he sold (x² + 6) litres of oil from the second tin. How much oil was left In the second tin?
Solution:
Quantity of oil in the second tin = 3x² + 6x – 5 litres.
Quantity of oil added = x + 7 litres
∴ Total quantity of oil in the second tin
= (3x² + 6x – 5) + (x + 7) litres
= 3x² + (6x + x) + (-5 + 7)
= 3x² + (6 + 1)x + 2
= 3x² + 7x + 2 litres
Quantity of oil sold = x + 6 litres
∴ Quantity of oil left in the second tin = (3x² + 7x + 2) – (x² + 6)(3x² – x² ) + 7x + (2 – 6)
= (3 – 1)x² + 7x + (-4) = 2x² + 7x – 4
Quantity of oil left = 2x² + 7x – 4 litres

Try this Page No. 70

Question 1.
Every algebraic expression is a polynomial. Is this statement true? Why?
Solution:
No, This statement is not true. Because Polynomials contain only whole numbers as the powers of their variables. But an algebraic expression may contains fractions and negative powers on their variables.
Eg. 2y² + 5y^-1 – 3 is a an algebraic expression. But not a polynomial.

Try this Page No. 71

Question 2.
-(5y² + 2y – 6) Is this correct? If not, correct the mistake.
Solution:
Taking -(5y² + 2y – 6) = 5y² + [(-)(+) 2y] + [(-) × (-)6]
= -5y² – 2y + 6
≠ -5y² – 2y + 6
∴ Correct answer is -5y² + 2y – 6 = -(5y² + 2y + 6)

Try this Page No 71

(i) 3ab², -2a²b³
(ii) 4xy, 5y²x, (-x²)
(iii) 2m, -5n, -3p
Solution:
(i) (3ab²) × (-2a²b²) = (+) × (-) × (3 × 2) × (a × a²) × (b² × b³) = -6a³ b⁵

(ii) (4xy) × (5y²x) × (-x²)
= (+) × (+) × (-) × (4 × 5 × 1) × (x × x × x²) × (y × y²)
= -20x⁴y³

(iii) (2m) × (-5n) × (-3p) = (+) × (-) × (-) × (2 × 5 × 3) × m × n × p
= + 30mnp = 30 mnp

Try this Page No. 71

Question 1.
Why 3 + (4x – 7y) ≠ 12x – 21y?
Solution:
Addition and multiplication are different
3 + (4x – 7y) = 3 + 4x – 7y
We can add only like terms.

Try this Page No. 72

Question 1.
Which is corrcet? (3a)² is equal to
(i) 3a²
(ii) 3²a
(iii) 6a²
(iv) 9a²
Solution:
(3a) =3²a² = 9a²
(iv) 9a² is the correct answer

Try These Page No.72

Question 1.
Multiply
(i) (5x² + 7x – 3) by 4x²
Solution:
(5x² + 7x – 3) × 4x²
= 4x²(5x² + 7x – 3) Multiplication is commutative
= 4x² (5x² + 4x² (7x) + 4x² (-3)
= (4 × 5)(x² × x²) + (4 × 7)(x² × x) + (4 × -3)(x²)
= 20x⁴ + 28x³ – 12x²

(ii) (10x – 7y + 5z) by 6xyz
Solution:
(10x – 7y + 5z) by 6xyz
(10x – 7y + 5z) × 6xyz = 6xyz (10x – 7y + 5z) [∵ Multiplication is commutative]
= 6xy (10x) + 6xyz (-7y) + 6xyz (5z)
= (6 × 10)(x × x × y × z) + (6 × -7) + (x × y × y × z) + (6 × 5)(x × y × z × z)
= 60x²yz + (-42xy²z) + 30xyz²
= 60x²yz – 42x²z + 30xyz²

(iii) (ab + 3bc – 5ca) by – 3abc
Solution:
(ab + 3bc – 5ca) × (- 3abc) = (-3abc) (ab + 3bc – 5ca)
[∵ Multiplication is commutativel
= (-3abc) (ab) + (-3abc) (3bc) + (-3abc) (5ca)
= (-3)(a × a × b × b × c) + (- 3 × 3) + (a × b × b × c × c)
= -3a²b²c – 9ab²c² – 30a²bc²

Try these Page No. 74

Question 1.
Multiply
(i) (a – 5) and (a + 4)
Solution:
(a – 5) (a + 4) = a(a + 4) – 5 (a + 4)
= (a × a) + (a × 4) + (-5 × a) + (-5 × 4)
= a² + 4a – 5a – 20 = a² – a – 20

(ii) (a + b) and (a – b)
Solution:
(a + b) (a – b) = a(a – b) + b (a – b)
= (a × a) + (a × -b)+(b × a) + b(-b)
= a² – ab + ab – b² = a² – b²

