Class 8

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 1.2 8th Std Maths Answers Solutions Chapter 1 Rational and Irrational Numbers.

Rational and Irrational Numbers Class 8 Maths Chapter 1 Practice Set 1.2 Solutions Maharashtra Board

Std 8 Maths Practice Set 1.2 Chapter 1 Solutions Answers

Question 1.
Compare the following numbers.
i. 7, -2
ii. 0, \(\frac { -9 }{ 5 }\)
iii. \(\frac { 8 }{ 7 }\), 0
iv. \(-\frac{5}{4}, \frac{1}{4}\)
v. \(\frac{40}{29}, \frac{141}{29}\)
vi. \(-\frac{17}{20},-\frac{13}{20}\)
vii. \(\frac{15}{12}, \frac{7}{16}\)
viii. \(-\frac{25}{8},-\frac{9}{4}\)
ix. \(\frac{12}{15}, \frac{3}{5}\)
x. \(-\frac{7}{11},-\frac{3}{4}\)
Solution:
i. 7, -2
If a and b are positive numbers such that a < b, then -a > -b.
Since, 2 < 7 ∴ -2 > -7

ii. 0, \(\frac { -9 }{ 5 }\)
On a number line, \(\frac { -9 }{ 5 }\) is to the left of zero.
∴ 0 > \(\frac { -9 }{ 5 }\)

iii. \(\frac { 8 }{ 7 }\), 0
On a number line, zero is to the left of \(\frac { 8 }{ 7 }\) .
∴ \(\frac { 8 }{ 7 }\) > 0

iv. \(-\frac{5}{4}, \frac{1}{4}\)
We know that, a negative number is always less than a positive number.
∴ \(-\frac{5}{4}<\frac{1}{4}\)

v. \(\frac{40}{29}, \frac{141}{29}\)
Here, the denominators of the given numbers are the same.
Since, 40 < 141
∴ \(\frac{40}{29}<\frac{141}{29}\)

vi. \(-\frac{17}{20},-\frac{13}{20}\)
Here, the denominators of the given numbers are the same.
Since, 17 < 13
∴ -17 < -13
∴ \(-\frac{17}{20}<-\frac{13}{20}\)

vii. \(\frac{15}{12}, \frac{7}{16}\)
Here, the denominators of the given numbers are not the same.
LCM of 12 and 16 = 48

Alternate method:
15 × 16 = 240
12 × 7 = 84
Since, 240 > 84
∴ 15 × 16 > 12 × 7

viii. \(-\frac{25}{8},-\frac{9}{4}\)
Here, the denominators of the given numbers are not the same.
LCM of 8 and 4 = 8

ix. \(\frac{12}{15}, \frac{3}{5}\)
Here, the denominators of the given numbers are not the same.
LCM of 15 and 5 = 15

x. \(-\frac{7}{11},-\frac{3}{4}\)
Here, the denominators of the given numbers are not the same.
LCM of 11 and 4 = 44

Maharashtra Board Class 8 Maths Solutions Chapter 1 Rational and Irrational Numbers Practice Set 1.2 Questions and Activities

Question 1.
Verify the following comparisons using a number line. (Textbook pg. no, .3)
i. 2 < 3 but – 2 > – 3
ii. \(\frac{5}{4}<\frac{7}{4}\) but \(\frac{-5}{4}<\frac{-7}{4}\)
Solution:

We know that, on a number line the number to the left is smaller than the other.
∴ 2 < 3 and -3 < -2
i.e. 2 < 3 and -2 > -3

Std 8 Maths Digest

• Practice Set 1.1 Class 8 Answers
• Practice Set 1.2 Class 8 Answers
• Practice Set 1.3 Class 8 Answers
• Practice Set 1.4 Class 8 Answers
• Practice Set 2.1 Class 8 Answers
• Practice Set 2.2Class 8 Answers
• Practice Set 2.3 Class 8 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 17.2 8th Std Maths Answers Solutions Chapter 17 Circle: Chord and Arc.

Circle: Chord and Arc Class 8 Maths Chapter 17 Practice Set 17.2 Solutions Maharashtra Board

Std 8 Maths Practice Set 17.2 Chapter 17 Solutions Answers

Question 1.
The diameters PQ and RS of the circle with centre C are perpendicular to each other at C. State, why arc PS and arc SQ are congruent. Write the other arcs which are congruent to arc PS.

Solution:
diameter PQ ⊥ diameter RS … [Given]
∴ m∠PCS = m∠SCQ = m∠PCR = m∠RCQ = 90°
The measure of the angle subtended at the centre by an arc is the measure of the arc.
∴ m(arc PS) = m∠PCS = 90° …(i)
m (arc SQ) = m∠SCQ = 90° …(ii)
∴ m(arc PS) = m(arc SQ) … [From (i) and (ii)]
∴ arc PS ≅ arc SQ … [If the measures of two arcs of a circle are same, then the two arcs are congruent]
m(arc PR) = m∠PCR = 90° .. .(iii)
m (arc RQ) = m∠RCQ = 90° … (iv)
∴ m(arc PS) = m(arc PR) = m(arc RQ) … [From (i), (iii) and (iv)]
∴ arc PS ≅ arc PR ≅ arc RQ
… [If the measures of two arcs of a circle are same, then the two arcs are congruent]
∴ arc PR and arc RQ are congruent to arc PS.

