Class 8

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.4

Students can Download Maths Chapter 3 Algebra Ex 3.4 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.
Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.2

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.4

Question 1.
Factorise the following by taking out the common factor
(i) 18xy – 12yz
(ii) 9x5y3 + 6x3y2 – 18x2y
(iii) x(b – 2c) + y(b – 2c)
(iv) (ax + ay) + (bx + by)
(v) 2x2(4x – 1) – 4x + 1
(vi) 3y(x – 2)2 – 2(2 – x)
(vii) 6xy – 4y2 + 12xy – 2yzx
(viii) a3 – 3a2 + a – 3
(ix) 3y3 – 48y
(x) ab2 – bc2 – ab + c2
Solution:
(i) 18xy – 12yz = (2 × 3 × 3 × y × x) – (2 × 2 × 3 × y × z)
Taking out the common factors 2, 3, y, we get
= 2 × 3 × y (3x – 2z) = 6y (3x- 2z)

i) 9x5y3 + 6x3y2 – 18x2y = (3 × 3 × x2 × x3 × y × y2) + (2 × 3 × x2 × x × y × y)
Taking out the common factors 3, x2, y, we get
= 3 × x2 × y (3x3 y2 + 2xy – 6)
= 3x2y (3x3 y2 + 2xy – 6)

(iii) x(b – 2c) + y(b – 2c)
Taking out the binomial factor (b – 2c) from each term, we have
= (b – 2c)(x + y)

(iv) (ax + ay) + (bx + by)
Taking at ‘a’ from the first term and ‘b’ from the second term we have
(ax + ay) + (bx + by) = a (x + y) + b (x + y)
Now taking out the binomial factor (x + y) from each term
= (x + y)(a + b)

(v) 2x2(4x – 1) – 4x + 1
Taking out -1 from last two terms
2x2 (4x – 1) – 4x + 1 = 2x2 (4x – 1) – 1 (4x- 1)
Taking out the binomial factor 4x – 1, we get
= (4x – 1)(2x2 – 1)

(vi) 3y(x – 2)2 – 2(2 – x)
3y(x – 2)2 – 2(2 – x) = 3y(x – 2)(x – 2)-2(-1) (x – 2) [∵ Taking out -1 from 2 – x]
= 3y (x – 2) (x – 2) + 2 (x – 2)
Taking out the binomial factor x – 2 from each term, we get
= (x – 2) [3y (x – 2) + 2]

(vii) 6xy – 4y2 + 12xy – 2yzx
= 6xy + 12xy – 4y2 – 2yzx [∵ Addition is commutative]
= (6 × x × y) + (2 × 6 × x × y) + (-1) (2) (2) y + y) + ((-1) (2) (y) (z) (x))
Taking out 6 × x × y from first two terms and (-1) × 2 × y from last two terms we get
= 6 × x × y(1 + 2) + (-1)(2)y[2y + zx]
= 6xy (3) -2y(2y + zx)
= (2 × 3 × 3 × x × y) – 2xy (2y + zx)
Taking out 2y from two terms
= 2y (9x – (2y + zx)) = 2y (9x – 2y – xz)

(viii) a2 – 3a2 + a – 3 = a2 (a – 3) + 1 (a – 3) [∵ Grouping the terms suitably]
= (a – 3) (a2 + 1)

(ix) 3y2 – 48y = 3 × y × y2 – 3 × 16 × y
Taking out 3 × y = 3y (y2 – 16) = 3y (y2 – 42)
Comparing y2 – 42 with a2 – b2
a = y, b = 4
a2 – b2 = (a + b) (a – b)
y2 – 42 = (y + 4) (y – 4)
∴ 3y (y2 – 16) = 3y (y + 4) (y – 4)

(x) ab2 – bc2 – ab + c2
Grouping suitably
ab2 – bc2 – ab + c2 = b ((ab – c2) – 1(ab – c2)
Taking out the binomial factor ab – c2 = (ab – c2) (b – 1)

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.2

Question 2.
Factorise the following expressions
(i) x2 + 14x + 49
(ii) y2 – 10y + 25
(iii) c2 – 4c – 12
(iv) m2 + m – 72
(v) 4x2 – 8x + 3
Solution:
x2 + 14x + 49 = x2 + 14x + 72
Comparing with a2 + 2ab + b2 = (a + b)2 we have a = x and b = 7
⇒ x2 + 2(x) (7) + 72 = (x + 7)2
∴ x2 + 14x + 49 = (x + 7)2

(ii) y2 – 10y + 25 = y2 – 10y + 52
Comparing with a2 – 2ab + b2 = (a – b)2 we get a = y ; b = 5
⇒ y2 – 2(y) (5) + 52 = (y – 5)2
∴ y2 – 10y + 25 = (y – 5)2

(iii) c2 – 4c – 12
This is of the form ax2 + bx + c
Where a = 1,b = – 4 c = – 12, x = c
Now the product ac = 1 × – 12 = – 12 and the sum b = -4
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Ex 3.4 1

(iv) m2 + m – 72
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Ex 3.4 2
This is of the form ax2 + bx + c
where a = 1, b = 1, c = -12
Product a × c = 1 × -72 = -72
Sum b = 1
The middle term m can be written as 9m – 8m
m2 + m – 72 = m2 + 9m – 8m – 12
= m (m + 9) – 8 (m + 9)
Taking out (m + 9)
= (m + 9) (m – 8)
∴ m2 + m – 72 = (m + 9) (m – 8)

(v) 4x2 – 8x + 3
This is of the form ax2 + bx + c with a = 4 b = -8 c = 3
Product ac = 4 × 3 = 12
Sum b = -8
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Ex 3.4 3

