Class 8

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 15.3 8th Std Maths Answers Solutions Chapter 15 Area.

Practice Set 15.3 8th Std Maths Answers Chapter 15 Area

Question 1.
In the given figure, ☐ABCD is a trapezium, side AB || side DC, l(AB) = 13 cm, l(DC) = 9 cm, l(AD) = 8 cm, find the area ☐ABCD.

Solution:
☐ABCD is a trapezium, side AB || side DC,
l(AB) = 13 cm, l(DC) = 9 cm, l(AD) = 8 cm,
Area of a trapezium = \(\frac { 1 }{ 2 }\) x sum of lengths of parallel sides x height
∴ A (☐ABCD) = \(\frac { 1 }{ 2 }\) x [l(AB) + l(DC)] x l(AD)
= \(\frac { 1 }{ 2 }\) x (13 + 9) x 8
= \(\frac { 1 }{ 2 }\) x 22 x 8
= 11 x 8
= 88 sq.cm
∴ The area of ☐ABCD is 88 sq. cm.
[Note: The question is modified.]

Question 2.
Length of the two parallel sides of a trapezium are 8.5 cm and 11.5 cm respectively and its height is 4.2 cm, find its area.
Solution:
Length of the two parallel sides of a trapezium are 8.5 cm and 11.5 cm and its height is 4.2 cm.
Area of a trapezium
= \(\frac { 1 }{ 2 }\) x sum of lengths of parallel sides x height
= \(\frac { 1 }{ 2 }\) x (8.5 + 11.5) x 4.2
= \(\frac { 1 }{ 2 }\) x 20 x 4.2
= 10 x 4.2
= 42 sq. cm
∴ The area of the trapezium is 42 sq. cm.

Question 3.
☐PQRS is an isosceles trapezium. l(PQ) = 7 cm, seg PM ⊥ seg SR, l(SM) = 3 cm. Distance between two parallel sides is 4 cm, find the area of ☐PQRS.

Solution:
☐PQRS is an isosceles trapezium.
l(PQ) = 7 cm, seg PM ⊥ seg SR,
l(SM) = 3 cm, l(PM) = 4cm
Draw seg QN ⊥ seg SR.
In ☐PMNQ,
seg PQ || seg MN
∠PMN = ∠QNM = 90°
∴ ☐PMNQ is a rectangle.

Opposite sides of a rectangle are congruent.
∴ l(PM) = l(QN) = 4 cm and
l(PQ) = l(MN) = 7 cm
In ∆PMS, m∠PMS = 90°
∴ [l(PS)]² = [l(PM)]² + [l(SM)]² … [Pythagoras theorem]
∴ [l(PS)]² = (4)² + (3)²
∴ [l(PS)]² = 16 + 9 = 25
∴ l(PS) = √25 = 5 cm
…[Taking square root of both sides]
☐PQRS is an isosceles trapezium.
∴ l(PS) = l(QR) = 5 cm
In ∆QNR, m ∠QNR = 90°
∴ [l(QR)]² = [l(QN)]² + [l(NR)]²
… [Pythagoras theorem]
∴ (5)² = (4)² + [l(NR)]²
∴ 25 = 16 + [l(NR)]²
∴ [l(NR)]² = 25 – 16 = 9
∴ l(NR) = √9 = 3 cm
…[Taking square root of both sides]
l(SR) = l(SM) + l(MN) + l(NR)
= 3 + 7 + 3
= 13 cm
Area of a trapezium
= \(\frac { 1 }{ 2 }\) x sum of lengths of parallel sides x height
∴ A(☐PQRS) = \(\frac { 1 }{ 2 }\) x [l(PQ) + l(SR)] x l(PM)
= \(\frac { 1 }{ 2 }\) x (7+ 13) x 4
= \(\frac { 1 }{ 2 }\) x 20 x 4
= 40 sq.cm
∴ The area of ☐PQRS is 40 sq. cm.

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 15.2 8th Std Maths Answers Solutions Chapter 15 Area.

Practice Set 15.2 8th Std Maths Answers Chapter 15 Area

Question 1.
Lengths of the diagonals of a rhombus are 15 cm and 24 cm, find its area.
Solution:
Lengths of the diagonals of a rhombus are 15 cm and 24 cm.
Area of a rhombus
= \(\frac { 1 }{ 2 }\) × product of lengths of diagonals
= \(\frac { 1 }{ 2 }\) × 15 × 24
= 15 × 12
= 180 sq.cm
∴ The area of the rhombus is 180 sq. cm.

Question 2.
Lengths of the diagonals of a rhombus are 16.5 cm and 14.2 cm, find its area.
Solution:
Lengths of the diagonals of a rhombus are 16.5 cm and 14.2 cm.
Area of a rhombus
= \(\frac { 1 }{ 2 }\) × product of lengths of diagonals
= \(\frac { 1 }{ 2 }\) × 16.5 × 14.2
= 16.5 × 7.1
= 117.15 sq cm
∴ The area of the rhombus is 117.15 sq. cm.

Question 3.
If perimeter of a rhombus is 100 cm and length of one diagonal is 48 cm, what is the area of the quadrilateral?
Solution:
Let ₹ABCD be the rhombus. Diagonals AC and BD intersect at point E.

l(AC) = 48 cm …(i)
l(AE) = \(\frac { 1 }{ 2 }l(AC)\) …[Diagonals of a rhombus bisect each other]
= \(\frac { 1 }{ 2 }\) × 48 …[From (i)]
= 24 cm …(ii)
Perimeter of rhombus = 100 cm …[Given]
Perimeter of rhombus = 4 × side
∴ 100 = 4 × l(AD)
∴ l(AD) = \(\frac { 100 }{ 4 }\) = 25 cm …(iii)
In ∆ADE,
m∠AED = 90° …[Diagonals of a rhombus are perpendicular to each other]
∴ [l(AD)]² = [l(AE)]² + [l(DE)]² … [Pythagoras theorem]
∴ (25)² = (24)² + l(DE)² … [From (ii) and (iii)]
∴ 625 = 576 + l(DE)²
∴ l(DE)² = 625 – 576
∴ l(DE)² = 49
∴ l(DE) = √49
… [Taking square root of both sides]
l(DE) = 7 cm …(iv)
l(DE) = \(\frac { 1 }{ 2 } l(BD)\) ….[Diagonals of a rhombus bisect each other]
∴ 7 = \(\frac { 1 }{ 2 } l(BD)\) …[From (iv)]
∴ l(BD) = 7 × 2
= 14 cm …(v)
Area of a rhombus
= \(\frac { 1 }{ 2 }\) × product of lengths of diagonals
= \(\frac { 1 }{ 2 }\) × l(AC) × l(BD)
= \(\frac { 1 }{ 2 }\) × 48 × 14 … [From (i) and (v)]
= 48 × 7
= 336 sq.cm
∴ The area of the quadrilateral is 336 sq.cm.

Question 4.
If length of a diagonal of a rhombus is 30 cm and its area is 240 sq.cm, find its perimeter.
Solution:
Let ₹ABCD be the rhombus.
Diagonals AC and BD intersect at point E.
l(AC) = 30 cm …(i)
and A(₹ABCD) = 240 sq. cm .. .(ii)

Area of the rhombus = \(\frac { 1 }{ 2 }\) × product of lengths of diagonal
∴ 240 = \(\frac { 1 }{ 2 }\) × l(AC) x l(BD) …[From (ii)]
∴ 240 = \(\frac { 1 }{ 2 }\) × 30 × l(BD) …[From (i)]
∴ l(BD) = \(\frac { 240\times 2 }{ 30 }\)
∴ l(BD) = 8 × 2 = 16 cm …(iii)
Diagonals of a rhombus bisect each other.
∴ l(AE) = \(\frac { 1 }{ 2 }l(AC)\)
= \(\frac { 1 }{ 2 }\) × 30 … [From (i)]
= 15 cm …(iv)
and l(DE) = \(\frac { 1 }{ 2 }l(BD)\)
= \(\frac { 1 }{ 2 }\) × 16
= 8 cm
In ∆ADE,
m∠AED = 90°
…[Diagonals of a rhombus are perpendicular to each other]
∴[l(AD)]² = [l(AE)]² + [l(DE)]²
…[Pythagoras theorem]
∴l(AD)² = (15)² + (8)² … [From (iv) and (v)]
= 225 + 64
∴l(AD)² = 289
∴l(AD) = √289
…[Taking square root of both sides]
∴l(AD) = 17 cm
Perimeter of rhombus = 4 × side
= 4 × l(AD)
= 4 × 17
= 68 cm
∴The perimeter of the rhombus is 68 cm.

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 12.1 8th Std Maths Answers Solutions Chapter 12 Equations in One Variable.