Question 2.
In the given figure, O is the centre of the circle whose diameter is MN. Measures of some central angles are given in the figure.
i. m∠AOB and m∠COD
ii. Show that arc AB ≅ arc CD
iii. Show that chord AB ≅ chord CD

Solution:
i. Seg MN is the diameter of the circle. … [Given]
∴ m∠AOM + m∠AON = 180° … [Angles in a linear pair]
∴ m∠AOM + (m∠AOB + m∠BON) = 180° … [Angle addition property]
∴ 100° + m∠AOB + 35° = 180°
…[∵ m∠AOM = 100°, m∠BON = 35°]
∴ m∠AOB + 135° = 180°
∴ m∠AOB = 180°- 135°
∴m∠AOB = 45° …(i)
Also, m∠DOM + m∠DON = 180° … [Angles in a linear pair]
∴ m∠DOM + (m∠COD + m∠CON) = 180° … [Angle addition property]
∴ 100° +m∠COD + 35°= 180°
…[∵ m∠DOM = 100°, m∠CON = 35° ]
∴ m∠COD + 135° = 180°
∴ m∠COD = 180°- 135°
∴ m∠COD = 45° …(ii)

ii. m(arc AB) = m∠AOB = 45° … [From (i)]
m(arc DC) = m∠DOC = 45° .. .[From (ii)]
∴ m(arc AB) = m(arc DC) …[From (i) and (ii)]
∴ arc AB ≅ arc CD
… [If the measures of two arcs of a circle are same, then the two arcs are congruent]

iii. arc AB ≅ arc CD
∴ chord AB ≅ chord CD ….[The chords corresponding to congruent arcs are congruent]

Maharashtra Board Class 8 Maths Chapter 17 Circle: Chord and Arc Practice Set 17.2 Intext Questions and Activities

Question 1.
If the measures of two arcs of a circle are same, then two arcs are congruent. Verify this property using tracing paper. (Textbook pg. no. 117)
Solution:
[Students should attempt the above activities on their own.]

Question 2.
With the help of following activity find out the properties of the chord and the corresponding arc.
i. a. Draw a circle with centre O.
b. Draw ∠COD and ∠AOB of same measure.
You will find that the arc AXB and arc CYD are congruent.
c. Draw chords AB and CD.
d. Using compass experience that the length of chord AB and chord CD is also same.
(Textbook pg. no. 117)

Solution:
[Students should attempt the above activities on their own.]

ii. a. Draw a circle with centre C.
b. Draw the congruent chords AB and DE of the circle. Draw the radii CA, CB, CD and CE.
c. Check that ∠ACB and ∠DCE are congruent.
d. Hence show that measure of arc AB and arc DE is equal. Hence these arcs are congruent. (Textbook pg. no. 117)

Solution:
[Students should attempt the above activities on their own.]

Std 8 Maths Digest

• Practice Set 15.1 Class 8 Answers
• Practice Set 15.2 Class 8 Answers
• Practice Set 15.3 Class 8 Answers
• Practice Set 15.4 Class 8 Answers
• Practice Set 15.5 Class 8 Answers
• Practice Set 15.6 Class 8 Answers
• Practice Set 16.1 Class 8 Answers
• Practice Set 16.2 Class 8 Answers
• Practice Set 16.3 Class 8 Answers
• Practice Set 17.1 Class 8 Answers
• Practice Set 17.2 Class 8 Answers

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.

Circle: Chord and Arc Class 8 Maths Chapter 17 Practice Set 17.1 Solutions Maharashtra Board

Std 8 Maths Practice Set 17.1 Chapter 17 Solutions Answers

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.]

Std 8 Maths Digest

• Practice Set 15.1 Class 8 Answers
• Practice Set 15.2 Class 8 Answers
• Practice Set 15.3 Class 8 Answers
• Practice Set 15.4 Class 8 Answers
• Practice Set 15.5 Class 8 Answers
• Practice Set 15.6 Class 8 Answers
• Practice Set 16.1 Class 8 Answers
• Practice Set 16.2 Class 8 Answers
• Practice Set 16.3 Class 8 Answers
• Practice Set 17.1 Class 8 Answers
• Practice Set 17.2 Class 8 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 16.3 8th Std Maths Answers Solutions Chapter 16 Surface Area and Volume.