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.2

Question 3.
Factorize the following expressions using a3 + b3 = (a + b)(a2 – ab + b2) identity
(i) h3 + k2
(ii) 2a3 + 16
(iii) x3y3 + 27
(iv) 64m3 + n3
(v) r4 + 27p3r
Solution:
(i) h3 + k3
Comparing h3 + k3 with a3 + b3 = (a + b)(a2 – ab + b2) we have a = h, b = k
∴ h3 + k3 = (h + k)(h2 – hk + k2)

(ii) 2a2 + 16
(2 × a3) + (2 × 8) = 2(a3 + 8) = 2 (a3 + 23)
∴ 2a3 + 16 = 2 (a3 + 23)
Comparing with a3 + b3 we have a = a and b = 2
a3 + b3 = (a + b)(a2 – ab + b2)
2(a3 + 23) = 2[(a + 2) (a2 – (a) (2) + 22)] = 2[(a + 2) (a2 – 2a + 4)]
2a3 +16 = 2 (a + 2) (a2 – 2a + 4)

(iii) x3 y3 + 27 = (xy)3 + 33
Comparing with a3 + b3 we have a = xy ;b = 3
a3 + b3 = (a + b) (a2 – ab + b2)
(xy)3 + 33 = (xy + 3) ((xy)2 – (xy) (3) + 32) = (xy + 3) (x2y2 – 3xy + 9)
∴ x3y3 + 27 = (xy + 3)(x2y2 – 3xy + 9)

(iv) 64m3 + n3 = (43m3) + n3
= (4m)3 + n3
Comparing this with a3 + b3 we have a = 4m; b = n
a3 + b3 = (a + b) (a2 – ab + b2)
(4m)3 + n3 = (4m + n) [(4m)2 – (4m) (n) + n2]
= (4m + n) [42m2 – 4mn + n2]
= (4m + n) [ 16m2 – 4mn + n2]
64m3 + n3 = (4m + n) (16m2 – 4mn + n2)

(v) r3 + 27p3r = r (r3 + 27p3) = r [r3 + 33p3]
Comparing r3 + (3p)3 with a3 + b3 we have a = r; b = 3p
a3 + b3 = (a + b)(a2 – ab + b2)
r[r3 + (3p)3] = r[(r + 3p)(r2 – r(3p) + (3p)2)]
= r[(r+ 3p) (r2 – 3rp + 32p2]
= r(r + 3p) (r2 – 3rp + 9p2)
r4 + 27p3r = r(r + 3p) (r2 – 3rp + 9p2)

Question 4.
Factorize the following expressions using a3 – b3 = (a – b)(a2 + ab + b2) identity
(i) y3 – 27
(ii) 3b3 – 192c3
(iii) -16y3 + 2x3
(iv) x3 y3 – 73
(v) c3 – 27b3 a3
Solution:
(i) y3 – 27 = y3 – 33
Comparing this with a3 – b3 , we have a = y and b = 3
a3 – b3 = (a + b) (a2 + ab + b2 )
y3 – 33 = (y – 3)(y2 + (y) (3) + 32 ) = (y – 3) (y2 + 3y + 9)
y3 – 27 = (y – 3) (y2 + 3y + 9)

(ii) 3b3 + 192c3 = (3 × b3) – (3 × 4 × 4 × 4 × c3) = 3(b3 – 43 c3)
= 3(b3 – (4c)3 )
Comparing b3 – (4c)3 with a3 – b3 we have a = b and b = 4c
a3 – b3 = (a – b) (a2 + ab + b2)
3(b3 – (4c)3) = 3[(b – 4c)(b2 + (b)(4c) + (4c)2)]
= 3[(b – 4c)(b2 + 4bc + 42 c2)]
3b3 – 192c3 = 3 [(b – 4c) (b2 + 4bc + 16c2)]

(iii) -16y3 + 2x3 = 2x3 – 16y3 [∵ Addition is commatative]
= 2(x3 – 8y3) = 2(x3 – 23y3)
= 2(x3 – (2y)3)
Comparing x3 – (2y)3 with a3 – b3 we have a = x and b = 2y
a3 – b3 = (a – b)(a2 + ab + b2)
2[x3 – (2y)3] = 2[(x – 2y) (x2 + (x) (2y) + (2y)2)]
= 2[(x – 2y) (x2 + 2xy + 22 y2)]
-16y3 + 2x3 = 2[(x – 2y)(x2 + 2xy + 4y2)]

(iv) x3y3 – 73 = (xy)3 – 73
Comparing with a3 – b3 we have a = xy and b = 7
a3 – b3 = (a – b)(a2 + ab + b2)
(xy)3 – 73 = (xy – 7) ((xy)2 + (xy) (7) + 72)
x3y3 – 73 = (xy – 7) (x2y2 + 7xy + 49)

(v) c3 – 27 b3 a3 = c3 – 33b3 a3 = c3 – (3ba)3
Comparing this with a3 – b3 we have a = x and b = 3ba
a3 – b3 = (a – b)(a2 + ab + b2)
∴ c3 – (3ba)3 = (c – 3ba) (c2 + (c) (3ba) + (3ba)2)
= (c – 3ba) (c2 + 3bac + 32 b2a2)
c3 – 27b3a3 = (c – 3ab) (c2 + 3bac + 9a2b2)

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.4 Read More »

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

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

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.
Samacheer Kalvi 8th Maths Term 1 Chapter 4 Geometry Additional Questions 1
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.
Samacheer Kalvi 8th Maths Term 1 Chapter 4 Geometry Additional Questions 40
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