Practice Set 12.1 8th Std Maths Answers Chapter 12 Equations in One Variable

Equation in One Variable Practice Set 12.1 Question 1. Each equation is followed by the values of the variable. Decide whether these values are the solutions of that equation.
i. x – 4 = 3, x = – 1, 7, – 7
ii. 9m = 81, m = 3, 9, -3
iii. 2a + 4 = 0, a = 2, – 2, 1
iv. 3 – y = 4, y = – 1, 1, 2
Solution:
i. x – 4 = 3 ….(i)
Substituting x = – 1 in L.H.S. of equation (i),
L.H.S. = (-1) – 4
= – 5
R.H.S. = 3
∴ L.H.S. ≠ R.H.S.
∴ x = – 1 is not the solution of the given equation.

Substituting x = 7 in L.H.S. of equation (i),
L.H.S. = (7) – 4
= 3
R.H.S. = 3
∴ L.H.S. = R.H.S.
∴ x = 7 is the solution of the given equation.

Substituting x = – 7 in L.H.S. of equation (i),
L.H.S. = (- 7) – 4
= -11
R.H.S. = 3
∴ L.H.S. ≠ R.H.S.
∴ x = – 7 is not the solution of the given equation.

ii. 9m = 81 …(i)
Substituting m = 3 in L.H.S. of equation (i),
L.H.S. = 9 × (3)
= 27
R.H.S. = 81
∴L.H.S. ≠ R.H.S.
∴m = 3 is not the solution of the given equation.

Substituting m = 9 in L.H.S. of equation (i),
L.H.S. = 9 × (9)
= 81
R.H.S. = 81
∴L.H.S. = R.H.S.
∴m = 9 is the solution of the given equation.

Substituting m = – 3 in L.H.S. of equation (i),
L.H.S. = 9 × (- 3)
= -27
R.H.S. = 81
∴L.H.S. ≠ R.H.S.
∴m = – 3 is not the solution of the given equation.

iii. 2a + 4 = 0 …..(i)
Substituting a = 2 in L.H.S. of equation (i),
L.H.S. = 2 (2) + 4
= 4 + 4
= 8
R.H.S. = 0
∴L.H.S. ≠ R.H.S.
∴a = 2 is not the solution of the given equation.

Substituting a = – 2 in L.H.S. of equation (i),
L.H.S. = 2 (-2)+ 4
= -4 + 4
= 0
R.H.S. = 0
∴L.H.S. = R.H.S.
∴a = – 2 is the solution of the given equation.

Substituting a = 1 in L.H.S. of equation (i),
L.H.S. = 2(1)+ 4
= 2 + 4
= 6
R.H.S. = 0
∴ L.H.S. ≠ R.H.S.
∴a = 1 is not the solution of the given equation.

iv. 3 – y = 4 …(i)
Substituting y = -1 in L.H.S. of equation (i),
L.H.S. = 3 – (- 1)
= 3 + 1
= 4
R.H.S. = 4
∴L.H.S. = R.H.S.
∴y = – 1 is the solution of the given equation.

Substituting y = 1 in L.H.S. of equation (i),
L.H.S. = 3-(1)
= 2
R.H.S. = 4
∴L.H.S. ≠ R.H.S.
∴y = 1 is not the solution of the given equation.

Substituting y = 2 in L.H.S. of equation (i),
L.H.S. = 3-(2)
= 1
R.H.S. = 4
∴L.H.S. ≠ R.H.S.
∴y = 2 is not the solution of the given equation.

Practice Set 12.1 Question 2.
Solve the following equations:
i. 17p – 2 = 49
ii. 2m + 7 = 9
iii. 3x + 12 = 2x – 4
iv. 5 (x – 3) = 3 (x + 2)
v. \(\frac { 9x }{ 8 }+1=10\)
vi. \(\frac{y}{7}+\frac{y-4}{3}=2\)
vii. 13x – 5 = \(\frac { 3 }{ 2 }\)
viii. 3 (y + 8) = 10 (y – 4) + 8
ix. \(\frac{x-9}{x-5}=\frac{5}{7}\)
x. \(\frac{y-4}{3}+3 y=4\)
xi. \(\frac{b+(b+1)+(b+2)}{4}=21\)
Solution:
i. 17p – 2 = 49
∴ 17p – 2 + 2 = 49 + 2
…[Adding 2 on both the sides]
∴ 17p = 51
∴ \(\frac{17 p}{17}=\frac{51}{17}\) …[Dividing both the sides by 17]
p = 3

ii. 2m + 7 = 9
∴ 2m + 7 – 7 = 9 – 7
…[Subtracting 7 from both the sides]
∴ 2m = 2
∴ \(\frac{2 m}{2}=\frac{2}{2}\) [Dividing both the sides by 2]
∴ m = 1

iii. 3x + 12 = 2x – 4
∴ 3x + 12 – 12 = 2x – 4 – 12
…[Subtracting 12 from both the sides]
∴ 3x = 2x – 16
∴ 3x – 2x = 2x – 16 – 2x
…[Subtracting 2x from both the sides]
∴ x = – 16

iv. 5 (x – 3) = 3 (x + 2)
∴ 5x – 15 = 3x + 6
∴ 5x – 15 + 15 = 3x + 6 + 15
…[Adding 15 on both the sides]
∴ 5x = 3x + 21
∴ 5x – 3x = 3x + 21 – 3x
…[Subtracting 3x from both the sides]
∴ 2x = 21
∴ \(\frac{2 x}{2}=\frac{21}{2}\) …[Dividing both the sides by 2]
∴ \(x=\frac{21}{2}\)

v. \(\frac { 9x }{ 8 }+1=10\)

vi. \(\frac{y}{7}+\frac{y-4}{3}=2\)

vii. 13x – 5 = \(\frac { 3 }{ 2 }\)

viii. 3 (y + 8) = 10 (y – 4) + 8
∴ 3y + 24 = 10y – 40 + 8
∴ 3y + 24 = 10y – 32
∴ 3y + 24 – 24 = 10y – 32 – 24
…[Subtracting 24 from both the sides]
∴ 3y = 10y – 56
∴ 3y – 10y = 10y – 56
…[Subtracting 10y from both the sides]
∴ – 7y = – 56
∴ \(\frac{-7 y}{-7}=\frac{-56}{-7}\)…[Dividing both the sides by – 7]
∴ y = 8

ix. \(\frac{x-9}{x-5}=\frac{5}{7}\)
∴\(\frac{x-9}{x-5} \times 7(x-5)=\frac{5}{7} \times 7(x-5)\)
…[Multiplying both the sides by 7 (x – 5)]
∴7 (x – 9) = 5 (x – 5)
∴7x – 63 = 5x – 25
∴7x – 63 + 63 = 5x – 25 + 63
…[Adding 63 on both the sides]
∴7x = 5x + 38
∴7x – 5x = 5x + 38 – 5x
…[Subtracting 5x from both the sides]
∴ 2x = 38
∴\(\frac{2 x}{2}=\frac{38}{2}\) …[Dividing both the sides by 2]
∴x = 19

x. \(\frac{y-4}{3}+3 y=4\)
∴\(\frac{y-4}{3} \times 3+3 y \times 3=4 \times 3\)
…[Multiplying both the sides by 3]
∴y – 4 + 9y = 12
∴10y – 4 = 12
∴10y – 4 + 4=12 + 4
…[Adding 4 on both the sides]
∴10y = 16
∴\(\frac{10 y}{10}=\frac{16}{10}\)…[Dividing both the sides by 10]
∴y = \(\frac { 8 }{ 5 }\)

xi. \(\frac{b+(b+1)+(b+2)}{4}=21\)
∴\(\frac{b+(b+1)+(b+2)}{4} \times 4=21 \times 4\)
…[Multiplying both the sides by 4]
∴b + b + 1 + b + 2 = 84
∴3b + 3 = 84
∴3b + 3 – 3 = 84 – 3
…[ Subtracting 3 from both the sides]
∴3b = 81
∴\(\frac{3 b}{3}=\frac{81}{3}[/latex …[Dividing both the sides by 3]
∴b = 27

Maharashtra Board Class 8 Maths Chapter 12 Equations in One Variable Practice Set 12.1 Intext Questions and Activities

Std 8 Maths Practice Set 12.1 Question 1.
Fill in the boxes to solve the following equations. (Textbook pg. no. 75)
i. x + 4 = 9
∴x + 4 – __ = 9 – __
… [Subtracting 4 from both the sides]
∴ x = __

ii. x – 2 = 7
∴x – 2 + __ = 7 + __
… [Adding 2 on both the sides]
∴x = __

iii. [latex]\frac { x }{ 3 }=4\)
∴\(\frac { x }{ 3 }\) × __ = 4 ×__
∴x = __

iv. 4x = 24
∴ __ = __
∴x = __
Solution:
i. x + 4 = 9
∴x + 4 – 4 = 9 – 4
… [Subtracting 4 from both the sides]
∴ x = 5

ii. x – 2 = 7
∴x – 2 + 2 = 7 + 2
… [Adding 2 on both the sides]
∴x = 9

iii. \(\frac { x }{ 3 }=4\)
∴\(\frac { x }{ 3 }\) × 3 = 4 × 3
… [Multiplying both the sides by 3]
∴x = 12

iv. 4x = 24
∴ \(\frac{4 x}{[4]}=\frac{24}{[4]}\)
… [Dividing both the sides by 4]
∴x = 6

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 1.1 8th Std Maths Answers Solutions Chapter 1 Rational and Irrational Numbers.

