KTBS Solutions

KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.3 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.3.

Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.3

Question 1.
State which pairs of triangles In the following figures are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:

Solution:
(i)

In ∆ABC and ∆PQR,
∠A = ∠P = 60°
∠B=∠Q = 80°
∠C =∠R = 40°
∴ SimilarIty criterion of two triangles is A.A.A.
∴ ∆ABC ~ ∆PQR.

(ii)

In ∆ABC and PQR, we have
\(\frac{A B}{Q R}=\frac{B C}{P R}=\frac{A C}{P Q}=\frac{1}{2}\)
i.e., sides of one triangle are proportional to the sides of the other triangle.
∴SImilarity criterion is S.S.S.
∴ ∆ABC ~ ∆PQR.

(iii)

∆LMP and ∆DEF are not similar triangles because one pair of sides are not proportional to each other.
\(\quad \frac{\mathrm{MP}}{\mathrm{DE}}=\frac{\mathrm{LP}}{\mathrm{DF}}=\frac{2}{4}=\frac{3}{6}=\frac{1}{2}\)
But, \(\frac{\mathrm{LM}}{\mathrm{EF}}=\frac{2.7}{5} \neq \frac{1}{2}\)
∴ These are not similar triangles

(iv)

In ∆MNL and ∆PQR, we have
∠M = ∠Q = 70°
But, \(\frac{\mathrm{MN}}{\mathrm{PQ}} \neq \frac{\mathrm{ML}}{\mathrm{QR}}\)
∴ These are not similar triangles

(v)

In ∆ABC and ∆DEF, we have
\(\frac{A B}{D F}=\frac{B C}{E F}=\frac{1}{2}\)
∠A = ∠F = 80°
∴ Similarly criterion for these triangles is S.A.S
∴ ∆ABC ~ ∆DEF

(vi)

Sum of three angles of is 180°, but in ∆DEF. sum of two angles. ∠F= 30°. In ∆PQR, ∠P = 70°.
In ∆DEF and ∆PQR. we have
∠D = ∠P = 70°
∠E =∠R = 80°
∠F = ∠R = 30°
∴ Three angles of these triangles are equal to each other.
∴ SimIlarity criterion here is A.A.A.
∴ ∆DEF ~ ∆PQR.

Question 2.
In the following figure. ∆OBA ~ ∆ODC, ∠BOC = 125° and ∠CDO = 70°. Find ∠DOC, ∠DCO and ∠OAB.

Solution:
∆OBA ~ ∆ODC (data)
In ∆OBA and ∆ODC, we have
∠ODC = ∠OBA = 70° (. Alternate angle)
∠AOB = 180°- 125° = 55° ( Adjacent angle)
∴ ∠AOB = ∠DOC = 55° (: VertIcally opposite angles)
∠OAB = ∠OCD = 55° (. Alternate angles)
∴ ∠DOC=55°
∠DCO = 55%
∠OAB= 55%.

Question 3.
Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that \(\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}\)

Solution:
Data: Diagonals AC and BD of a trapezium ABCD with AB || DC. intersect each other at the point O.
To Prove: \(\frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}\)
In the trapezium ABCD AB || DC.
∴ ∠OCD = ∠OAB (Alternate angles)
∠ODC = ∠OBA (Alternate angles)
∠DOC = ∠AOB (Vertically opposite angles)
∴ These are equlangular triangles.
∴ These triangles are similar.
∴ ∆ODC ~ ∆OAB
\(\quad \frac{\mathrm{OA}}{\mathrm{OC}}=\frac{\mathrm{OB}}{\mathrm{OD}}\)
∵ Corresponding sides are in proportional.

Question 4.
In the following figure. \(\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PR}}\) and ∠1 = ∠2. Show that ∆PQS ~ ∆TQR.

Solution:
Data: Here \(\frac{\mathrm{QR}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PR}}\)and ∠1 = ∠2
To prove: ∆PQS ~ ∆TQR.
In ∆PQS and ∆TQR, we have
\(\frac{\mathrm{Q} \mathrm{R}}{\mathrm{QS}}=\frac{\mathrm{QT}}{\mathrm{PR}}\)
∠PQR = ∠PRQ (∵ ∠1 = ∠2)
Here, Similarity criterion used here is side, angle, side (SAS).
∴ ∆PQS ~ ∆TQR.

Question 5.
S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS. Show that ∆RPQ ~ ∆RTS.

Solution:
Data: S and T are points on sides PR and QR of ∆PQR such that ∠P = ∠RTS.
To Prove: ∆RPQ ~ ∆RTS.
In ∆RPQ and ∆RTS,
∠P = ∠RTS (Data)
∠PRQ = ∠SRT (Common)
∴ 3rd angle ∠PRQ = ∠SRT
∴ These are equiangularangular triangles.
∴ Here A.A.A. similarity criterion.
∴ ∆RPQ ~ ∆RTS.

Question 6.
In the given fIgure. ∆ABE ≅ ∆ACD. show that ∆ADE ~ ∆ABC.