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

Short Answers [3 Marks]

Question 1.
In the figure with respect to ABEP and CPD prove that BP × PD = EP × PC.
Samacheer Kalvi 8th Maths Term 1 Chapter 4 Geometry Additional Questions 51
Solution:
Proof:
In ∆EPB and ∆DPC
∠PEB = ∠PDC = 90° [given]
∠EPB = ∠DPC [Vertically opposite angles]
∠EPB = ∠PCD [∵ Remaining angles]
Thus,
Samacheer Kalvi 8th Maths Term 1 Chapter 4 Geometry Additional Questions 50

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:
Samacheer Kalvi 8th Maths Term 1 Chapter 4 Geometry Additional Questions 52
AB = AP + PB
= 3 + 6 cm = 9 cm
AC = AQ + QC = 510 cm = 15
Samacheer Kalvi 8th Maths Term 1 Chapter 4 Geometry Additional Questions 53
Samacheer Kalvi 8th Maths Term 1 Chapter 4 Geometry Additional Questions 54

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 4 Geometry Additional Questions Read More »

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 4 Geometry Ex 4.3

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.
Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 4 Geometry Ex 4.3

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
Samacheer Kalvi 8th Maths Term 1 Chapter 4 Geometry Ex 4.3 1
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:
Samacheer Kalvi 8th Maths Term 1 Chapter 4 Geometry Ex 4.3 2

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.
Samacheer Kalvi 8th Maths Term 1 Chapter 4 Geometry Ex 4.3 80
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:
Samacheer Kalvi 8th Maths Term 1 Chapter 4 Geometry Ex 4.3 81

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 4 Geometry Ex 4.3

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
Samacheer Kalvi 8th Maths Term 1 Chapter 4 Geometry Ex 4.3 3
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:
Samacheer Kalvi 8th Maths Term 1 Chapter 4 Geometry Ex 4.3 5

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
Samacheer Kalvi 8th Maths Term 1 Chapter 4 Geometry Ex 4.3 6
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:
Samacheer Kalvi 8th Maths Term 1 Chapter 4 Geometry Ex 4.3 10

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 4 Geometry Ex 4.3

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°
Samacheer Kalvi 8th Maths Term 1 Chapter 4 Geometry Ex 4.3 11
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:
Samacheer Kalvi 8th Maths Term 1 Chapter 4 Geometry Ex 4.3 16Area 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°
Samacheer Kalvi 8th Maths Term 1 Chapter 4 Geometry Ex 4.3 17
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:
Samacheer Kalvi 8th Maths Term 1 Chapter 4 Geometry Ex 4.3 20
∴ Area of the quadrilateral = 22.87 cm2

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 4 Geometry Ex 4.3

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°
Samacheer Kalvi 8th Maths Term 1 Chapter 4 Geometry Ex 4.3 21
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:
Samacheer Kalvi 8th Maths Term 1 Chapter 4 Geometry Ex 4.3 22
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°
Samacheer Kalvi 8th Maths Term 1 Chapter 4 Geometry Ex 4.3 23
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:
Samacheer Kalvi 8th Maths Term 1 Chapter 4 Geometry Ex 4.3 25
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°
Samacheer Kalvi 8th Maths Term 1 Chapter 4 Geometry Ex 4.3 26
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:
Samacheer Kalvi 8th Maths Term 1 Chapter 4 Geometry Ex 4.3 27

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 4 Geometry Ex 4.3

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°
Samacheer Kalvi 8th Maths Term 1 Chapter 4 Geometry Ex 4.3 28
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:
Samacheer Kalvi 8th Maths Term 1 Chapter 4 Geometry Ex 4.3 30
Area of the quadrilateral YOGA = 28.08 cm2

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 4 Geometry Ex 4.3 Read More »

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

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

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.

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

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?
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Intext Questions 1
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.

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

Try these Page No. 38

Question 1.
Fill the central angle of the shaded sector (each circle is divided into equal sectors)
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Intext Questions 50

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.

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

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.

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

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
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Intext Questions 55
∴ 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.

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

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.
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Intext Questions 51
Solution:
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Intext Questions 52
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Intext Questions 53

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

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.
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Intext Questions 85
Solution:
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Intext Questions 54
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Intext Questions 59
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Intext Questions 89

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.

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

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:
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Intext Questions 60
From the table F + V – E = 2 for all the solid shapes.

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

Try this Page No. 58

Question 1.
Find the area of the given nets.
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Intext Questions 62
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

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Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.3

Students can Download Maths Chapter 3 Algebra Ex 3.3 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.

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.2

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.3

Question 1.
Expand
(i) (3m + 5)2
(ii) (5p – 1)2
(iii) (2n – 1)(2n + 3)
(iv) 4p2 – 25q2
Solution:
(i) (3m + 5)2
Comparing (3m + 5)2 with (a + b)2 we have a = 3m and b = 5
(a + b)2 = a2 + 2 ab + b2
(3m + 5)2 = (3m)2 + 2 (3m) (5) + 52
= 32m2 + 30m + 25 = 9m2 + 30m +25

(ii) (5p – 1)2
Comparing (5p – 1)2 with (a – b)2 we have a = 5p and b = 1
(a – b)2 = a2 – 2ab + b2
(5p – 1)2 = (5p)2 – 2 (5p) (1) + 12
= 52p2 – 10p + 1 = 25p2 – 10p + 1