Practice Set 1.1 8th Std Maths Answers Chapter 1 Rational and Irrational Numbers

Question 1.
Show the following numbers on a number line. Draw a separate number line for each example.
i. \(\frac{3}{2}, \frac{5}{2},-\frac{3}{2}\)
ii. \(\frac{7}{5}, \frac{-2}{5}, \frac{-4}{5}\)
iii. \(\frac{-5}{8}, \frac{11}{8}\)
iv. \(\frac{13}{10}, \frac{-17}{10}\)
Solution:
i. \(\frac{3}{2}, \frac{5}{2},-\frac{3}{2}\)

Here, the denominator of each fraction is 2.
∴ Each unit will be divided into 2 equal parts.

ii. \(\frac{7}{5}, \frac{-2}{5}, \frac{-4}{5}\)

Here, the denominator of each fraction is 5.
∴ Each unit will be divided into 5 equal parts.

iii. \(\frac{-5}{8}, \frac{11}{8}\)

Here, the denominator of each fraction is 8.
∴ Each unit will be divided into 8 equal parts.

iv. \(\frac{13}{10}, \frac{-17}{10}\)

Here, the denominator of each fraction is 10.
∴ Each unit will be divided into 10 equal parts.

Question 2.
Observe the number line and answer the questions.

i. Which number is indicated by point B?
ii. Which point indicates the number \(1\frac { 3 }{ 4 }\) ?
iii. State whether the statement, ‘the point D denotes the number \(\frac { 5 }{ 2 }\) is true or false.
Solution:
Here, each emit is divided into 4 equal parts.
i. Point B is marked on the 10th equal part on the left side of O.
∴ The number indicated by point B is \(\frac { -10 }{ 4 }\).

ii.

Point C is marked on the 7th equal part on the right side of O.
∴ The number \(1\frac { 3 }{ 4 }\) is indicated by point C.

iii. True
Point D is marked on the 10th equal part on the right side of O.
∴ D denotes the number \(\frac{10}{4}=\frac{5 \times 2}{2 \times 2}=\frac{5}{2}\)

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 8.2 8th Std Maths Answers Solutions Chapter 8 Quadrilateral: Constructions and Types.

Practice Set 8.2 8th Std Maths Answers Chapter 8 Quadrilateral: Constructions and Types

Question 1.
Draw a rectangle ABCD such that l(AB) = 6.0 cm and l(BC) = 4.5 cm.
Solution:

Question 2.
Draw a square WXYZ with side 5.2 cm.
Solution:

Question 3.
Draw a rhombus KLMN such that its side is 4 cm and m∠K = 75°.
Solution:

Question 4.
If diagonal of a rectangle is 26 cm and one side is 24 cm, find the other side.
Solution:

Let ₹ABCD be the rectangle.
l(BC) = 24cm, l(AC) = 26cm
In ∆ABC,
m∠ABC = 90° …[Angle of a rectangle]
∴[l(AC)]² = [l(AB)]2 + [l(BC)]²
…[Pythagoras theorem]
∴ (26 )² = [l(AB)]² + (24)²
∴(26)² – (24)² = [l(AB)]²
∴(26 + 24) (26 – 24) = [l(AB)]²
…[∵ a² – b² = (a + b)(a – b)]
∴50 x 2 = [l(AB)]²
∴100 = [l(AB)]²
i.e. [l(AB)]² = 100
∴l(AB) = √100
…[Taking square root of both sides]
∴l(AB) =10 cm
∴The length of the other side is 10 cm.

Question 5.
Lengths of diagonals of a rhombus ABCD are 16 cm and 12 cm. Find the side and perimeter of the rhombus.
Solution:
In rhombus ABCD,

l(AC) = 16 cm and l(BD) = 12 cm.
Let the diagonals of rhombus ABCD intersect at point O.
l(AO) = \(\frac { 1 }{ 2 }\) l(AC)
…[Diagonals of a rhombus bisect each other]
∴l(AO) = \(\frac { 1 }{ 2 }\) × 16
∴l(AO) = 8 cm
Also, l(DO) = \(\frac { 1 }{ 2 }\) l(BD)
…[Diagonals of a rhombus bisect each other]
∴l(DO) = \(\frac { 1 }{ 2 }\) × 12
∴l(DO) = 6 cm
In ∆DOA,
m∠DOA = 90°
..[Diagonals of a rhombus are perpendicular to each other]
[l(AD)]² = [l(AO)]² + [l(DO)]²
…[Pythagoras theorem]

= (8)² + (6)²
= 64 + 36
∴[l(AD)]² = 100
∴l(AD) = √100
… [Taking square root of both sides]
∴l(AD) = 10 cm
∴l(AB) = l(BC) = l(CD) = l(AD) = 10 cm
…[Sides of a rhombus are congruent]
Perimeter of rhombus ABCD
= l(AB) + l(BC) + l(CD) + l(AD)
= 10+10+10+10
= 40 cm
∴The side and perimeter of the rhombus are 10 cm and 40 cm respectively.

Question 6.
Find the length of diagonal of a square with side 8 cm.
Solution:
Let ₹XYWZ be the square of side 8cm.

seg XW is a diagonal.
In ∆ XYW,
m∠XYW = 90°
… [Angle of a square]
∴ [l(XW)]² = [l(XY)]² + [l(YW)]²
…[Pythagoras theorem]
= (8)² + (8)²
= 64 + 64
∴ [l(XW)]² = 128
∴ l(XW) = √128
…[Taking square root of both sides]
= √64 × 2
= 8 √2 cm
∴ The length of the diagonal of the square is 8 √2 cm.

Question 7.
Measure of one angle of a rhombus is 50°, find the measures of remaining three angles.
Solution:
Let ₹ABCD be the rhombus.

m∠A = 50°
m∠C = m∠A
….[Opposite angles of a rhombus are congruent]
∴ m∠C = 50°
Also, m∠D = m∠B …(i)
….[Opposite angles of a rhombus are congruent]
In ₹ABCD,
m∠A + m∠B + m∠C + m∠D = 360°
….[Sum of the measures of the angles of a quadrilateral is 360°]
∴ 50° + m∠B + 50° + m∠D = 360°
∴ m∠B + m∠D + 100° = 360°
∴ m∠B + m∠D = 360° – 100°
∴ m∠B + m∠B = 260° …[From (i)]
∴ 2m∠B = 260°
∴ m∠B = \(\frac { 260 }{ 2 }\)
∴ m∠B = 130°
∴ m∠D = m∠B = 130° …[From (i)]
∴ The measures of the remaining angles of the rhombus are 130°, 50° and 130°.

Maharashtra Board Class 8 Maths Chapter 8 Quadrilateral: Constructions and Types Practice Set 8.2 Intext Questions and Activities

Question 1.
Construct a rectangle PQRS by taking two convenient adjacent sides. Name the point of intersection of diagonals as T. Using divider and ruler, measure the following lengths.
i. lengths of opposite sides, seg QR and seg PS.
ii. lengths of seg PQ and seg SR.
iii. lengths of diagonals PR and QS.
iv. lengths of seg PT and seg TR, which are parts of the diagonal PR.
v. lengths of seg QT and seg TS, which are parts of the diagonal QS.
Observe the measures. Discuss about the measures obtained by your classmates. (Textbook pg. no. 44)
Solution:
Draw a rectangle PQRS such that, l(PQ) = 3 cm and l(QR) = 4 cm.

Steps of construction:
i. As shown in the rough figure, draw seg QR of length 4 cm.
ii. Placing the centre of the protractor at point Q, draw ray QW making an angle of 90° with seg QR.
iii. By taking a distance of 3 cm on the compass and placing it at point Q, draw an arc on ray QW. Name the point as P.
iv. Draw ray PV and ray RU making an angle of 90° with seg PQ and seg QR respectively.
v. Name the point of intersection of ray PV and ray RU as S.
₹PQRS is the required rectangle.

From the figure,
i. l(QR) = l(PS) = 4 cm
ii. l(PQ) = l(SR) = 3 cm
iii. l(PR) = l(QS) = 5 cm
iv. l(PT) = l(TR) = 2.5 cm
v. l(QT) = l(TS) = 2.5 cm

From the above measures, we can say that for any rectangle,
i. Opposite sides are congruent.
ii. Diagonals are congruent.
iii. Diagonals bisect each other.

Question 2.
Draw a square by taking convenient length of side. Name the point of intersection of its diagonals as E. Using the apparatus in a compass box, measure the following lengths.
i. lengths of diagonal AC and diagonal BD.
ii. lengths of two parts of each diagonal made by point E.
iii. all the angles made at the point E.
iv. parts of each angle of the square made by each diagonal, (e.g. ∠ADB and ∠CDB).
Observe the measures. Also observe the measures obtained by your classmates and discuss about them. (Textbook pg. no. 44)
Solution:
Draw a square ABCD such that its side is 5cm

Steps of construction:
i. As shown in the rough figure, draw seg BC of length 5 cm.
ii. Placing the centre of the protractor at point B, draw ray BP making an angle of 90° with seg BC.
iii. By taking a distance of 5 cm on the compass and placing it at point B, draw an arc on ray BP. Name the point as A.
iv. Placing the centre of the protractor at point C, draw ray CQ making an angle of 90° with seg BC.
v. By taking a distance of 5 cm on the compass and placing it at point C, draw an arc on ray CQ. Name the point as D.
vi. Draw seg AD.
₹ABCD is the required square.

