Class 8

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.

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)

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

(iv) m² + m – 72

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

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)

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 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 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.

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

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

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

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.

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

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)

Solution:

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


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

Question 5.
From the measures given below, find the area of the sectors.

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 = \(\frac{22}{7}\))

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

∴ Central angle of the sector = 120°

(ii) Radius of the sector = 8.4 cm
Area of the sector = 18.48 cm²

(iii) Radius of the sector = 35 m
Length of the arc l = 44 m

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

(ii) Radius of the sector r = 70 cm
Number of equal sectors = 5

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

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)

Solution:
Central angle of the quadrant = 90°
Radius of the circle = 2 feet

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).

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 = \(\frac{22}{7}\))

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)

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)

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(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.

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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

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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:

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Question 1.
Divide

Solution:

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Question 1.
Are the following divisions correct ?

Solution:

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Question 1.

Solution:

Exercise 3.3

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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 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.

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

Question 1.
Fill in the blanks:

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:

Question 4.

Solution:

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:

Question 6.
Find Adirai’s percentage of marks who scored 25m³n²p out of 100m²np
Solution:
Total marks = 100 m²np
Adirai’s score = 25 m³n²p

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

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

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

Solution:

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

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

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

(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

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

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