(iii) (2n – 1)(2n + 3)
Comparing (2n – 1) (2n + 3) with (x + a) (x + b) we have a = -1; b = 3
(x + a) (x + b) = x2 + (a + b)x + ab
(2n +(- 1)) (2n + 3) = (2n)2 + (-1 + 3)2n + (-1) (3)
= 22n2 + 2 (2n) – 3 = 4n2 + 4n – 3

(iv) 4p2 – 25q2 = (2p)2 – (5q)2
Comparing (2p)2 – (5q)2 with a2 – b2 we have a = 2p and b = 5q
(a2 – b2) = (a + b)(a – b) = (2p + 5q) (2p – 5q)

Question 2.
Expand
(i) (3 + m)3
(ii) (2a + 5)3
(iii) (3p + 4q)3
(iv) (52)3
(v) (104)3
Solution:
(i) (3 + m)3
Comparing (3 + m)3 with (a + b)3 we have a = 3; b = m
(a + b)3 = a2 + 3a2b + 3 ab2 + b3
(3 + m)3 = 33 + 3(3)2 (m) + 3 (3) m2 + m3
= 27 + 27m + 9m2 + m3 = m3 + 9 m2 + 27m + 27

(ii) (2a + 5)3
Comparing (2a + 5)3 with (a + b)3 we have a = 2a, b = 5
(a + b)3 = a3 + 3a2b + 3ab2 + b3 = (2a)3 + 3(2a)2 5 + 3 (2a) 52 + 53
= 23a3 + 3(22a2) 5 + 6a (25) + 125
= 8a3+ 60a2 + 150a + 125

(iii) (3p + 4q)3
Comparing (3p + 4q)3 with (a + b)3 we have a = 3p and b = 4q
(a + b) 3 = a3 + 3a2b + 3ab2 + b3
(3p + 4q)3 = (3p)3 + 3(3p)2 (4q) + 3(3p)(4q)2 + (4q)3
= 33p3 +3 (9p2) (4q) + 9p (16q2) + 43q3
= 27p3 + 108p2q + 144pq2 + 64q3

(iv) (52)3 = (50 + 2)3
Comparing (50 + 2)3 with (a + b)3 we have a = 50 and b = 2
(a + b)3 = a3 + 3 a2b + 3 ab2 + b3
(50 + 2)3 = 503 + 3 (50)22 + 3 (50)(2)2 + 23
523 = 125000 + 6(2,500) + 150(4) + 8
= 1,25,000 + 15,000 + 600 + 8
523 = 1,40,608

(v) (104)3 = (100 + 4)3
Comparing (100 + 4)3 with (a + b)3 we have a = 100 and b = 4
(a + b)3 = a3 + 3 a2b + 3 ab2 + b3
(100 + 4)3 = (100)3 + 3 (100)2 (4) + 3 (100) (4)2 + (4)3
= 10,00,000 + 3(10000) 4 + 300 (16) + 64
= 10,00,000 + 1,20,000 + 4,800 + 64 = 11,24,864

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.2

Question 3.
Expand
(i) (5 – x)3
(ii) (2x – 4y)3
(iii) (ab – c)3
(iv) (48)3
(v) (97xy)3
Solution:
(i) (5 – x)3
Comparing (5 – x)3 with (a – b)3 we have a = 5 and b = x
(a – b)3 = a3 – 3a2b + 3ab2 – b3
(5 – x)3 = 53 – 3 (5)2 (x) + 3(5)(x2) – x3
= 125 – 3(25)(x) + 15x2 – x3 = 125

(ii) (2x – 4y)3
Comparing (2x – 4y)3 with (a – b)3 we have a = 2x and b = 4y
(a – b)3 = a3 – 3a2b + 3ab3 – b3
(2x – 4y)3 = (2x)3 – 3(2x)2 (4y) + 3(2x) (4y)2 – (4y)3
= 23x3 – 3(22x2) (4y) + 3(2x) (42y2) – (43y3)
= 8x3 – 48x2y + 96xy2 – 64y3

(iii) (ab – c)3
Comparing (ab – c)3 with (a – b)3 we have a = ab and b = c
(a – b)3 = a3 – 3a2b + 3ab2 – b3
(ab – c)3 = (ab)3 – 3 (ab)2 c + 3 ab (c)2 – c3
= a3b3 – 3(a2b2) c + 3abc2 – c3
= a3b3 – 3a2b2 c + 3abc2 – c3

(iv) (48)3 = (50 – 2)3
Comparing (50 – 2)3 with (a – b)3 we have a = 50 and b = 2
(a – b)3 = a3 – 3a2b + 3ab2 – b3
(50 – 2)3 = (50)3 – 3(50)2(2) + 3 (50)(2)2 – 23
= 1,25,000 – 15000 + 600 – 8 = 1,10,000 + 592
= 1,10,592

(v) (97xy)3
= 973 x3 y3 = (100 – 3)3 x3y3
Comparing (100 – 3)3 with (a – b)3 we have a = 100, b = 3
(a – b)3 = a3 – 3a2b + 3ab2 – b3
(100 – 3)3 = (100)3 – 3(100)2 (3) + 3 (100)(3)2 – 33
973 = 10,00,000 – 90000 + 2700 – 27
973 = 910000 + 2673
973 = 912673
97x3y3 = 912673x3y3

Question 4.
Simplify (i) (5y + 1)(5y + 2)(5y + 3)
(ii) (p – 2)(p + 1)(p – 4)
Solution:
(i) (5y + 1) (5y + 2) (5y + 3)
Comparing (5y + 1) (5y + 2) (5y + 3) with (x + a) (x + b) (x + c) we have x = 5y ; a = 1; b = 2 and c = 3.
(x + a) (x + b) (x + c) = x3 + (a + b + c) x2 (ab + bc + ca) x + abc
= (5y)3 + (1 + 2 + 3) (5y)2 + [(1) (2) + (2) (3) + (3) (1)] 5y + (1)(2) (3)
= 53y3 + 6(52y2) + (2 + 6 + 3)5y + 6
= 1253 + 150y2 + 55y + 6