From the figure,
i. l(AC) = l(BD) ≅ 7cm
ii. l(AE) = l(EC) ≅ 3.5cm,
l(BE) = l(ED) ≅ 3.5cm
iii. m∠AED = m∠BEC = m∠CED = m∠BEA = 90°
iv. Angles made by diagonal AC:
m∠BAC = m∠DAC = 45°
m∠BCA = m∠DCA = 45°
Angles made by diagonal BD:
m∠ABD = m∠CBD = 45°
m∠ADB = m∠CDB = 45°

From the above measures, we can say that for any square,
i. Diagonals are congruent.
ii. Diagonals bisect each other.
iii. Diagonals are perpendicular to each other.
iv. Diagonals bisect the opposite angles.

Question 3.
Draw a rhombus EFGH by taking convenient length of side and convenient measure of an angle.
Draw its diagonals and name their point of Intersection as M.
i. Measure the opposite angles of the quadrilateral and angles at the point M.
ii. Measure the two parts of every angle made by the diagonal.
iii. Measure the lengths of both diagonals. Measure the two parts of diagonals made by point M.
Observe the measures. Also observe the measures obtained by your classmates and discuss about them. (Textbook pg. no. 45)
Solution:
Draw a rhombus EFGH such that its side is 5 cm and m∠F = 60°.

Steps of construction:
i. As shown in the rough figure, draw seg FG of length 5 cm.
ii. Placing the centre of the protractor at point F, draw ray FX making an angle 60° with seg FG.
iii. By taking a distance of 5 cm on the compass and placing it at point F, draw an arc on ray FX. Name the point as E.
iv. By taking a distance of 5 cm on the compass and placing it at point E and point G, draw arcs. Name the point of intersection of arcs as H. ₹EFGH is the required rhombus.

From the figure,
i. Opposite angles:
m∠EFG = m∠GHE = 60°,
m∠FEH = m∠HGF = 120°
Angles at the point M:
m∠EMF = m∠FMG = m∠GMH = m∠HME = 90°

ii. Angles made by diagonal FH:
m∠EFH = m∠GFH = 30° m∠EHF = m∠GHF = 30°
Angles made by diagonal EG:
m∠FEG = m∠HEG = 60° m∠FGE = m∠HGE = 60°

iii. l(FH) ≈ 8.6 cm
l(EG) = 5 cm
l(FM) = l(HM) ≈ 4.3 cm
l(EM) = l(GM) ≈ 2.5 cm

From the above measures, we can say that for any rhombus,
i. Opposite angles are congruent.
ii. Diagonals bisect the opposite angles.
iii. Diagonals bisect each other and they are perpendicular to each other.

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 17.1 8th Std Maths Answers Solutions Chapter 17 Circle: Chord and Arc.

Practice Set 17.1 8th Std Maths Answers Chapter 17 Circle: Chord and Arc

Question 1.
In a circle with centre P, chord AB is drawn of length 13 cm, seg PQ ⊥ chord AB, then find l(QB)

Solution:
seg PQ ⊥ chord AB … [Given]
∴l(QB) = \(\frac { 1 }{ 2 }\) l(AB)… [Perpendicular drawn from the centre of a circle to its chord bisects the chord]
∴l(QB) = \(\frac { 1 }{ 2 }\) x 13 …[∵ l(AB) = 13 cm]
∴l(QB) = 6.5 cm

Question 2.
Radius of a circle with centre O is 25 cm. Find the distance of a chord from the centre if length of the chord is 48 cm.

Solution:
seg OP ⊥ chord CD … [Given]
∴l(PD) = \(\frac { 1 }{ 2 }\) l(CD) … [Perpendicular drawn from the centre of a circle to its chord bisects the chord]
∴l(PD) = \(\frac { 1 }{ 2 }\) x 48 …[∵ l(CD) = 48 cm]
∴l(PD) = 24 cm …(i)
In ∆OPD, m∠OPD = 90°
∴[l(OD)]² = [l(OP)]² + [l(PD)]² … [Pythagoras theorem]
∴(25)² = [l(OP)]² + (24)² … [From (i) and l(OD) = 25 cm]
∴(25)² – (24)² = [l(OP)]²
∴(25 + 24) (25 – 24) = [l(OP)]² …[∵ a² – b² = (a + b) (a – b)]
∴49 x 1 = [l(OP)]²
∴[l(OP)]² = 49
∴l(OP) = √49 …[Taking square root of both sides]
∴l(OP) = 7 cm
∴The distance of the chord from the centre of the circle is 7 cm.

Question 3.
O is centre of the circle. Find the length of radius, if the chord of length 24 cm is at a distance of 9 cm from the centre of the

Solution:
Let seg OP ⊥ chord AB
∴ l(AP) = \(\frac { 1 }{ 2 }\) l(AB) … [Perpendicular drawn from the centre of a circle to its chord bisects the chord]

∴l(AP) = \(\frac { 1 }{ 2 }\) x 24 …[∵ l(AB) = 24 cm]
∴l(AP) = 12 cm …(i)
In ∆OPA, m∠OPA = 90°
∴[l(AO)]² = [l(OP)]² + [l(AP)]² … [Pythagoras theorem]
∴[l(AO)]² = (9)² + (12)² … [From (i) and l(OP) = 9 cm]
= 81 + 144
∴[l(AO)]² = 225
∴l(AO) = √225 …[Taking square root of both sides]
∴l(AO) = 15 cm
∴The length of radius of the circle is 15 cm.

Question 4.
C is the centre of the circle whose radius is 10 cm. Find the distance of the chord from the centre if the length of the chord is 12 cm.
Solution:
Let seg AB be the chord of the circle with centre C.
Draw seg CD ⊥ chord AB.

∴l(AD) = \(\frac { 1 }{ 2 }\) l(AB) …[Perpendicular drawn from the centre of a circle to its chord bisects the chord]
= \(\frac { 1 }{ 2 }\) x 12 …[∵ l(AB) = 12 cm]
∴l(AD) = 6 cm …(i)
∴In ∆ACD, m∠ADC = 90°
∴[l(AC)]² = [l(AD)]² + [l(CD)]² … [Pythagoras theorem]
∴(10)² = (6)² + [l(CD)]² … [From (i) and l(AC) = 10 cm]
∴(10)² – (6)² = [l(CD)]²
∴100 – 36 = [l(CD)]²
∴64 = [l(CD)]²
i. e. [l(CD)]² = 64
∴l(CD) = √64 …[Taking square root of both sides]
∴l(CD) = 8 cm
∴The distance of the chord from the centre of the circle is 8 cm.

Maharashtra Board Class 8 Maths Chapter 17 Circle: Chord and Arc Practice Set 17.1 Intext Questions and Activities

Question 1.
In the given figure, O is the centre of the circle. With reference to the figure fill in the blanks. (Textbook pg. No. 114)

Solution:

1. Seg OD is radius of the circle.
2. Seg AB is diameter of the circle.
3. Seg PQ is chord of the circle.
4. ∠DOB is the central angle.
5. Minor arc : arc AXD, arc BD, arc AP, arc PQ, arc BQ, etc.
6. Major arc : arc PAB, arc PDQ, arc PDB, arc ADQ, etc.
7. Semicircular arc : arc ADB, arc AQB.
8. m (arc DB) = m∠DOB
9. m (arc DAB) = 360° – m∠DOB

Question 2.
Draw chord AB of a circle with centre O. Draw perpendicular OP to chord AB. Measure seg AP and seg PB. What do you observe. (Textbook pg. no. 114)
Solution:
l(AP) = l(PB) = 0.9 cm
∴the perpendicular drawn from the centre of the circle to its chord bisects the chord.

Question 3.
Draw five circles with different radii. Draw a chord and perpendicular from the centre to each chord in each circle. Verify with a divider that the two parts of the chords are equal. (Textbook pg. no. 114)
Solution:
[Students should attempt the above activities on their own.]

Question 4.
Draw five circles of different radii on a paper. Draw a chord in each circle. Find the midpoint of each chord. Join the centre of the circle and midpoint of the chord as shown in the figure. Name the chord as AB and midpoint of the chord as P. Check with set-square or protractor that ∠APO or ∠BPO are right angles.
Check whether the same result is observed for the chord of each circle. (Textbook pg, no. 115)
Solution:
[Students should attempt the above activities on their own.]

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 14.2 8th Std Maths Answers Solutions Chapter 14 Compound Interest.