(iii) (m⁴ + n⁴) and (m – n)
Solution:
(m⁴ + n⁴)(m – n) = m⁴(m – n) + n⁴(m – n)
= (m⁴ × m) + (m⁴ × (-n)) + (n⁴ × m (n⁴ × (-n))
= m⁵ – m⁴n + mn⁴ – n⁵

(iv) (2x + 3)(x – 4)
Solution:
(2x + 3)(x – 4) = 2x(x – 4) + 3(x – 4)
= (2x² × x) – (2x × 4) + (3 × x) – (3 × 4)
= 2x² – 8x + 3x – 12 = 2x² – 5x – 12

(v) (x – 5)(3x + 7)
Solution:
(x – 5)(3x + 7) = x(3x + 7) – 5(3x + 7)
= (x × 3x) + (x × 7) + (-5 × 3x) + (-5 × 7)
= 3x² + 7x – 15x – 35
= 3x² – 8x – 35

(vi) (x – 2)(6x – 3)
Solution:
(x – 2)(6x – 3) × (6x – 3) – 2(6x – 3)
= (x × 6x)+(x × (-3) × (2 × 6x) – (2 × 3)
= 6x² – 3x – 12x + 6
= 6x² – 15x + 6

Try this Page No. 74

Question 2.
3x² (x⁴ – 7x³ + 2), what is the highest power in the expression.
Solution:
3x²(x⁴ – 7x³ + 2) = (3x²) (x⁴) + 3x² (-7x³)+ (3x²)²
= 3x⁶ – 21x⁵ + 6x²
Highest power is 6 in x⁶.

Exercise 3.2

Try this Page No. 77

Question 1.
Are the following correct?
(i) \(\frac{x^{3}}{x^{8}}=x^{8-3}=x^{5}\)
(ii) \(\frac{10 m^{4}}{10 m^{4}}=0\)
(iii) When a monomial is divided by itself, we will get I?
Solution:

Try this Page No. 77

Question 1.
Divide

Solution:

Try this Page No. 78

Question 1.
Are the following divisions correct ?

Solution:

Try this Page No. 78

Question 1.

Solution:

Exercise 3.3

Try these Page No. 81

Question 1.
1. (p + 2)² = …….
2. (3 – a)² = …….
3. (6² – x²) = ………
4. (a + b)² – (a – b)² = …….
= a² + 2ab + b² – a² – 2ab – b²
= (1 – 1)a² + (2 + 2)ab + (+1 – 1 )b² = 4ab
5. (a + b)² = (a + b) × (a + b)
6. (m + n)( m – n) = m² – n²
7. (m + 7)² = m² + 14m + 49
8. (k² – 36) ≡ k² – 6² = (k + 6)(k – 6)
9. m² – 6m + 9 = (m – 3)²
10. (m – 10)(m + 5) = m² + (-10 + 5)m + (-10)(5) = m² – 5m – 50
Solution:

Try these page No. 83

Question 1.
Expand using appropriate identities
Question 1.
(3p + 2q)²
Solution:
(3p + 2q)²
Comparing (3p + 2q)² with (a + b)², we get a = 3p and b = 2q.
(a + b)² = a² + 2ab + b²
(3p + 2q)² = (3p)²+ 2(3p) (2q) + (2q)²
= 9p² + 12pq + 4q²

Question 2.
(105)²
Solution:
(105)² = (100 + 5)²
Comparing (100 + 5)² with (a + b)², we get a = 100 and b = 5.
(a + b)² = a² + 2ab + b²
(100 + 5)² = (100)² + 2(100)(5) + 5² = 1oooo + 1000 + 25
105² = 11,025

Question 3.
( 2x – 5d)²
Solution:
(2x – 5d)²
Comparing with (a – b)², we get a = 2x b = 5d.
(a – b)² = a² – 2ab + b²
(2x – 5d)² = (2x)² – 2(2x)(5d) + (5d)²
= 2x² – 20 xd + 5²d² = 4x² – 20 xd + 25d²

Question 4.
(98)²
Solution:
(98)² = (100 – 2)²
Comparing (100 – 2)² with (a – b)² we get
a = 100, b = 2
(a – b)² = a² – 2ab + b²
(100 – 2)² = 100² – 2(100)(2) + 2²
= 10000 – 400 + 4 = 9600 + 4 = 9604

Question 5.
(y – 5)(y + 5)
Solution:
(y – 5)(y + 5)
Comparing (y – 5) (y + 5) with (a – b) (a + b) we get
a = y; b = 5
(a – b)(a + b) = a² – b²
(y – 5)(y + 5) = y² – 5² = y² – 25