(ii) (p – 2)(p + 1)(p – 4) = (p + (-2))0 +1)(p + (-4))
Comparing (p – 2) (p + 1) (p – 4) with (x + a) (x + b) (x + c) we have x = p ; a = -2; b = 1 ; c = -4.
(x + a) (x + b) (x + c) = x3 + (a + b + c) x2 + (ab + be + ca) x + abc
= p3 + (-2 + 1 + (-4))p2 + ((-2) (1) + (1) (-4) (-4) (-2)p + (-2) (1) (-4)
= p3 + (-5 )p2 + (-2 + (-4) + 8)p + 8
= p3 – 5p2 + 2p + 8

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.2

Question 5.
Find the volume of the cube whose side is (x + 1) cm.
Solution:
Given side of the cube = (x + 1) cm
Volume of the cube = (side)3 cubic units = (x + 1)3 cm3
We have (a + b)3 = (a3 + 3a2b + 3ab2 + b3) cm3
(x + 1)3 = (x3 + 3x2 (1) + 3x (1)2 + 13) cm3
Volume = (x3 + 3x2 + 3x + 1) cm3

Question 6.
Find the volume of the cuboid whose dimensions are (x + 2),(x – 1) and (x – 3).
Solution:
Given the dimensions of the cuboid as (x + 2), (x – 1) and (x – 3)
∴ Volume of the cuboid = (l × b × h) units3
= (x + 2) (x – 1) (x – 3) units3
We have (x + a) (x + b) (x + c) = x3 + (a + b + c) x2 + (ab + bc+ ca)x + abc
∴ (x + 2)(x – 1) (x – 3) = x3 + (2 – 1 – 3)x2 + (2 (-1) + (-1) (-3) + (-3) (2)) x + (2)(-1) (-3)
x3 – 2x2 + (-2 + 3 – 6)x + 6
Volume = x3 – 2x2 – 5x + 6 units3

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Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.1

Students can Download Maths Chapter 2 Measurements Ex 2.1 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.
Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.1

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.1

Question 1.
Fill in the blanks:
(i) The ratio between the circumference and diameter of any circle is _________.
(ii) A line segment which joins any two points on a circle is a ______.
(iii) The longest chord of a circle is _______.
(iv) The radius of a circle of diameter 24 cm is ______.
(v) A part of circumference of a circle is called as _____.
Solution:
(i) π
(ii) chord
(iii) diameter
(iv) 12 cm
(v) an arc

Question 2.
Match the following
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Ex 2.1 1
Solution:
(i) 3
(ii) 4
(iii) 5
(iv) 2
(v) 1

Question 3.
Find the central angle of the shaded sectors (each circle is divided into equal sectors)
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Ex 2.1 17
Solution:
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Ex 2.1 3

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.1

Question 4.
For the sectors with given measures, find the length of the arc, area and perimeter, (π = 3.14)
(i) central angle 45°, r = 16 cm
(ii) central angle 120°, d = 12.6 cm
(iii) central angle 60°, r = 36 cm
(iv) central angle 72°, d = 10 cm
Solution:
(i) Central angle 45°, r = 16 cm
Length of the arc l = \(\frac{\theta^{\circ}}{360^{\circ}}\) × 2πr units
l = \(\frac{45^{\circ}}{360^{\circ}}\) × 2 × 3.14 × 16 cm
l = \(\frac{1}{8}\) × 2 × 3.14 × 16 cm
l = 12.56 cm
Area of the sector = \(\frac{\theta^{\circ}}{360^{\circ}}\) × πr2
A = \(\frac{45^{\circ}}{360^{\circ}}\) × 3.14 × 16 × 16
A = 100.48 cm2
Perimeter of the sector
P = l + 2r units
P = 12.56 + 2(16) cm
P = 44.56 cm

(ii) Central angle 120°, d = 12.6 cm
∴ r = \(\frac{12.6}{2}\) cm
r = 6.3 cm
Length of the arc
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Ex 2.1 4
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Ex 2.1 5
Area of the sector missing
Perimeter of the sector
P = l + 2r units
P = 6.28 + 2(5) cm
P = 6.28 + 10 cm
P = 16.28 cm

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.1

Question 5.
From the measures given below, find the area of the sectors.
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Ex 2.1 18
Solution:
(i) Area of the sector
A = \(\frac{l r}{2}\) sq. units
l = 48 m
r = 10 m
= \(\frac{48 \times 10}{2}\) m2
= 24 × 10 m2
= 240 m2
Area of the sector = 240 m2

(ii) length of the arc l = 12.5 cm
Radius r = 6 cm
Area of the sector
A = \(\frac{l r}{2}\) sq. units
= \(\frac{12.5 \times 6}{2}\)
= 12.5 × 3 cm2 cm2
= 37.5 cm2
Area of the sector = 37.5 cm2

(iii) length of the arc l = 50 cm
Radius r = 13.5 cm
Area of the sector
A = \(\frac{l r}{2}\) sq. units
= \(\frac{50 \times 13.5}{2}\)
= 25 × 13.5 cm2 cm2
= 337.5 cm2
Area of the sector = 337.5 cm2