Practice Set 14.2 8th Std Maths Answers Chapter 14 Compound Interest

Compound Interest class 8 practice set 14.2 Question 1. On the construction work of a flyover bridge there were 320 workers initially. The number of workers were increased by 25% every year. Find the number of workers after 2 years.
Solution:
Here, P = Initial number of workers = 320
R = Increase in the number of workers per year = 25%
N = 2 years
A = Number of workers after 2 years

∴ The number of workers after 2 years would be 500.

Question 2.
A shepherd has 200 sheep with him. Find the number of sheeps with him after 3 years if the increase in number of sheeps is 8% every year.
Solution:
Here, P = Present number of sheeps = 200
R = Increase in number of sheeps per year = 8%
N = 3 years
A = Number of sheeps after 3 years

= \(\frac{0.32}{25} \times 27 \times 27 \times 27\)
= 0.0128 × 27 × 27 × 27
= 251.9424
= 252
∴ The number of sheeps with the shepherd after 2 years would be 252 (approx).

8th Class Math Practice Set 14.2 Question 3.
In a forest there are 40,000 trees. Find the expected number of trees after 3 years if the objective is to increase the number at the rate 5% per year.
Solution:
Here, P = Present number of trees in the forest = 40,000
R = Increase in the number of trees per year = 5%
N = 3 years
A = Number of trees after 3 years

= 5 × 21 × 21 × 21
= 5 × 9261
= 46,305
∴ The expected number of trees in the forest after 3 years is 46,305.

Std 8 Maths Practice Set 14.2 Question 4.
The cost price of a machine is Rs 2,50,000. If the rate of depreciation is 10% per year, find the depreciation in price of the machine after two years.
Solution:
Here, P = Cost price of machine = Rs 2,50,000
R = Rate of depreciation per year = 10%
N = 2 years
A = Depreciated price of the machine after 2 years

= 2,500 × 81
= Rs 2,02,500
Depreciation in price = Cost price (P) – Depreciated price (A)
= 2,50,000 – 2,02,500
= Rs 47,500
∴ The depreciation in price of the machine after 2 years would be Rs 47,500.

Question 5.
Find the compound interest if the amount of a certain principal after two years is Rs 4036.80 at the rate of 16 p.c.p.a.
Solution:
Here, A = Rs 4036.80, R = 16 p.c.p.a. and N = 2 years
i. \(\mathbf{A}=\mathbf{P}\left[1+\frac{\mathbf{R}}{100}\right]^{N}\)

ii. Interest = Amount (A) – Principal (P)
= 4036.80 – 3000
= Rs 1036.80
∴ The compound interest after 2 years would be Rs 1036.80.

Question 6.
A loan of Rs 15,000 was taken on compound interest. If the rate of compound interest is 12 p.c.p.a. find the amount to settle the loan after 3 years.
Solution:
Here, P = Rs 15,000, R = 12 p.c.p.a, and
N = 3 years

∴ The amount required to settle the loan after 3 years is Rs 21,073.92.

Practice Set 14.2 Class 8 Question 7.
A principal amounts to Rs 13,924 in 2 years by compound interest at 18 p.c.p.a. Find the principal.
Solution:
Here, A = Rs 13,924, R = 18 p.c.p.a., and N = 2 years

∴ P = 4 x 50 x 50
∴ P = Rs 10,000
∴ The principal is Rs 10,000.

Question 8.
The population of a suburb is 16,000. Find the rate of increase in the population if the population after two years is 17,640.
Solution:
Here, P = Population of a suburb = 16,000
N = 2 years
A = Increase in the population after 2 years = 17,640
R = Rate of increase in population


∴5 = R
i.e., R = 5%
∴The rate of increase in the population is 5 p.c.p.a.

Compound Interest Practice Set 14.2 Question 9.
In how many years Rs 700 will amount to Rs 847 at a compound interest rate of 10 p.c.p.a.
Solution:
Here, P = Rs 700, R = 10 p.c.p.a., A = Rs 847

∴Rs 700 will amount to Rs 847 in 2 years.

Practice Set 14.2 Question 10.
Find the difference between simple interest and compound interest on Rs 20,000 in 2 years at 8 p.c.p.a.
Solution:
Here, P = Rs 20,000, R = 8 p.c.p.a.,
N = 2 years
i. Simple interest (I)

Simple interest (I) = Rs 3200

ii. Compound Interest (I):


= 32 × 27 × 27
= Rs 23,328
Compound interest (I)
= Amount (A) – Principal (P)
= 23,328 – 20,000
= Rs 3328 ,..(ii)

iii. Difference
= Compound interest – Simple interest
= 3328 – 3200 … [Form (i) and (ii)]
= Rs 128
∴ The difference between compound interest and simple interest is Rs 128.
[Note: The question is modified as per the answer given in the textbook.]

Maharashtra Board Class 8 Maths Chapter 14 Compound Interest Practice Set 14.2 Intext Questions and Activities

8th Standard Maths Practice Set 14.2 Question 1.
Visit the bank nearer to your house and get the information regarding the different schemes and rates of interests. Make a chart and display in your class. (Textbook pg. no. 90)
Solution:
(Students should attempt this activity at their own.)

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 13.1 8th Std Maths Answers Solutions Chapter 13 Congruence of Triangles.

Practice Set 13.1 8th Std Maths Answers Chapter 13 Congruence of Triangles

Congruence of Triangles Practice Set 13.1 Question 1.
In each pair of triangles in the following figures, parts bearing identical marks are congruent. State the test and correspondence of vertices by which triangles in each pair are congruent.
i.

ii.

iii.

iv.

v.

Solution:
i.

The two triangles are congruent by SAS test in the correspondence XWZ ↔ YWZ.

ii.

The two triangles are congruent by hypotenuse-side test in the correspondence KJI ↔ LJI.

iii.

The two triangles are congruent by SSS test in the correspondence HEG ↔ FGE.

iv.
The two triangles are congruent by ASA test is the correspondence SMA ↔ OPT.

v.

The two triangles are congruent by ASA test or SAS test or SAA test in the correspondence MTN ↔ STN.

Maharashtra Board Class 8 Maths Chapter 13 Congruence of Triangles Practice Set 13.1 Intext Questions and Activities

Practice Set 13.1 Question 1.
Write answers to the following questions referring to the given figure.

1. Which is the angle opposite to the side DE?
2. Which is the side opposite to ∠E?
3. Which angle is included by side DE and side DF?
4. Which side is included by ∠E and ∠F?
5. State the angles adjacent to side DE. (Textbook pg, no. 81)

Solution:

1. ∠DFE i.e. ∠F is the angle opposite to side DE.
2. Side DF is the side opposite to ∠E.
3. ∠EDF i.e. ∠D is included by side DE and side DF.
4. Side EF is included by ∠E and ∠F.
5. ∠DEF and ∠EDF i.e. ∠E and ∠D are adjacent to side DE.

Congruence of Triangles Class 8th Practice Set 13.1 Question 2.
In the given figure, parts of triangles indicated by identical marks are congruent.
a. Identify the one-to-one correspondence of vertices in which the two triangles are congruent and write the congruence.
b. State with reason, whether the statement, ∆XYZ ≅ ∆STU is right or wrong. (Textbook pg. no. 82)

Solution:
a. From the figure,
S ↔ X, T ↔ Z, U ↔ Y i.e.,
STU ↔ XZY, or SUT ↔ XYZ, or
TUS ↔ ZYX, or TSU ↔ ZXY, or
UTS ↔ YZX, or UST ↔ YXZ

∴ ∆STU ≅ ∆XZY, or ∆SUT ≅ ∆XYZ, or
∆TUS ≅ ∆ZYX, or ∆TSU ≅ ∆ZXY, or
∆UTS ≅ ∆YZX, or ∆UST ≅ ∆YXZ

b. If ∆XYZ ≅ ∆STU, then
∠Y ≅ ∠T, ∠Z ≅ ∠U,
seg XY ≅ seg ST, seg XZ ≅ seg SU
∴ But, all the above statements are wrong. The statement AXYZ ≅ ASTU is wrong.

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 13.2 8th Std Maths Answers Solutions Chapter 13 Congruence of Triangles.

Practice Set 13.2 8th Std Maths Answers Chapter 13 Congruence of Triangles

Congruence of Triangles Class 8th Practice Set 13.2 Question 1.
In each pair of triangles given below, parts shown by identical marks are congruent. State the test and the one-to-one correspondence of vertices by which triangles in each pair are congruent. Also state the remaining congruent parts.