Question 6.
(3x)² – 5²
Solution:
(3x)² – 5²
Comparing (3x)² – 5² with a² – b² we have
a = 3x; b = 5
(a² – b²) = (a + b)(a – b)
(3x)² – 5² = (3x + 5)(3x – 5) = 3x(3x – 5) + 5(3x – 5)
= (3x) (3x) – (3x)(5) + 5(3x) – 5(5)
= 9x² – 15x + 15x – 25 = 9x² – 25

Question 7.
(2m + n)(2m +p)
Solution:
(2m + n) (2m + p)
Comparing (2m + n) (2m + p) with (x + a) (x + b) we have
x = 2n; a = n ;b = p
(x – a)(x + b) = x² + (a + b)x + ab
(2m +n) (2m +p) = (2m²) + (n +p)(2m) + (n) (p)
= 2²m² + n(2m) + p(2m) + np
= 4m² + 2mn + 2mp + np

Question 8.
203 × 197
Solution:
203 × 197 = (200 + 3)(200 – 3)
Comparing (a + b) (a – b) we have
a = 200, b = 3
(a + b)(a – b) = a² – b²
(200 + 3)(200 – 3) = 200² – 3²
203 × 197 = 40000 – 9
203 × 197 = 39991

Question 9.
Find the area of the square whose side is (x – 2)
Solution:
Side of a square = x – 2
∴ Area = Side × Side
= (x – 2) (x – 2) = x(x – 2) – 2(x – 2)
= x(x) + (x)(-2) + (-2)(x) + (-2)(-2)
= x – 2x – 2x + 4x² – 4x + 4

Question 10.
Find the area of the rectangle whose length and breadth are (y + 4) and (y – 3).
Solution:
Length of the rectangle = y+ 4
breadth of the rectangle = y – 3
Area of the rectangle = length × breadth
= (y + 4)(y – 3) = y² + (4 +(-3))y + (4)(-3)
= y² + y – 12

Try these Page No. 88

Question 1.
Expand :
Question 1.
(x + 4)³
Solution:
Comparing (x + 4)³ with (a + b)³, we have a = x and b = 4.
(a + b)³ = a³ + 3a²b + 3ab² + b³
(x + 4)³ = x³ + 3x²(4) + 3(x)(4)² + 4³
= x³ + 12x² + 48x + 64

Question 2.
( y – 2)²
Solution:
Comparing (y – 2) with (a – b)³ we have a = y b = z
(a – b)³ = a³ – 3a²b + 3ab² – b³
(y – 2)² = y³ – 3y(2) + 3y(2)² + 2³
= y³ – 6y² + 12y + 8

Question 3.
(x + 1)(x + 3)(x + 5)
Solution:
Comparing (x + 1) (x + 3) (x + 5) with (x + a) (x + b) (x + c) we have
a = 1
b = 3
and c = 5

Exercise 3.4

Try These Page No.92

Question 1.
Factorize the following:
Question 1.
3y + 6
Solution:
3y + 6
3y + 6 = 3 × y + 2 × 3
Taking out the common factor 3 from each term we get 3 (y + 2)
∴ 3y + 6 = 3(y + 2)

Question 2.
10x² + 15y
Solution:
10x² + 15y²
10x² + 15y² = (2 × 5 × x × x) + (3 × 5 × y × y)
Taking out the common factor 5 we have
10x² + 15y² = 5(2x² + 3y²)

Question 3.
7m(m – 5) + 1(5 – m)
Solution:
7m(m – 5) + 1(5 – m)
7m(m – 5) + 1(5 – m) = 7m(m – 5) + (-1)(-5 + m)
= 7m(m – 5) – 1 (m – 5)
Taking out the common binomial factor (m – 5) = (m – 5)(7m – 1)

Question 4.
64 – x²
Solution:
64 – x²
64 – x² = 8² – x²
This is of the form a² – b²
Comparing with a² – b² we have a = 8, b = x
a² – b² = (a + b)(a – b)
64 – x² = (8 + x)(8 – x)

Students can Download Maths Chapter 2 Measurements Additional Questions and Answers, Notes Pdf, Samacheer Kalvi 8th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 2 Measurements Additional Questions

Additional Questions And Answers

Exercise 2.1

Very Short Answers [2 Marks]

Question 1.
Find the length of arc if the perimeter of sector is 45 cm and radius is 10 cm.
Solution:
Given Radius of the sector = 10 cm
Perimeter of the sector P = 45 cm
l + 2r = 45
l + 2(10) = 45
l + 20 = 45
l = 45 – 20
l = 25 cm
Length of the arc l = 25 cm