Surface Area and Volume Class 8 Maths Chapter 16 Practice Set 16.3 Solutions Maharashtra Board

Std 8 Maths Practice Set 16.3 Chapter 16 Solutions Answers

Question 1.
Find the volume of the cylinder if height (h) and radius of the base (r) are as given below.
i. r = 10.5 cm, h = 8 cm
ii. r = 2.5 m, h = 7 m
iii. r = 4.2 cm, h = 5 cm
iv. r = 5.6 cm, h = 5 cm
Solution:
i. Given: r = 10.5 cm and h = 8 cm
To find: Volume of the cylinder
Volume of the cylinder = πr²h
= \(\frac { 22 }{ 7 }\) x 10.5 x 10.5 x 8
= 22 x 1.5 x 10.5 x 8
= 2772 cc
∴ The volume of the cylinder is 2772 cc.

ii. Given: r = 2.5 m and h = 7 m
To find: Volume of the cylinder
Volume of the cylinder = πr²h
= \(\frac { 22 }{ 7 }\) x 2.5 x 2.5 x 7
= 22 x 2.5 x 2.5
= 137.5 cu.m
∴ The volume of the cylinder is 137.5 cu.m.

iii. Given: r = 4.2 cm and h = 5 cm
To find: Volume of the cylinder
Volume of the cylinder = πr²h
= \(\frac { 22 }{ 7 }\) x 4.2 x 4.2 x 5
= 22 x 0.6 x 4.2 x 5
= 277.2 cc
∴ The volume of the cylinder is 277.2 cc.

iv. Given: r = 5.6 cm and h = 5 cm
To find: Volume of the cylinder
Volume of the cylinder = πr²h
= \(\frac { 22 }{ 7 }\) x 5.6 x 5.6 X 5 7
= 22 x 0.8 x 5.6 x 5
= 492.8 cc
∴ The volume of the cylinder is 492.8 cc.

Question 2.
How much iron is needed to make a rod of length 90 cm and diameter 1.4 cm?
Solution:
Given: For cylindrical rod: length of rod (h) = 90 cm, and
diameter (d) = 1.4 cm
To find: Iron required to make a rod
diameter (d) = 1.4 cm
∴ radius (r) = \(\frac{\mathrm{d}}{2}=\frac{1.4}{2}\) = 0.7 cm
Volume of rod = πr²h
= \(\frac { 22 }{ 7 }\) x 0.7 x 0.7 x 90
= 22 x 0.1 x 0.7 x 90
= 138.60 cc
∴ 138.60 cc of iron is required to make the rod.

Question 3.
How much water will a tank hold if the interior diameter of the tank is 1.6 m and its depth is 0.7 m?
Solution:
Given: interior diameter of the tank (d) = 1.6 m
and depth (h) = 0.7 m
To find: Capacity of the tank
interior diameter of the tank (d) = 1.6 m
∴ Interior radius (r) = \(\frac{\mathrm{d}}{2}=\frac{1.6}{2}\)
= 0.8 m
= 0.8 x 100
…[∵ 1m = 100cm]
= 80cm
h = 0.7 m = 0.7 x 100 = 70 cm
Capacity of the tank = Volume of the tank = πr²h
= \(\frac { 22 }{ 7 }\) x 80 x 80 x 70
= 22 x 80 x 80 x 10
= 1408000 cc
= \(\frac { 1408000 }{ 1000 }\)
…[∵1 litre = 1000 cc]
= 1408 litre
∴The tank can hold 1408 litre of water.

Question 4.
Find the volume of the cylinder if the circumference of the base of cylinder is 132 cm and height is 25 cm.
Solution:
Given: Circumference of the base of cylinder = 132 cm and height (h) = 25 cm
To find: Volume of the cylinder
i. Circumference of base of cylinder = 2πr
∴132 = 2 x \(\frac { 22 }{ 7 }\) x r
∴\(\frac{132 \times 7}{2 \times 22}=r\)
∴\(\frac{6 \times 7}{2}=r\)
∴3 x 7 = r
∴r = 21 cm

ii. Volume of the cylinder = πr²h
= \(\frac { 22 }{ 7 }\) x 21 x 21 x 25
= 22 x 3 x 21 x 25
= 34650 cc
∴ The volume of the cylinder is 34650 cc.

Maharashtra Board Class 8 Maths Chapter 16 Surface Area and Volume Practice Set 16.3 Intext Questions and Activities

Question 1.
Leonard Euler, discovered an interesting formula regarding the faces, vertices and edges of solid figures.
Count and write the faces, vertices and edges of the following figures and complete the table. From the table verify Euler’s formula, F + V = E + 2. (Textbook pg. No. 113)

Solution:

From the above table, F + V = E + 2 i.e. Euler’s formula is verified.

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 16.2 8th Std Maths Answers Solutions Chapter 16 Surface Area and Volume.