Solution:
i. In ∆MST and ∆TBM,
∴ side MS ≅ side TB … [Given]
m∠MST = m∠TBM = 90° … [Given]
hypotenuse MT ≅ hypotenuse MT
…[Common side]
∴ ∆MST ≅ ∆TBM …[by hypotenuse-side test]
∴ side ST ≅ side BM …[Corresponding sides of congruent triangles]
∠SMT ≅ ∠BTM …[Corresponding sides of congruent triangles]
∠STM ≅ ∠BMT …[Corresponding sides of congruent triangles]

ii. In ∆PRQ and ∆TRS,
side PR ≅ side TR … [Given]
∠PRQ ≅ ∠TRS …[Vertically opposite angles]
side RQ ≅ side RS … [Given]
∴ ∆PRQ ≅ ∆TRS …[by SAS test]
∴ side PQ ≅ side TS …[Corresponding sides of congruent triangles]
∠RPQ ≅ ∠RTS …[Corresponding sides of congruent triangles]
∠PQR ≅ ∠TSR …[Corresponding sides of congruent triangles]

iii. In ∆DCH and ∆DCF,
∠DCH ≅ ∠DCF …[Given]
∠DHC ≅ ∠DFC …[Given]
side DC ≅ side DC …[Common side]
∴ ∆DCH ≅ ∆DCF …[by AAS test]
∴ side HC ≅ side FC …[Corresponding sides of congruent triangles]
side DH ≅ side DF…[Corresponding sides of congruent triangles]
∠HDC ≅ ∠FDC ….[Corresponding sides of congruent triangles]

Congruence of Triangles Practice Set 13.2 Question 2.
In the given figure, seg AD ≅ seg EC. Which additional information is needed to show that ∆ABD and ∆EBC will be congruent by AAS test?

Solution:
In ∆ABD and ∆CBE,
∴ seg AD ≅ seg CE …[Given]
∠ABD ≅ ∠CBE …[Vertically opposite angles]
∴ The necessary condition for the two triangles to be congruent by AAS test is
∠ADB ≅ ∠CEB, or
∠DAB ≅ ∠ECB

Maharashtra Board Class 8 Maths Chapter 13 Congruence of Triangles Practice Set 13.2 Intext Questions and Activities

Practice Set 13.2 Class 8 Question 1.
Draw ∆ABC and ∆LMN such that two pairs of their sides and the angles included by them are congruent.
Draw ∆ABC and ∆LMN, l(AB) = l(LM), l(BC) = l(MN), m∠ABC = m∠LMN.
Copy ∆ABC on a tracing paper. Place the paper on ∆LMN in such a way that point A coincides with point L, side AB overlaps side LM. What do you notice?(Textbook pg. no. 83)

Solution:
We notice that ∆ABC ≅ ∆LMN.

Congruence of Triangles Class 8 Solutions Question 2.
Draw ∆PQR and ∆XYZ such that l(PQ) = l(X Y), l(Q R) = l(YZ), l(RP) = l(ZX). Copy ∆PQR on a tracing paper. Place it on ∆XYZ observing the correspondence P ↔ X, Q ↔ Y, R ↔ Z. What do you notice? (Textbook pg. no. 84)

Solution:
We notice that ∆PQR ≅ ∆XYZ.

Congruence of Triangles Class 8 Question 3.
Draw ∆XYZ and ∆DEF such that, l(XZ) = l(DF), ∠X ≅ ∠D and ∠Z ≅ ∠F.
Copy ∆XYZ on a tracing paper and place it over ∆DEF. What do you notice?(Textbook pg. no. 84)

Solution:
We notice that ∆XYZ ≅ ∆DEF in the correspondence X ↔ D, Y ↔ E, Z ↔ F.

Question 4.
Draw two right angled triangles such that a side and the hypotenuse of one is congruent with the corresponding parts of the other. Copy one triangle on tracing paper and place it over the other. What do you notice? (Textbook pg. no. 84)

Solution:
We notice that the two triangles are congruent.
(Students should draw figures and verify the answers.)

Balbharti Maharashtra State Board Class 8 Marathi Solutions Sulabhbharati Chapter 4 आपण सारे एक Notes, Textbook Exercise Important Questions and Answers.

Maharashtra State Board Class 8 Marathi Solutions Chapter 4 आपण सारे एक

Marathi Sulabhbharti Class 8 Solutions Chapter 4 आपण सारे एक Textbook Questions and Answers

1. एका शब्दांत उत्तरे लिहा.

प्रश्न 1.
एका शब्दांत उत्तरे लिहा.

उत्तर:
(अ) माणसाच्या सौंदर्यात भर घालणारे – [नयनकुमार]
(आ) माणसाच्या जीवनव्यवहारात मदत करणारी- [कर्णिका]
(इ) वास घेण्याचे जास्तीचे काम करणारी – [नासिका]
(ई) जिव्हाताई यांना संपात सामील करून घेणारे – [दंतराज]
(उ) सर्वांनी ज्यांच्याकडे तक्रार केली ते – [मेंदूराजे]
(ऊ) अजिबात काम न करण्याचा आरोप असणारे – [पोटोबा]

2. आकृती पूर्ण करा.

प्रश्न 1.
आकृती पूर्ण करा.

उत्तर:
अ.

आ.

इ.

3. कारणे शोधा व लिहा.

अ. जिव्हाताई गप्प आहे, कारण…………………..
आ. पोटोबा मधून मधून गुरगुरतो, कारण ……………..

प्रश्न 1.
अ. जिव्हाताई गप्प आहे, कारण…………………..
आ. पोटोबा मधून मधून गुरगुरतो, कारण ……………..
उत्तर:
अ. जिव्हाताई गप्प आहे, कारण पोटोबा खवय्येची मोठी चीड यायला लागली आहे म्हणून.
आ. पोटोबा मधून मधून गुरगुरतो, कारण त्याला गुरगुरण्याची सवय आहे.

4. स्वमत स्पष्ट करा.

प्रश्न अ.
‘आपण सगळे शरीररूपी राज्याचे सेवक आहोत’ या विधानाबाबत तुमचे मत.
उत्तरः
उतारा 3 मधील ‘कृती 4 – स्वमत’ चे उत्तर पहा.

प्रश्न आ.
पोटाबद्दलची तक्रार सांगून सर्व इंद्रियांनी ती तक्रार करण्याची कारणे.
उत्तरः
उतारा 2 मधील ‘कृती 4 – स्वमत’ चे उत्तर पहा.

खेळूया शब्दांशी.

(अ) खालील वाक्यात योग्य विरामचिन्हे घाला.

प्रश्न अ.
तो म्हणेल तेवढचं खायची सक्ती असते माझ्यावर
उत्तर:
“तो म्हणेल तेवढचं खायची सक्ती असते माझ्यावर!”

प्रश्न आ.
हो हो आमची तयारी आहे
उत्तर:
“हो, हो आमची तयारी आहे.”

(आ) वर्गीकरण करुन तक्ता पूर्ण करा.

प्रश्न 1.
भराभर, सावकाश, पोटोबाविरुद्ध, बापरे, आणि, सतत, किंवा, कशासाठी, पोटोबामुळे, स्वयंपाकघरापर्यंत, तुमच्याबद्दल, अथवा, अबब
उत्तर:

क्रियाविशेषण अव्यय	शब्दयोगी अव्यय	उभयान्वयी अव्यय	केवलप्रयोगी अव्यय
भरभर, सावकाश, सतत	पोटाबाविरुद्ध, कशासाठी, पोटोबामुळे, स्वयंपाक घरापर्यंत, तुमच्याबद्दल	आणि, किंवा, बापरे, अबब अथवा	बापरे, अबब

(इ) खालील शब्दांसाठी पाठात वापरलेले शब्द लिहा.

प्रश्न 1.
खालील शब्दांसाठी पाठात वापरलेले शब्द लिहा.

उत्तर:
(अ) कान – [कर्णिका]
(आ) नाक – [नासिका]
(इ) हात – [हस्तकराज]
(ई) पाय – [पदकुमार]
(उ) जीभ – [जिव्हाताई]

उपक्रम :
1. ‘आधी पोटोबा मग विठोबा’ ही, म्हण या पाठात आली आहे. याप्रमाणे शरीर अवयवांशी संबंधित असणाऱ्या इतर म्हणी शोधा व लिहा.
2. या पाठाचे नाट्यीकरण वर्गात सादर करा.

Marathi Sulabhbharti Class 8 Solutions Chapter 4 आपण सारे एक Important Additional Questions and Answers

पुढील उताऱ्याच्या आधारे दिलेल्या सूचनेनुसार कृती करा.
कृती 1 : आकलन कृती

प्रश्न 1.
खालील कृती पूर्ण करा.
उत्तर:
i.

ii.

प्रश्न 2.
चौकटी पूर्ण करा.
उत्तर:
i.

ii.

प्रश्न 3.
जोड्या जुळवा.

‘अ’ गट (नाटिकेतील नावे)	‘ब’ गट (अवयवांची नावे)
1.  नासिका	(अ) डोळे
2.  कर्णिका	(आ) जीभ
3.  नयनकुमार	(इ) पाय
4. जिव्हाताई	(ई) नाक
5.  हस्तकराज	(उ) कान
6.  पदकुमार	(ऊ) हात

उत्तर:

‘अ’ गट (नाटिकेतील नावे)	‘ब’ गट (अवयवांची नावे)
1.  नासिका	(ई) नाक
2.  कर्णिका	(उ) कान
3.  नयनकुमार	(अ) डोळे
4. जिव्हाताई	(आ) जीभ
5.  हस्तकराज	(ऊ) हात
6.  पदकुमार	(इ) पाय

प्रश्न 4.
वेब पूर्ण करा.
उत्तर:

प्रश्न 5.
असे कोण कोणास म्हणाले ते लिहा.