Question 6.
Find the central angle of each of the sectors whose measures are given below (π = \(\frac{22}{7}\))
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Ex 2.1 19
Solution:
(i) Radius of the sector = 21 cm
Area of the sector = 462 cm2
\(\frac{l r}{2}\) = 462
\(\frac{l \times 21}{2}\) = 462
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Ex 2.1 8
∴ Central angle of the sector = 120°

(ii) Radius of the sector = 8.4 cm
Area of the sector = 18.48 cm2
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Ex 2.1 9

(iii) Radius of the sector = 35 m
Length of the arc l = 44 m
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Ex 2.1 10
Question 7.
Answer the following questions:
(i) A circle of radius 120 m is divided into 8 equal sectors. Find the length of the arc of each of the sectors.
(ii) A circle of radius 70 cm is divided into 5 equal sectors. Find the area of each of the sectors.
Solution:
(i) Radius of the circle r = 120 m
Number of equal sectors = 8
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Ex 2.1 11

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.1

(ii) Radius of the sector r = 70 cm
Number of equal sectors = 5
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Ex 2.1 12
Note: We can solve this problem using A = \(\frac{1}{n}\) πr2 sq. units also.

Question 8.
Find the area of a sector whose length of the arc is 50 mm and radius is 14 mm.
Solution:
Length of the arc of the sector l = 50 mm
Radius r = 14 mm
Area of the sector = \(\frac{l r}{2}\) sq. units = \(\frac{50 \times 14}{2}\) mm2 = 50 × 7 mm2 = 350 mm2
Area of the sector = 350 mm2

Question 9.
Find the area of a sector whose perimeter is 64 cm and length of the arc is 44 cm.
Solution:
Length of the arc of the sector l = 44 cm
Perimeter of the sector P = 64 cm
l + 2r = 64 cm
44 + 2 r = 64 .
2 r = 64 – 44
2 r = 20
r = \(\frac{20}{2}\) = 10 cm2
Area of the sector = \(\frac{l r}{2}\) sq. units = \(\frac{44 \times 10}{2}\) cm2 = 22 × 10 cm2 = 220 cm2
Area of the sector = 220 cm2

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 2 Measurements Ex 2.1

Question 10.
A sector of radius 4.2 cm has an area 9.24 cm2. Find its perimeter
Solution:
Radius of the sector r = 4.2 cm
‘ Area of the sector = 9.24 cm2
\(\frac{l r}{2}\) = 9.24
\(\frac{l \times 4.2}{2}\) = 9.24
l × 2.1 = 9.24
l = \(\frac{9.24}{2.1}\)
l = 4.4 cm
Perimeter of the sector = 1 + 2r units = 4.4 + 2(4.2) cm
= 4.4 + 8.4 cm = 12. 8 cm
Perimeter of the sector = 12.8 cm

Question 11.
Infront of a house, flower plants are grown in a circular quadrant shaped pot whose radius is 2 feet. Find the area of the pot in which the plants grow. (π = 3.14)
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Ex 2.1 20
Solution:
Central angle of the quadrant = 90°
Radius of the circle = 2 feet
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Ex 2.1 14
Area of the quadrant = 3.14 sq. feet (approximately)

Question 12.
Dhamu fixes a square tile of 30 cm on the floor. The tile has a sector design on it as shown in the figure. Find the area of the sector, (π = 3.14).
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Ex 2.1 21
Solution:
Side of the square = 30 cm
∴ Radius of the sector design = 30 cm
Given design in the design of a circular quadrant.
Area of the quadrant = \(\frac{1}{4}\) πr2 sq. units
= \(\frac{1}{4}\) × 3.14 × 30 × 30 cm2
= 3.14 × 15 × 15 cm2
∴ Area of the sector design = 706.5 cm2 (approximately)

Question 13.
A circle is formed with 8 equal granite stones as shown in the figure each of radius 56 cm and whose central angle is 45°. Find the area of each of the granite. (π = \(\frac{22}{7}\))
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Ex 2.1 22
Solution:
Number of equal sectors ‘n’ = 8
Radius of the sector ‘r’ = 56 cm
Area of the each sector = \(\frac{1}{n}\) πr2 sq. units
= \(\frac{1}{8} \times \frac{22}{7}\) × 56 × 56 cm2 = 1232 cm2
Area of each sector = 1232 cm2 (approximately)

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

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

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

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

Question 2.
\(\frac{x}{3}+\frac{2 y}{5}-x y+7\)
Solution:
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Intext Questions 1
Numerical co efficient in -xy is -1
Numerical co efficient in 7 is 7

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

Question 3.
Pick out the like terms from the following.
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Intext Questions 6
Solution:
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Intext Questions 7

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

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

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

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

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

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

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
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Intext Questions 70
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

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

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)

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

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.

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

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

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

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.

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

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:
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Intext Questions 50

Try this Page No. 77

Question 1.
Divide
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Intext Questions 61
Solution:
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Intext Questions 625

Try this Page No. 78

Question 1.
Are the following divisions correct ?
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Intext Questions 51
Solution:
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Intext Questions 52

Try this Page No. 78

Question 1.
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Intext Questions 600
Solution:
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Intext Questions 53
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Intext Questions 54

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:
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Intext Questions 90

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

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

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

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

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

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

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

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

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

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

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

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

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

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
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Intext Questions 63

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)

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

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)

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

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)

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Intext Questions Read More »

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.2

Students can Download Maths Chapter 3 Algebra Ex 3.2 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.
Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.2

Tamilnadu Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.2

Question 1.
Fill in the blanks:
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Ex 3.2 1
Solution:
(i) \(\frac{18 m^{4}\left(n^{8}\right)}{2 m^{(3)} n^{3}}\) = 9 mn5
(ii) \(\frac{l^{4} m^{5} n^{(7)}}{2 l m^{(3)} n^{6}}=\frac{l^{3} m^{2} n}{2}\)
(iii) \(\frac{42 a^{4} b^{5}\left(c^{2}\right)}{6(a)^{4}(b)^{2}}\) = (7)b3c2