Surface Area and Volume Class 8 Maths Chapter 16 Practice Set 16.2 Solutions Maharashtra Board

Std 8 Maths Practice Set 16.2 Chapter 16 Solutions Answers

Question 1.
In each example given below, radius of base of a cylinder and its height are given. Then find the curved surface area and total surface area.
i. r = 7 cm, h = 10 cm
ii. r = 1.4 cm, h = 2.1 cm
iii. r = 2.5 cm, h = 7 cm
iv. r = 70 cm, h = 1.4 cm
v. r = 4.2 cm, h = 14 cm
Solution:
i. Given: r = 7 cm and h = 10 cm
To find: Curved surface area of cylinder and total surface area
Curved surface area of the cylinder = 2πrh
= 2 x \(\frac { 22 }{ 7 }\) x 7 x 10
= 2 x 22 x 10
= 440 sq.cm
Total surface area of the cylinder:
= 2πr(h + r)
= 2 x \(\frac { 22 }{ 7 }\) x 7(10 + 7)
= 2 x \(\frac { 22 }{ 7 }\) x 7 x 17
= 2 x 22 x 17
= 748 sq.cm
The curved surface area of the cylinder is 440 sq.cm and its total surface area is 748 sq.cm.

ii. Given: r = 1.4 cm and h = 2.1 cm
To find: Curved surface area of cylinder and total surface area
Curved surface area of the cylinder = 2πrh
= 2 x \(\frac { 22 }{ 7 }\) x 1.4 x 2.1
= 2 x 22 x 0.2 x 2.1
= 18.48 sq.cm
Total surface area of the cylinder = 2πr (h + r)
= 2 x \(\frac { 22 }{ 7 }\) x 1.4 (2.1 + 1.4)
= 2 x \(\frac { 22 }{ 7 }\) x 1.4 x 3.5
= 2 x 22 x 0.2 x 3.5
= 30.80 sq.cm
∴ The curved surface area of the cylinder is 18.48 sq.cm and its total surface area is 30.80 sq.cm.

iii. Given: r = 2.5 cm and h = 7 cm
To find: Curved surface area of cylinder and total surface area
Curved surface area of the cylinder = 2πrh
= 2 x \(\frac { 22 }{ 7 }\) x 2.5 x 7
= 2 x 22 x 2.5
= 110 sq.cm
Total surface area of the cylinder = 2πr(h + r)
= 2 x \(\frac { 22 }{ 7 }\) x 2.5 (7+ 2.5)
= 2 x \(\frac { 22 }{ 7 }\) x 2.5 x 9.5
= \(\frac { 1045 }{ 7 }\)
= 149.29 sq.cm
∴ The curved surface area of the cylinder is 110 sq.cm and its total surface area is 149.29 sq.cm.

iv. Given: r = 70 cm and h = 1.4 cm
To find: Curved surface area of cylinder and total surface area
Curved surface area of the cylinder = 2πrh
= 2 x \(\frac { 22 }{ 7 }\) x 70 x 1.4
= 2 x 22 x 10 x 1.4
= 616 sq.cm
Total surface area of the cylinder = 2πr(h + r)
= 2 x \(\frac { 22 }{ 7 }\) x 70(1.4 + 70)
= 2 x \(\frac { 22 }{ 7 }\) x 70 x 71.4
= 2 x 22 x 10 x 71.4
= 2 x 22 x 714
= 31416 sq.cm
∴ The curved surface area of the cylinder is 616 sq.cm and its total surface area is 31416 sq.cm.

v. Given: r = 4.2 cm and h = 14 cm
To find: Curved surface area of cylinder and total surface area
Curved surface area of the cylinder = 2πrh
= 2 x \(\frac { 22 }{ 7 }\) x 4.2 x 14 = 2 x 22 x 4.2 x 2
= 369.60 sq.cm
Total surface area of the cylinder = 2πr (h + r)
= 2 x \(\frac { 22 }{ 7 }\) x 4.2 (14+ 4.2)
= 2 x \(\frac { 22 }{ 7 }\) x 4.2 x 18.2
= 2 x 22 x 0.6 x 18.2
= 480.48 sq.cm
∴ The curved surface area of the cylinder is 369.60 sq.cm and its total surface area is 480.48 sq.cm.

Question 2.
Find the total surface area of a closed cylindrical drum if its diameter is 50 cm and height is 45 cm. (π = 3.14)
Given: For cylindrical drum:
Diameter (d) = 50 cm
and height (h) = 45 cm
To find: Total surface area of the cylindrical drum
Solution:
Diameter (d) = 50 cm
∴ radius (r) = \(\frac{\mathrm{d}}{2}=\frac{50}{2}\) = 25 cm
Total surface area of the cylindrical drum = 2πr (h + r)
= 2 x 3.14 x 25 (45 + 25)
= 2 x 3.14 x 25 x 70
= 10,990 sq.cm
∴ The total surface area of the cylindrical drum is 10,990 sq.cm.