1. “या! या!! रामराव, आज स्वारी कशी आली इकडे?”
2. “अरे, अरे, एका दमात किती प्रश्न विचारता शामराव?”

उत्तर:

1. असे शामराव रामरावांना म्हणाले.
2. असे रामराव शामरावांना म्हणाले.

कृती 2 : आकलन कृती

प्रश्न 1.
खालील कृती पूर्ण करा.
उत्तर:
(i)

(ii)

प्रश्न 2.
एका शब्दात उत्तरे लिहा.
उत्तर:

1. पोटोबाची चीड येणारी – [जिव्हाताई]
2. पोटोबाला दिलेली उपमा – [खवय्या]
3. मानसन्मान याला मिळतो – [पोटोबा]
4. सारी धडपड कशासाठी – [पोटासाठी]
5. त्या पोटोबामुळे आपली होते – [फरफट]

प्रश्न 3.
आकृतिबंध पूर्ण करा.
उत्तर:

i.

ii.

iii.

प्रश्न 4.
खालील वेब पूर्ण करा.
उत्तर:
i.

ii.

कृती 3: व्याकरण कृती

प्रश्न 1.
खालील वाक्यातील विरामचिन्हे ओळखून त्यांची नावे लिहा. अरे, अरे, एका दमात किती प्रश्न विचारता शामराव ?
उत्तरः

विरामचिन्हे	विरामचिन्हाचे नाव
(,)	स्वल्पविराम
(?)	प्रश्नचिन्ह

प्रश्न 2.
खालील वाक्यांत योग्य विरामचिन्हे वापरुन वाक्य पुन्हा लिहा.

1. या या शामराव आज स्वारी कशी आली इकडे
2. चला चला कामाचं बघू नंतर आधी पोटोबा मग विठोबा

उत्तर:

1. “या! या!! शामराव, आज स्वारी कशी आली इकडे?”
2. “चला, चला, कामाचं बघू नंतर. आधी पोटोबा, मग विठोबा.”

प्रश्न 3.
खालील शब्दांचे लिंग बदलून लिहा.

1. बाबा
2. कुमार
3. ताई
4. बाई
5. तो

उत्तर:

1. आई
2. कुमारी
3. दादा
4. पुरुष
5. ती

प्रश्न 4.
खालील वाक्यांचे प्रकार ओळखून लिहा.

1. काय सांगू बाई!
2. आज स्वारी कशी आली इकडे ?

उत्तर:

1. उद्गारार्थी वाक्य
2. प्रश्नार्थी वाक्य

कृती 4 : स्वमत

प्रश्न 1.
‘आधी पोटोबा मग विठोबा’ ही म्हण या पाठात आली आहे. याप्रमाणे शरीर अवयवांशी संबंधित असणाऱ्या इतर म्हणी लिहा. (विदयार्थी यापेक्षा वेगळ्या म्हणी लिहू शकतात.)
उत्तर:

1. हातच्या कंकणाला आरसा कशाला?
2. कानामागून आली अन् तिखट झाली.
3. अंथरुण पाहून पाय पसरावेत.
4. आठ हात लाकूड, नऊ हात ढलपी.
5. आपला हात जगन्नाथ,
6. आपले नाक कापून दुसऱ्याला अपशकुन.

पुढील उताऱ्याच्या आधारे दिलेल्या सूचनेनुसार कृती करा

कृती 1 : आकलन कृती

प्रश्न 1.
आकृतिबंध पूर्ण करा.
उत्तर:
i.

ii.

iii.

iv.

प्रश्न 2.
खालील चौकटी पूर्ण करा.
उत्तर:

1. माणसाच्या जीवनव्यवहाराला मदत करणारी – [कर्णिका]
2. माणसाला श्वासाशिवाय जगू न देणारी – [नासिका]
3. माणसांच्या जीवनाला अर्थ देणारी – [जिव्हाताई]

कृती 2 : आकलन कृती

प्रश्न 2.
खालील वेब पूर्ण करा.
उत्तर:
i.

ii.

iii.

प्रश्न 2.
एका वाक्यात उत्तरे लिहा.

प्रश्न i.
सर्व अवयव कोणाविरुद्ध संप करणार आहेत?
उत्तर:
सर्व अवयव पोटोबाविरुद्ध संप करणार आहेत.

प्रश्न ii.
नाना तहेची चव मानव कोणामुळे चाखू शकतो?
उत्तर:
नाना त-हेची चव मानव जिव्हाताईमुळे चाखू शकतो.

प्रश्न 3.
कंसात दिलेल्या पर्यायांपैकी योग्य पर्याय निवडून रिकाम्या जागा भरा.

1. जो उठतो तो, ‘कशासाठी? पोटासाठी’, म्हणत आम्हांला ………………….. घेतो. (बजावून, राबवून, दामटवून, धाकात)
2. म्हणजे मग त्या पोटोबाला चांगलीच ………………………. घडेल. (शक्कल, नक्कल, अक्कल, अद्दल)

उत्तर:

1. राबवून
2. अद्दल

कृती 3: व्याकरण कृती

प्रश्न 1.
खालील वाक्यांतील अव्यये ओळखून लिहा.
i. श्वास घेतल्याशिवाय माणूस जगू शकेल का?
ii. तिचं म्हणणं आहे, की पोटोबा खवय्ये काहीच काम करत नाहीत.
उत्तर:
i. शिवाय
ii. की

प्रश्न 2.
खालील वाक्प्रचारांचा अर्थ सांगून वाक्यात उपयोग करा.
उत्तर:
1. राबवून घेणे – दुसऱ्यांकडून जबरदस्तीने काम करून घेणे.
वाक्य : मालक नोकरांना राबवून घेतात.

2. अद्दल घडणे – शिक्षा होणे, कानउघडणी होणे.
वाक्यः सतत खात राहणाऱ्या दादाला आज काहीच खाण्यास न मिळाल्याने चांगलीच अद्दल घडली.

प्रश्न 3.
खालील वाक्यांतील क्रियाविशेषण अव्यये शोधून लिहा.

1. अगं ही बडबडी, पोटाबद्दल तक्रार करतीय.
2. रात्रंदिवस श्वास घेण्याचं कार्य सतत चालूच असतं.
3. मी नेहमी बघण्याचं काम करतो.
4. आपण सारेजण अनेक कामे करतो.

उत्तर:

1. बडबडी
2. सतत
3. नेहमी
4. अनेक

प्रश्न 4.
खालील शब्दांना ‘यात’ प्रत्यय जोडून शब्द पुन्हा लिहा.

1. म्हणणे
2. बोलणे
3. माझं
4. गुरगुरणे
5. सौंदर्य
6. ऐकणे
7. तुझा
8. जगणे
9. खाणारे
10. घरटे
11. देणे
12. विचारणे

उत्तर:

1. म्हणण्यात
2. बोलण्यात
3. माझ्यात
4. गुरगुरण्यात
5. सौंदर्यात
6. ऐकण्यात
7. तुझ्यात
8. जगण्यात
9. खाणाऱ्यात
10. घरट्यात
11. देण्यात
12. विचारण्यात

प्रश्न 5.
खालील वाक्यांतील अधोरेखित शब्दांची जात ओळखा.

1. मी नेहमी बघण्याचं काम करतो.
2. आमची तयारी आहे संप करण्याची.
3. माणसाच्या सौंदर्यात भर घालतो, तो मीच.
4. पोटोबा मात्र आयते बसून खातात.

उत्तर:

1. मी – सर्वनाम, काम – नाम
2. आमची – सर्वनाम, संप – नाम
3. माणूस – नाम, मी – सर्वनाम
4. पोटोबा – नाम, आयते – विशेषण

कृती 4 : स्वमत

प्रश्न 1.
पोटोबाबद्दलची तक्रार सांगून सर्व इंद्रियांनी ती तक्रार करण्याची कारणे सांगा.
उत्तरः
‘पोटोबा हे फक्त खवय्ये आहेत ते काहीच काम करत नाहीत उलट सर्वांवर गुरगुरतात’ अशी सर्व इंद्रियांची तक्रार आहे. कारण प्रत्येक इंद्रिय काही ना काही काम करत आहे. नासिका रात्रंदिवस श्वास घेण्याचं कार्य करते. नयनकुमार सतत बघण्याचं काम करतात, तसेच माणसाच्या सौंदर्यात भर घालण्याचे काम करतात. कर्णिका ऐकण्याचं काम करते. जिव्हाताईमुळे माणसाच्या जीवनात अर्थ आहे. ती गोड बोलते, नाना त-हेच्या पदार्थांची चव चाखते. यांपैकी पोटोबा कोणतेच काम करत नसल्याने व सर्वांवर फक्त गुरगुरण्याचे काम करत असल्याने सर्व मेंदूराजेसाहेबांकडे तक्रार करणार आहेत.

पुढील उताऱ्याच्या आधारे दिलेल्या सूचनेनुसार कृती करा.