Question 2.
Say True or False:
(i) 5x3y ÷ 4x2 = 2xy
(ii) 7ab2 ÷ 14ab = 2b2
Solution:
(i) True
(ii) False

Question 3.
(i) 27y3 ÷ 3y
(ii) x3y2 ÷ x2y
(iii) 45x3y2z4 ÷ (-15xyz)
(iv) (3xy)2 ÷ 9xy
Solution:
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Ex 3.2 2

Question 4.
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Ex 3.2 3
Solution:
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Ex 3.2 4

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.2

Question 5.
Divide
(i) 32y2 – 8yz by 2y
(ii) (4m2 n3 + 16m4 n2 – mn) by 2 mn
(iii) 10 (4x – 8y) by 5 (x – 2y)
(iv) 81 (94q2r3 + 2p3q3r2 – 5p2q2r2) by (3pqr)2
Solution:
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Ex 3.2 5

Question 6.
Find Adirai’s percentage of marks who scored 25m3n2p out of 100m2np
Solution:
Total marks = 100 m2np
Adirai’s score = 25 m3n2p
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Ex 3.2 6

Question 7.
Identify the error and correct them.
(i) 7y2 – y2 + 3y2 = 10y2
(ii) 6xy + 3xy = 9x2y2
(iii) m (4m – 3) = 4m2 – 3
(iv) (4n)2 – 2n + 3 = 4n2 – 2n + 3
(v) (x – 2) (x + 3) = x2 – 6
Solution:
(i) 7y2 – y2 + 3y2 = (7 – 1 + 3)y2 = (6 + 3)y2 = 9y2
(ii) 6xy + 3xy = (6 + 3)xy = 9xy
(iii) m (4m – 3) = m (4m) + m (-3) = 4m2 – 3m
(iv) (4n)2 – 2n + 3 = 16n2 – 2n + 3
(v) (x – 2) (x + 3) = x (x + 3) – 2 (x + 3) = x (x) + (x) × 3 + (-2) (x) + (-2) (3)
= x2 + 3x – 2x – 6 = x2 + x – 6

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.2 Read More »

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.1

Students can Download Maths Chapter 3 Algebra Ex 3.1 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 Ex 3.1

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.1

Question 1.
Multiply a monomial by a monomial.
(i) 6x, 4
(ii) -3x, 7y
(iii) -2m2, (-5m)3
(iv) a3, – 4a2b
(v) 2p2q3, -9pq2
Solution:
(i) 6x × 4 = (6 × 4) (x) = 24x
(ii) -3x × 7y = (-3 × 7) (x × y) = -21xy
(iii) (-2m2) × (-5m)3 = -2m2 × (-)3 (53 (m)3) = -2m2 × (-125m3)
= (-) × (-)(2 × 125)(m2 × m3) = + 250m5 = 250 m
(iv) a3 × (-4a2 b) = (-4) × (a3 × a2) × (b) = -4a5b
(v) (2p2q3) × (-9pq2) = (+) × (-) × (2 × 9) (p2 × p(q3 × q2)) = -18p3q5

Question 2.
Complete the table
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Ex 3.1 1
Solution:
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Ex 3.1 2

Question 3.
Find the product of the terms.
(i) -2mn, (2m)2, -3mn
(ii) 3x2y, -3xy3, x2y2
Solution:
(i) (-2mn) × (2m)2 × (-3mn) = (-2mn) × 22 m2 × (-3mn) = (-2mn) × 4m2 × (-3mn)
= (-) (+)(-) (2 × 4 × 3) (m × m2 × m) (n × n)
= + 24 m4 n2

(ii) (32y) × (-31xy3) × (x2y2) = (+) × (-) × (+) × (3 × 3 × 1) (x2 × x × x2) x (y × y3 × y2)
= -9x5y6

Question 4.
If l = 4pq2, b = -3p 2q, h = 2p3q3 then, find the value of 1 × b × h.
Solution:
Given l = 4pq2
b = -3p2q
h = 2p3q3
l × b × h = (4pq2) × (-3p2 q) × (2p3q3)
= (+) (-) (+) (4 × 3 × 2) (p × p2 × p3) (q2 × q × q3)
= -24p6q6

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.1

Question 5.
Expand
(i) 5x (2y – 3)
(ii) -2p (5p2 – 3p + 7)
(iii) 3mn (m3n3 – 5m2n + 7mn2)
(iv) x2 (x + y + z) y2 (x + y + z) + z2 (x – y – z)
Solution:
(i) 5x(2y – 3) = (5x) (2y) – (5x) (3)
= (5 × 2) (x × y) – (5 × 3) x
= 10xy – 15x

(ii) -2p (5p2 – 3p + 7) = (-2p) (5p2) + (-2p) (-3p) + (-2p) (7)
= [(-) (+) (2 × 5) (p × p2)] + [(-) (+) (2 × 3) (p × q)] + (-) (+) (2 × 7) p
= -10p3 + 6p2 – 14p

(iii) 3mn(m3n3 – 5m2n + 7mn2)
= (3mn) (m3n3) + (3mn) (-5m2n) + (3mn)(7mn2)
= (3) (m × m3) (n × n3) + (+) (-) (3 × 5) (m × m2) (n × n) + (3 × 7) (m × m)(n × n2)
= 3m4n4 – 15m3 n2 + 21m2n3