Question 3.
Find the area of base and radius of a cylinder if its curved surface area is 660 sq.cm and height is 21 cm.
Given: Curved surface area = 660 sq.cm, and height = 21 cm
To find: area of base and radius of a cylinder
Solution:
i. Curved surface area of cylinder = 2πrh
∴ 660 = 2 x \(\frac { 22 }{ 7 }\) x r x 21
∴ 660 = 2 x 22 x r x 3
∴ \(\frac{660}{2 \times 22 \times 3}=r\)
∴ \(\frac{660}{2 \times 66}=r\)
∴ 5 = r
i.e., r = 5 cm

ii. Area of a base of the cylinder = πr²
= \(\frac { 22 }{ 7 }\) x 5 x 5
= \(\frac { 550 }{ 7 }\)
= 78.57 sq.cm
∴The radius of the cylinder is 5 cm and the area of its base is 78.57 sq.cm.

Question 4.
Find the area of the sheet required to make a cylindrical container which is open at one side and whose diameter is 28 cm and height is 20 cm. Find the approximate area of the sheet required to make a lid of height 2 cm for this container.
Given: For cylindrical container:
diameter (d) = 28 cm, height (h₁) = 20 cm
For cylindrical lid: height (h₂) = 2 cm
To find: i. Surface area of the cylinder with one side open
ii. Area of sheet required to made a lid
Solution:
diameter (d) = 28 cm
∴ radius (r) = \(\frac{\mathrm{d}}{2}=\frac{28}{2}\) = 14 cm
i. Surface area of the cylinder with one side open = Curved surface area + Area of a base
= 2πrh₁ + πr²
= πr (2h₁ + r)
= \(\frac { 22 }{ 7 }\) x 14 x (2 x 20 + 14)
= 22 x 2 x (40 + 14)
= 22 x 2 x 54
= 2376 sq.cm

ii. Area of sheet required to made a lid = Curved surface area of lid + Area of upper surface
= 2πrh₂ + πr²
= πr (2h₂ + r)
= \(\frac { 22 }{ 7 }\) x 14 x (2 x 2 + 14)
= 22 x 2 x (4 + 14)
= 22 x 2 x 18
= 792 sq cm
∴ The area of the sheet required to make the cylindrical container is 2376 sq. cm and the approximate area of a sheet required to make the lid is 792 sq. cm.

Std 8 Maths Digest

• Practice Set 15.1 Class 8 Answers
• Practice Set 15.2 Class 8 Answers
• Practice Set 15.3 Class 8 Answers
• Practice Set 15.4 Class 8 Answers
• Practice Set 15.5 Class 8 Answers
• Practice Set 15.6 Class 8 Answers
• Practice Set 16.1 Class 8 Answers
• Practice Set 16.2 Class 8 Answers
• Practice Set 16.3 Class 8 Answers
• Practice Set 17.1 Class 8 Answers
• Practice Set 17.2 Class 8 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 16.1 8th Std Maths Answers Solutions Chapter 16 Surface Area and Volume.

Surface Area and Volume Class 8 Maths Chapter 16 Practice Set 16.1 Solutions Maharashtra Board

Std 8 Maths Practice Set 16.1 Chapter 16 Solutions Answers

Question 1.
Find the volume of a box if its length, breadth and height are 20 cm, 10.5 cm and 8 cm respectively.
Given: For cuboid shaped box,
length (l) = 20 cm, breadth (b) = 10.5 cm and height (h) = 8cm
To find: Volume of a box
Solution:
Volume of a box = l x b x h
= 20 x 10.5 x 8
= 1680 cc
∴ The volume of the box is 1680 cc.

Question 2.
A cuboid shaped soap bar has volume 150 cc. Find its thickness if its length is 10 cm and breadth is 5 cm.
Given: For cuboid shaped soap bar,
length (l) = 10 cm, breadth (b) = 5 cm and volume = 150 cc
To find: Thickness of the soap bar (h)
Solution:
Volume of soap bar = l x b x h
∴ 150 = 10 x 5 x h
∴ 150 = 50h
∴ \(\frac { 150 }{ 50 }=h\)
∴ 3 = h
i.e., h = 3 cm
∴ The thickness of the soap bar is 3 cm.

Question 3.
How many bricks of length 25 cm, breadth 15 cm and height 10 cm are required to build a wall of length 6 m, height 2.5 m and breadth 0.5 m?
Given: For the cuboidal shape brick:
length (l₁) = 25 cm,
breadth (b₁) = 15 cm,
height (h₁) = 10 cm
For the cuboidal shape wall:
length (l₂) = 6 m,
height (h₂) = 2.5 m,
breadth (b₂) = 0.5 m
To find: Number of bricks required
Solution:
When all the bricks are arranged to build a wall, the volume of all the bricks is equal to volume of wall.
∴ \(\text { Number of bricks }=\frac{\text { volume of the wall }}{\text { volume of a brick }}\)

i. Volume of a brick = l₁ x b₁ x h₁
= 25 x 15 x 10 cc

ii. l₂ = 6m = 6 x 100 …[∵ 1m = 100cm]
= 600 cm
h₂ = 2.5 m = 2.5 x 100 = 250 cm
b₂ = 0.5 m = 0.5 x 100 = 50 cm
Volume of the wall = l₂ x b₂ x h₂
= 600 x 50 x 250 cc

iii. \(\text { Number of bricks }=\frac{\text { volume of the wall }}{\text { volume of a brick }}\)
= \(\frac{600 \times 50 \times 250}{25 \times 15 \times 10}\)
= 40 x 2 x 25
= 2000 bricks
∴ 2000 bricks are required to build the wall.