कृती 1: आकलन कृती

प्रश्न 1.
चौकटी पूर्ण करा.
उत्तरः

1. सर्वजण मेंदूराजाकडे जाताच – [मेंदूराजे तातडीने सभा घेतात.]
2. जीवनरस न मिळाल्याने – [सर्वांचे चेहरे सुकून गेले.]
3. पोटोबांनी खोटं ठरवलेली तक्रार – [मी काम करत नाही.]

प्रश्न 2.
कोण ते लिहा.
उत्तरः

1. मेंदूराजांचा जयजयकार करणारे – [सर्वजण]
2. सर्वांना बसायला सांगणारे – [मेंदूराजे]
3. पोटोबांना दिलेली उपाधी – [प्रधान गुरगुरणे]
4. पोटोबांना ‘प्रधान’ बोलणारे – [मेंदूराजे]
5. केव्हाच हजर झालेले – [पोटोबा]

प्रश्न 3.
खालील वेब पूर्ण करा.
उत्तरः
i.

ii.

iii.

कृती 2 : आकलन कृती

प्रश्न 1.
पुढील कृती पूर्ण करा.

i.

ii.

प्रश्न 2.
एका वाक्यात उत्तरे लिहा.

प्रश्न i.
सर्व इंद्रियांनी कोणाचा जयजयकार केला?
उत्तरः
सर्व इंद्रियांनी मेंदूराजेसाहेबांचा जयजयकार केला.

प्रश्न ii.
पोटोबा ‘महाराज’ असा उल्लेख कोणाचा करतात?
उत्तर:
पोटोबा ‘महाराज’ असा उल्लेख ‘मेंदूराजेसाहेबांचा’ करतात.

प्रश्न 3.
आकृतिबंध पूर्ण करा.

i.

ii.

iii.

कृती 3 : व्याकरण कृती

प्रश्न 1.
खालील शब्दांचे विरुद्धार्थी शब्द लिहा.

1. बसा
2. हजर
3. खोटं
4. सेवक

उत्तर:

1. उठा
2. गैरहजर
3. खरं
4. मालक

प्रश्न 2.
खालील वाक्यांतील अधोरेखित शब्दांचे विरोधी शब्द वापरून वाक्ये पुन्हा लिहा.

प्रश्न i.
मी केव्हाच हजर आहे महाराज.
उत्तर:
मी कधीच गैरहजर नसतो महाराज.

प्रश्न ii.
आपण सारेच या शरीररूपी राज्याचे सेवक आहोत.
उत्तर:
आपण सारेच या शरीररूपी राज्याचे मालक आहोत.

प्रश्न 3.
खालील वाक्यांतील अधोरेखित शब्दांचे वचन बदलून वाक्ये पुन्हा लिहा.

प्रश्न i.
जीवनरस मिळाला नाही म्हणून त्यांचे चेहरे बघा कसे सुकून गेलेत.
उत्तरः
जीवनरस मिळाला नाही म्हणून त्यांचा चेहरा बघा कसा सुकून गेलाय.

प्रश्न ii.
ती माझी सवय आहे.
उत्तर:
त्या माझ्या सवयी आहेत.

प्रश्न 4.
खालील वाक्यांत विरामचिन्हांचा वापर करून वाक्य पुन्हा लिहा.

प्रश्न i.
पोटोबा या सर्व मंडळींची तुमच्याबद्दल तक्रार आहे की तुम्ही अजिबात काम करत नाही.
उत्तर:
“पोटोबा, या सर्व मंडळींची तुमच्याबद्दल तक्रार आहे, की तुम्ही अजिबात काम करत नाही.”

प्रश्न ii.
आपण सारे एक आपण सारे एक.
उत्तर:
आपण सारे एक! आपण सारे एक!!

प्रश्न 5.
खालील वाक्यांतील वाक्प्रचार ओळखून लिहा.

प्रश्न i.
महाराज बोलताना जरा बेअदबी होतेय,
उत्तर:
बेअदबी होणे.

प्रश्न ii.
आता आम्ही तुमच्याविषयी कधीही कुरकुर करणार नाही.
उत्तर:
कुरकुर करणे.

कृती 4 : स्वमत

प्रश्न 1.
‘आपण सारे शरीररूपी राज्याचे सेवक आहोत’ या विधानाबाबत तुमचे मत स्पष्ट करा.
उत्तरः
आपले शरीर ही एक संस्था आहे. या संस्थेतील सर्व इंद्रिये एकमेकांवर अवलंबून आहेत. प्रत्येकाचे काम वेगवेगळे आहे. प्रत्येकाने केलेल्या कार्यावर आपले शरीर चालते. पण प्रत्येक इंद्रियाचे कार्य करण्यास लागणारी ऊर्जा ही अन्नाद्वारे मिळते. आपणास भूक लागली की आपण जेवतो, जेवल्यानंतर पोटातील अन्नावर प्रक्रिया होऊन त्याचे रूपांतर ऊर्जेत होऊन त्यावर वेगवेगळी इंद्रिये चालतात व शरीराचे काम चालते.

ज्याप्रमाणे राज्याचा राजा असेल तरच राज्याचे कार्य वेगवेगळ्या अधिकाऱ्यांकडून केले जाते. त्याचप्रमाणे पोटाला भूक लागली की आपण जेवतो. जेवल्यावर अन्नाच्या रूपाने प्रत्येक इंद्रियांना ऊर्जा मिळून शरीराचे कार्य चालते. पोटाला भूकच नाही लागली, आपण जेवलोच नाही, तर कार्य होण्यास लागणारी ऊर्जा न मिळाल्याने शरीराचे कार्य चालणारच नाही. म्हणूनच म्हणावेसे वाटते की, सर्व इंद्रिये ही साऱ्या शरीररूपी राज्याचे सेवक आहेत.

आपण सारे एक Summary in Marathi

पाठपरिचय :

आपले शरीर ही एक परिसंस्था आहे. त्यातील सर्व इंद्रिये या परिसंस्थेचे घटक आहेत. ही सर्व इंद्रिये आपले स्वत:चे नेमून दिलेले काम चोख करत असतात. या सर्व इंद्रियांवर आपले शरीर चालते. या शरीररूपी राज्यातील ही सर्वच इंद्रिये परस्परांना अनुकूल व महत्त्वाची असतात. या सर्व इंद्रियांनी परस्परांची काळजी घेतली तर आरोग्य कसे उत्तम राहते, हे या छोट्या नाटिकेतून सहज सोप्या शब्दांतून लेखिकेने पटवून दिले आहे. पर्यायाने आपले कुटुंब, परिसर, देश यांच्यासाठी हा पाठ एकात्मतेचा संदेश देतो.

Our whole body is an organisation and organs are its elements. All organs function properly. Our body works because of these organs. All organs are very important. If all organs work hand in hand and take care of each other then health will be perfect, this message has been given through this play. Ultimately the message that has been given is for unity of family, locality & Nation.

शब्दार्थ :

1. फेरी – फेरफटका, प्रदक्षिणा – a round, a trip
2. दम – श्वास, धाप – breath, gasping
3. फिरस्ती – प्रवास – travel
4. अवयव – शरीराचे भाग – body part
5. गमतीदार – मनोरंजक, मजेशीर – funny, joyous
6. संवाद – संभाषण – conversation
7. उदास – खिन्न, निराश – gloomy, sad
8. सारी – सर्व – all
9. धडपड – खटपट, जोरदार – struggle
10. प्रयत्न सक्ती – जबरदस्ती, जुलूम – compulsion
11. गुरगुरणे – (पोटात) गुरगुर आवाज होणे – to rumble (in the belly)
12. खवय्ये – खाणारे, खादाड – very greedy
13. कार्य – काम – work
14. राबवणे – एखादयाकडून – to force selfishly
15. जबरदस्तीने काम a person to work
16. करून घेणे – hard
17. नाना – वेगवेगळ्या – different
18. चाखणे – चव घेणे, आस्वाद घेणे – to test
19. दंत – दात – tooth
20. बेअदबी – अपमान – disrespect, insuct
21. कसूर – निष्काळजीपणा – negligence

वाक्प्रचार :

1. पोटोबाची पूजा करणे – खाणे, जेवणे
2. पोटासाठी वणवण हिंडणे – अन्न मिळवण्यासाठी कष्ट करणे
3. फरफट होणे – गैरसोय होणे राबवून घेणे – दुसऱ्याकडून जबरदस्तीने काम
4. करवून घेणे आयते बसून खाणे – काही काम न करता बसून खाणे
5. अद्दल घडणे – शिक्षा होणे बेअदबी होणे – अवमान होणे
6. कसूर माफ करणे – गुन्हा माफ करणे
7. कुरकुर करणे – तक्रार करणे

टिपा :

• आधी पोटोबा, मग विठोबा (एक म्हण) – प्रथम भूक लागली म्हणून जेवणे व नंतर देवाची पूजा करणे.
• संप – अधिक मजुरी, अधिक सवलती मिळवण्यासाठी किंवा अन्यायाविरुद्ध कर्मचारी वर्गाने पुकारलेले कामबंद आंदोलन.