(iv) x2 (x + y + z) + y2 (x + y + z) + z2 (x – y – z)
= (x2 × x) + (x2 × y) + (x2 × z) + (y2 × x) + (y2 × y) + (y2 × z) + (z2 × x) + z2 (-y) + z2 (-z)
= x3 + x2y + x2z + xy2 + y3 + y2z + xz2 – yz2 – z3
= x3 + y3 – z2 + x2y + x2z + xy2 + zy2 + xz2 – yz2

Question 6.
Find the product of
(i) (2x + 3)(2x – 4)
(ii) (y2 – 4) (2y2 + 3y)
(iii) (m2 – m) (5m2n2 – n2)
(iv) 3(x – 5) × 2(x – 1)
Solution:
(2x) (2x – 4) + 3 (2x – 4) = (2x) (2x – 4) + 3 (2x – 4)
= (2x × 2x) – 4 (2x) + 3(2x) – 3 (4)
= 4x2 – 8x + 6x – 12
= 4x2 + (- 8 + 6)x – 12
= 4x2 – 2x – 12

(ii) (y2 -4) (2y2 + 3y) = y2 (2y2 + 3y) – 4 (2y2 + 37)
= y2(2y2) + y2(3y) – 4(2y2) -4 (3y)
= 2y4 + 3y3 – 8y2 – 12y

(iii) (m2 – n) (5m2n2 – n2) = m2 (5m2n2 – n2) – n (5m2n2 – n2)
= m2 (5m2n2) + m2 (-n2) – n (5m2n2) + (-) (-) n (n2)
= 5m4n2 – m2n2 – 5m2n3 + n3

(iv) 3(x – 5) × 2(x – 1) = (3 × 2) (x – 5) (x – 1)
= 6 × [x (x – 1) – 5 (x- 1)]
= 6 [x.x – x . 1 – 5.x + (-1) (-) 5 1]
= 6 [x2 – x – 5x + 5] = 6 [x2 + (-1 – 5)x + 5]
= 6 [x2 – 6x + 5] = 6x2 – 36x + 30

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.1

Question 7.
Find the missing term.
(i) 6xy – × ______ = -12x3y
(ii) ________ × (-15m2n3p) = 45m3n3p2
(iii) 2y(5x2y – ___ + 3 ___) = 10x2y2 – 2xy + 6y3
Solution:
(i) 6xy – × (-2x2) = -12x3y
(ii) -3mp × (-15m2n3p) = 45m3n3p2
(iii) 2y(5x2y – x + 3 y2) = 10x2y2 – 2xy + 6y3

Question 8.
Match the following
Samacheer Kalvi 8th Maths Term 1 Chapter 3 Algebra Ex 3.1 3
(A) iv, v, ii, i, iii
(B) v, iv, iii, ii, i
(C) iv, v, ii, iii, i
(D) iv, v, ii, iii, i
Solution:
(a) iv
(b) v
(c) ii
(d) iii
(e) i

Question 9.
A car moves at a uniform speed of (x + 30) km/hr. Find the distance covered by the car in (y + 2)hours. (Hint: distance = speed × time).
Solution:
Sppeed of the car = (x + 30) km / hr.
Time = (y + 2) hours
Distance = Speed × time = (x + 30) (y + 2) = x(y + 2) + 30 (y + 2) = x (y + 2) + 30 (y + 2)
= (x) (y) + (x) (2) + (30) (y) + (30) (2)
= xy + 2x + 30y + 60
Distance covered = (xy + 2x + 30y + 60) km

Objective Type Questions

Question 10.
The product of 7p3 and (2p2)2 is
(A) 14p2
(B) 28p7
(C) 9p7
(D) 11p12
Solution:
(B) 28p7

Question 11.
The missing terms in the product -3m3 n × 9(- -) = ____ m4n3 are
(A) mn2, 27
(B) m2n, 27
(C) m2n2, -27
(D) mn2, -27
Solution
(A) mn2 ,27

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.1

Question 12.
If the area of a square is 36x4y2 then, its side is ______ .
(A) 6x4y2
(B) 8x2y2
(C) 6x2y
(D) -6x2y
Solution:
(C) 6x2y

Question 13.
If the area of a rectangle is 48m2n3 and whose length is 8mn2 then, its breadth is ____ .
(A) 6 mn
(B) 8m2n
(C) 7m2n2
(D) 6m2n2
Solution:
(A) 6mn

Question 14.
If the area of a rectangular land is (a2 – b2) sq.units whose breadth is (a – b) then, its length is _____
(A) a – b
(B) a + b
(C) a2 – b
(D) (a + b)2
Solution:
(B) a + b

Samacheer Kalvi 8th Maths Solutions Term 1 Chapter 3 Algebra Ex 3.1 Read More »

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

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

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

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

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:
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Additional Questions 1
∴ 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
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Additional Questions 2
∴ 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
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Additional Questions 3
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.
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Additional Questions 4
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.
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Additional Questions 5

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 = \(\frac{22}{7}\))
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Additional Questions 6
Solution:
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Additional Questions 7
∴ 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
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Additional Questions 8
Solution:
Radius of the big semicircle = 14 cm
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Additional Questions 9
∴ Required area = 308 + 154 cm2 = 462 cm2

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

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

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

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.
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Additional Questions 10
(b) The following shows a net with areas of faces. What can be the shape?
Samacheer Kalvi 8th Maths Term 1 Chapter 2 Measurements Additional Questions 11
Solution:
(a) 2 + 5 = 6 + 1 = 3 + 4 = 7
∴ It can be a die.
(b) It is a cuboid

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