Question 2.
Find the radius of sector whose perimeter and length of arc are 30 cm and 16 cm respectively.
Solution:
Given length of the arc = 16 cm
Perimeter of the arc = 30 cm
l + 2r = 30
16 + 2 r = 30
2 r = 30 – 16
2 r = 14
r = \(\frac{14}{2}\)
r = 7 cm
Radius of the sector = 7 cm

Question 3.
Find the length of arc whose radius is 7 cm and central angle 90°.
Solution:
Here θ = 90°; radius r = 7cm
Length of the arc = \(\frac{\theta^{\circ}}{360^{\circ}}\) × 2πr units
= \(\frac{90^{\circ}}{360^{\circ}}\) × 2 × \(\frac{22}{7}\) × 7 = 11 cm
∴ Length of the arc = 11 cm

Short Answers [3 Marks]

Question 1.
Find the length arc whose radius is 42 cm and central angle is 60°.
Solution:
Length of the arc = \(\frac{\theta^{\circ}}{360^{\circ}}\) × 2πr units
Given central angle 0 = 60°
Radius of the sector r = 42 cm
l = \(\frac{60^{\circ}}{360^{\circ}}\) × 2 × \(\frac{22}{7}\) × 42 = 44 cm
∴ Length of the arc = 44 cm

Question 2.
Find the length of the arc whose radius is 10.5 cm and central angle is 36°.
Solution:

∴ Length of the arc = 6.6 cm

Long Answers [5 Marks]

Question 1.
A sector is cut from a circle of radius 21 cm. The angle of the sector is 150°. Find the length of its arc and area of the sector.
Solution:
Radius of the sector = 21 cm
Length of the arc

∴ Length of the arc = 55 cm
Area of the sector = 577.5 cm²

Question 2.
Find the perimeter of sector whose area is 324 sq. cm and radius is 27 cm.
Solution:
Radius of the sector = 27 cm
Area of the sector =324 cm²

Perimeter of the sector P = (l + 2r) units = 24 + 2(27) cm = (24 + 54) cm = 78 cm

Exercise 2.2

Question 1.
PQRS is a diameter of a circle of radius 6 cm. The lengths PQ, QR and RS are equal semi-circles drawn on PQ and Question as diameters. Find the p perimeter and area of the shaded region.

Solution:
PS = Diameter of a circle of radius 6 cm = 12 cm
PQ = QR = RS = \(\frac{12}{3}\) = 4 cm Question = QR + RS = 4 + 4 = 8 cm
∴ Perimeter of the shaded part = Arc length of semi-circle of radius 6 cm + Arc length of semicircle of radius 4 cm + Arc length of semi-circle of radius 2 cm.
= (π × 6) + (π × 4) + (π × 2) cm
P = 12 π cm
Area required = Area of semicircle with PS as diameter + Area of semi circle with PQ as diameter – Area of semi-circle with Question as diameter.

Question 2.
In the figure AOBCA represents a quadrant of a circle of radius 3.5cm with center ‘O’ calculate the area of the shaded portion (π = \(\frac{22}{7}\))

Solution:

∴ Area of shaded region = Area of the quadrant – Area of triangle
= 9.625 – 3.5 cm² = 6.125 cm²

Question 3.
Find the area of the shaded region in the figure

Solution:
Radius of the big semicircle = 14 cm

∴ Required area = 308 + 154 cm² = 462 cm²

Exercise 2.3

Very Short Answers [2 Marks]

Question 1.
What is the least number of planes that can enclose a solid? What is the name of the solid?
Solution:
Least number of planes = 4, the solid is tetrahedron.

Question 2.
Can a polyhedron have for its faces = 12 edges = 16 and vertices = 6.
Solution:
Verifying Euler’s formula
F + V – E = 12 + 6 – 16 = 18 – 16 = 2
Yes, the polyhedron can have F = 12, E = 16 and V = 6

Short Answers [3 Marks]

Question 1.
Verify Euler’s formula for a pyramid.
Solution:
A pyramid has faces = 5, Vertices = 5, Edges = 8
By Euler’s formula F + V – E = 5 + 5 – 8 = 10 – 8 = 2

Question 2.
Verify Eulers formula for a triangular prism.
Solution:
For a triangular prism
Faces = 5, Edges = 9, Vertices = 6
By Euler’s formula F + V – E = 5 + 6 – 9 = 11 – 9 = 2

Long Answers [5 Marks]

Question 1.
(a) Dice are cubes where the numbers on the opposite faces must total 7. Is the following a die.

(b) The following shows a net with areas of faces. What can be the shape?

Solution:
(a) 2 + 5 = 6 + 1 = 3 + 4 = 7
∴ It can be a die.
(b) It is a cuboid