Question 4.
For rain water harvesting a tank of length 10 m, breadth 6 m and depth 3 m is built. What is the capacity of the tank? How many litre of water can it hold?
Given: For a cuboidal tank,
Length (l) = 10 m, breadth (b) = 6 m, depth (h) = 3 m
To find: Capacity of the tank and litre of water tank can hold.
Solution:
i. l = 10m = 10 x 100 …[∵ 1m = 100cm]
= 1000 cm,
b = 6 m = 6 x 100 = 600 cm,
h = 3 m = 3 x 100 = 300 cm
Volume of the tank = l x b x h
= 1000 x 600 x 300
= 18,00,00,000 cc

ii. Capacity of the tank = Volume of the tank
= 18,00,00,000 cc
= \(\frac{18,00,00,000}{1000}\)
…[∵ 1 litre =1000 cc]
= 1,80,000 litre
∴ The capacity of the tank is 18,00,00,000 cc and it can hold 1,80,000 litre of water.

Std 8 Maths Digest

• Practice Set 15.1 Class 8 Answers
• Practice Set 15.2 Class 8 Answers
• Practice Set 15.3 Class 8 Answers
• Practice Set 15.4 Class 8 Answers
• Practice Set 15.5 Class 8 Answers
• Practice Set 15.6 Class 8 Answers
• Practice Set 16.1 Class 8 Answers
• Practice Set 16.2 Class 8 Answers
• Practice Set 16.3 Class 8 Answers
• Practice Set 17.1 Class 8 Answers
• Practice Set 17.2 Class 8 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 15.6 8th Std Maths Answers Solutions Chapter 15 Area.

Area Class 8 Maths Chapter 15 Practice Set 15.6 Solutions Maharashtra Board

Std 8 Maths Practice Set 15.6 Chapter 15 Solutions Answers

Question 1.
Radii of the circles are given below, find their areas.
i. 28 cm
ii. 10.5 cm
iii. 17.5 cm
Solution:
i. Radius of the circle (r) = 28 cm … [Given]
Area of the circle = πr²
= \(\frac { 22 }{ 7 }\) x (28)²
= \(\frac { 22 }{ 7 }\) x 28 x 28
= 22 x 4 x 28
= 2464 sq. cm

ii. Radius of the circle (r) = 10.5 cm … [Given]
Area of the circle = πr²
= \(\frac { 22 }{ 7 }\) x (10.5)²
= \(\frac { 22 }{ 7 }\) x 10.5 x 10.5
= 22 x 1.5 x 10.5
= 346.5 sq. cm

iii. Radius of the circle (r) = 17.5 cm … [Given]
Area of the circle = πr²
= \(\frac { 22 }{ 7 }\) x(17.5)²
= \(\frac { 22 }{ 7 }\) x 17.5 x 17.5
= 22 x 2.5 x 17.5
= 962.5 sq. cm

Question 2.
Areas of some circles are given below, find their diameters.
i. 176 sq.cm
ii. 394.24 sq. cm
iii. 12474 sq. cm
Solution:
i. Area of the circle =176 sq. cm .. .[Given]
Area of the circle = πr²
∴ 176 = \(\frac { 22 }{ 7 }\) x r²
∴ r² = 176 x \(\frac { 7 }{ 22 }\)
∴ r² = 56
∴ r = √56 … [Taking square root of both sides]
Diameter = 2r = 2√56 CM

ii. Area of the circle = 394.24 sq. cm … [Given]
Area of the circle = πr²

∴ Diameter = 2r = 2 x 11.2 = 22.4 cm

iii. Area of the circle = 12474 sq. cm …[Given]
Area of the circle = πr²
∴ 12474 = \(\frac { 22 }{ 7 }\) x r²
∴ r² = 12474 x \(\frac { 7 }{ 22 }\)
∴ r² = 567 x 7
∴ r² = 3969
∴ r = 63 …[Taking square root of both sides]
∴ Diameter = 2r = 2 x 63 = 126cm

Question 3.
Diameter of the circular garden is 42 m. There is a 3.5 m wide road around the garden. Find the area of the road.

Solution:

Diameter of the circular garden is 42 m. … [Given]
∴ Radius of the circular garden (r) = \(\frac { 42 }{ 2 }\) = 21 m
Width of the road = 3.5 m …[Given]
Radius of the outer circle (R)
= radius (r) + width of the road
= 21 + 3.5
= 24.5 m
Area of the road = area of outer circle – area of circular garden
= πR² – πr²
= π (R² – r²)
= \(\frac { 22 }{ 7 }\) [(24.5)² – (21)²]
= \(\frac { 22 }{ 7 }\) (24.5 + 21) (24.5 – 21)
…..[∵ a²-b² = (a+b)(a-b)]
= \(\frac { 22 }{ 7 }\) x 45.5 x 3.5
= 22 x 45.5 x 0.5
= 500.50 sq. m
∴ The area of the road is 500.50 sq. m.

Question 4.
Find the area of the circle if its circumference is 88 cm.
Solution:
Circumference of the circle = 88 cm …[Given]
Circumference of the circle = 2πr
∴ 88 = 2 x \(\frac { 22 }{ 7 }\) x r
∴ \(r=\frac{88 \times 7}{2 \times 22}\) ∴ r = 14cm
Area of the circle = πr² = \(\frac { 22 }{ 7 }\) x (14)²
= \(\frac { 22 }{ 7 }\) x 14 x 14 = 22 x 2 x 14 = 616 sq. cm
∴ The area of circle is 616 Sq cm

Maharashtra Board Class 8 Maths Chapter 15 Area Practice Set 15.6 Intext Questions and Activities

Question 1.
Draw a circle of radius 28mm. Draw any one triangle and draw a trapezium on the graph paper. Find the area of these figures by counting the number of small squares on the graph paper. Verify your answers using formula for area of these figures.
Observe that smaller the squares of graph paper, better is the approximation of area. (Textbook pg. no. 105)
Solution:
(Students should do this activity on their own.)

Std 8 Maths Digest

• Practice Set 15.1 Class 8 Answers
• Practice Set 15.2 Class 8 Answers
• Practice Set 15.3 Class 8 Answers
• Practice Set 15.4 Class 8 Answers
• Practice Set 15.5 Class 8 Answers
• Practice Set 15.6 Class 8 Answers
• Practice Set 16.1 Class 8 Answers
• Practice Set 16.2 Class 8 Answers
• Practice Set 16.3 Class 8 Answers
• Practice Set 17.1 Class 8 Answers
• Practice Set 17.2 Class 8 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 15.5 8th Std Maths Answers Solutions Chapter 15 Area.

Area Class 8 Maths Chapter 15 Practice Set 15.5 Solutions Maharashtra Board

Std 8 Maths Practice Set 15.5 Chapter 15 Solutions Answers

Question 1.
Find the areas of given plots. (All measures are in meters.)

Solution:
i. Here, ∆QAP, ∆RCS are right angled triangles and ☐QACR is a trapezium.
In ∆QAP, l(AP) = 30 m, l(QA) = 50 m
A(∆QAP)
= \(\frac { 1 }{ 2 }\) x product of sides forming the right angle
= \(\frac { 1 }{ 2 }\) x l(AP) x l(QA)
= \(\frac { 1 }{ 2 }\) x 30 x 50
= 750 sq. m
In ☐QACR, l(QA) = 50 m, l(RC) = 25 m,
l(AC) = l(AB) + l(BC)
= 30 + 30 = 60 m
A(☐QACR)
= \(\frac { 1 }{ 2 }\) x sum of lengths of parallel sides x height
= \(\frac { 1 }{ 2 }\) x [l(QA) + l(RC)] x l(AC)
= \(\frac { 1 }{ 2 }\) x (50 + 25) x 60
= \(\frac { 1 }{ 2 }\) x 75 x 60
= 2250 sq.m
In ∆RCS, l(CS) = 60 m, l(RC) = 25 m A(∆RCS)
= \(\frac { 1 }{ 2 }\) x product of sides forming the right angle
= \(\frac { 1 }{ 2 }\) x l(CS) x l(RC)
= \(\frac { 1 }{ 2 }\) x 60 x 25
= 750 sq. m
In ∆PTS, l(TB) = 30 m,
l(PS) = l(PA) + l(AB) + l(BC) + l(CS)
= 30 + 30 + 30 + 60
= 150m
A(∆PTS) = \(\frac { 1 }{ 2 }\) x base x height
= \(\frac { 1 }{ 2 }\) x l(PS) x l(TB)
= \(\frac { 1 }{ 2 }\) x 150 x 30
= 2250 sq. m
∴ Area of plot QPTSR = A(∆QAP) + A(☐QACR) + A(∆RCS) + A(∆PTS)
= 750 + 2250 + 750 + 2250
= 6000 sq. m
∴ The area of the given plot is 6000 sq.m.

ii. In ∆ABE, m∠BAE = 90°, l(AB) = 24 m, l(BE) = 30 m
∴ [l(BE)]² = [l(AB)]² + [l(AE)]²
…[Pythagoras theorem]
∴ (30)² = (24)² + [l(AE)]²
∴ 900 = 576 + [l(AE)]²
∴ [l(AE)]² = 900 – 576
∴ [l(AE)]² = 324
∴ l(AE) = √324 = 18 m
…[Taking square root of both sides]
A(∆ABE)
= \(\frac { 1 }{ 2 }\) x product of sides forming the right angle
= \(\frac { 1 }{ 2 }\) x l(AE) x l(AB)
= \(\frac { 1 }{ 2 }\) x 18 x 24
= 216 sq. m
In ∆BCE, a = 30m, b = 28m, c = 26m

∴ Area of plot ABCDE
= A(∆ABE) + A(∆BCE) + A(∆EDC)
= 216 + 336 + 224
= 776 sq. m
∴ The area of the given plot is 776 sq.m.
[Note: In the given figure, we have taken l(DF) = 16 m]

Std 8 Maths Digest

• Practice Set 15.1 Class 8 Answers
• Practice Set 15.2 Class 8 Answers
• Practice Set 15.3 Class 8 Answers
• Practice Set 15.4 Class 8 Answers
• Practice Set 15.5 Class 8 Answers
• Practice Set 15.6 Class 8 Answers
• Practice Set 16.1 Class 8 Answers
• Practice Set 16.2 Class 8 Answers
• Practice Set 16.3 Class 8 Answers
• Practice Set 17.1 Class 8 Answers
• Practice Set 17.2 Class 8 Answers

Balbharti Maharashtra State Board Class 8 Maths Solutions covers the Practice Set 15.4 8th Std Maths Answers Solutions Chapter 15 Area.

Area Class 8 Maths Chapter 15 Practice Set 15.4 Solutions Maharashtra Board

Std 8 Maths Practice Set 15.4 Chapter 15 Solutions Answers

Question 1.
Sides of a triangle are 45 cm, 39 cm and 42 cm, find its area.
Solution:
Sides of a triangle are 45 cm, 39 cm and 42 cm.
Here, a = 45cm, b = 39cm, c = 42cm
Semi perimeter of triangle = s = \(\frac { 1 }{ 2 }(a+b+c)\)
= \(\frac { 1 }{ 2 }(45+39+42)\)
= \(\frac { 126 }{ 2 }\)
= 63
Area of a triangle

∴ The area of the triangle is 756 sq.cm.

Question 2.
Look at the measures shown in the given figure and find the area of ☐PQRS.

Solution:
A (☐PQRS) = A(∆PSR) + A(∆PQR)
In ∆PSR, l(PS) = 36 m, l(SR) = 15 m
A(∆PSR)
= \(\frac { 1 }{ 2 }\) x product of sides forming the right angle
= \(\frac { 1 }{ 2 }\) x l(SR) x l(PS)
= \(\frac { 1 }{ 2 }\) x 15 x 36
= 270 sq.m
In ∆PSR, m∠PSR = 90°
[l(PR)]² = [l(PS)]² + [l(SR)]²
…[Pythagoras theorem]
= (36)² + (15)²
= 1296 + 225
∴ l(PR)² = 1521
∴ l(PR) = 39m
…[Taking square root of both sides]
In ∆PQR, a = 56m, b = 25m, c = 39m

A(☐PQRS) = A(∆PSR) + A(∆PQR)
= 270 + 420
= 690 sq. m
∴ The area of ☐PQRS is 690 sq.m

Question 3.
Some measures are given in the figure, find the area of ☐ABCD.

Solution:
A(☐ABCD) = A(∆BAD) + A(∆BDC)
In ∆BAD, m∠BAD = 90°, l(AB) = 40m, l(AD) = 9m
A(∆BAD) = \(\frac { 1 }{ 2 }\) x product of sides forming the right angle
= \(\frac { 1 }{ 2 }\) x l(AB) x l(AD)
= \(\frac { 1 }{ 2 }\) x 40 x 9
= 180 sq. m
In ∆BDC, l(BT) = 13m, l(CD) = 60m
A(∆BDC) = \(\frac { 1 }{ 2 }\) x base x height
= \(\frac { 1 }{ 2 }\) x l(CD) x l(BT)
= \(\frac { 1 }{ 2 }\) x 60 x 13
= 390 sq. m
A (☐ABCD) = A(∆BAD) + A(∆BDC)
= 180 + 390
= 570 sq. m
∴ The area of ☐ABCD is 570 sq.m.

Std 8 Maths Digest

• Practice Set 15.1 Class 8 Answers
• Practice Set 15.2 Class 8 Answers
• Practice Set 15.3 Class 8 Answers
• Practice Set 15.4 Class 8 Answers
• Practice Set 15.5 Class 8 Answers
• Practice Set 15.6 Class 8 Answers
• Practice Set 16.1 Class 8 Answers
• Practice Set 16.2 Class 8 Answers
• Practice Set 16.3 Class 8 Answers
• Practice Set 17.1 Class 8 Answers
• Practice Set 17.2 Class 8 Answers