Solution:
Data: ∆ABE ≅ ∆ACD.
To Prove: ∆ADE ~ ∆ABC.
∆ABE ≅ ∆ACD
AB = AC
AD = AE.
Then DC = BE
AB = AC
AD + DB = AE + EC
∴ DB = EC
(∵ DA = AE)
In ∆ADE and ∆ABC,
\(\frac{\mathrm{AD}}{\mathrm{AB}}=\frac{\mathrm{AE}}{\mathrm{AC}}=\frac{\mathrm{DE}}{\mathrm{BC}}\)
Here, corresponding sides are in proportion.
∴ Similarity criterion for ∆ is side, side, side
∴ ∆ADE ~ ∆ABC

Question 7.
In following figure. altitudes AD and CE of ∆ABC Intersect each other at the point P. Show that:

i) ∆AEP ~ ∆CDP
ii) ∆ABD ~ ∆CBE
iii) ∆AEP ~ ∆AÐB
iv) ∆PDC ~ ∆BEC.
Solution:
Data: altitudes AD and CE of ABC
Intersect each other at the point P.
To Prove: i) ∆AEP ~ ∆CDP
ii) ∆ABD ~ ∆CBE
iii) ∆AEP ~ ∆ADB
iv) ∆PDC ~ ∆BEC

i) In ∆AEP and ∆CDP.
∠AEP = ∠CDP = 90° (data)
∠APE = ∠CPD (Vertically opposite angle)
∴ ∠PAE = ∠PCD
These are equiangular triangles.
Similarity criterion for ∆ is A.A.A
∴ ∆AEP ~ ∆CDP

(ii) In ∆ABD and ∆CBE.
∠ADB = ∠CEB = 90° (data)
∠ABD = ∠CBE (common)
∴ ∠DAB = ∠BCE
These are equiangular triangles.
∴ Similarity criterion for is A.A.A.
∴ ∆ABD ~ ∆CBE

(iii) In ∆AEP and ∆ADB.
∠AEB = ∠ADB = 90° (data)
∠PAE = ∠DAB (common)
∴ ∠APE = ∠ABD.
∴ These are equiangular triangles.
∴ Similarity criterion for is A.A.A.
∴ ∆AEP ~ ∆ADB

(iv) In ∆PDC and ∆BEC.
∠PDC = ∠BEC = 90° (data)
∠PCD = ∠BCE (common)
∴ ∠CPD = ∠CBE
∴ These are equiangularangular triangles.
Similarity criterion for ∆ is A.A.A.
∴ ∆PDC ~ ∆BEC.

Question 8.
E is a point on the side AD produced of a parallelogram ABCD and BE Intersects CD at F. Show that ∆ABE ~ ∆CFB.

Solution:
Data: E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F.
To Prove: ∆ABE ~ ∆CFB
In \(\) ABCD Adjacent angles are equal.
Let ∠DAB = ∠BCD = 70°
∠DAB = ∠EAF = 70° (∵ Corresponding angle)
In ∆EDF, ∠DEF = 30° then,
∠EFD = 80°.
∠EFD = ∠BFC = 80° (vertIcally opposite angles)
In ∆FBC, ∠FBC = 30°. ,
Now in ∆ABE and ∆CFB,
∆EAB = ∆BCF = 70°
∆AEB = ∆FBC = 30°
∆ABE = ∆BFC = 80°
∴ Similarity criterion for ∆ is A.A.A.
∴ ∆ABE ~ ∆CFB

Question 9.
In the given figure. ∆ABC and ∆AMP are two right triangles, right angled at B and M respectively. Prove that:
i) ∆ABC ~ ∆AMP
ii) \(\frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}}\)

Solution:
Data: ∆ABC and ∆AMP are two right triangles, right angled at B and M respectvely.
To Prove: i) ∆ABC ~ ∆AMP
ii) \(\frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}}\)
(i) In ∆ABC and ∆AMP,
∠ABC = ∠AMP = 90° (data)
∠CAB = ∠MAP (common)
∴ ∠ACB = ∠MPA
∴ These are equiangular triangles.
Similarity criterion for ∆ is A.A.A.
∴ ∆ABC ~ ∆AMP

(ii) ∆ABC ~ ∆AMP(Proved)
∴ Corresponding sides are in proportion.
LC and LP are corresponding angles.
∴ Adjacent sides are CA, PA.
Similarly BC and MP are adjacent sides.
\(\quad \frac{\mathrm{CA}}{\mathrm{PA}}=\frac{\mathrm{BC}}{\mathrm{MP}}\)

Question 10.
CD and OH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ∆ABC and ∆EFG respectively.

If ∆ABC ~ ∆EFG, show that
i) \(\frac{C D}{G H}=\frac{A C}{F G}\)
ii) ∆DCB ~ ∆HGE
iii) ∆DCA ~ ∆HGF
Solution:
Data : CD and OH are respectively the bisectors of ∠ACB and ∠EGF such that D and H 11e on sides AB and FE of ∆ABC and ∆EFG respectively.
∆ABC ~ ∆EFG
To Prove: i) \(\frac{C D}{G H}=\frac{A C}{F G}\)
ii) ∆DCB ~ ∆HGE
iii) ∆DCA ~ ∆HGF
Proof: ∆ABC ~ ∆EFG (data given)
∴ Their corresponding sides are in proportion.
\(\quad \frac{\mathrm{AB}}{\mathrm{EF}}=\frac{\mathrm{BC}}{\mathrm{FG}}=\frac{\mathrm{AC}}{\mathrm{EG}}\)
∠B = ∠F, ∠A = ∠E, ∠C = ∠G.

(i) In ∆ADC and ∆EHG,
∠A = ∠E .
∠ACD = ∠EGH
∴ Their sides are in proportion.
\(\frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{AC}}{\mathrm{FG}}\)

(ii) In ∆DCB and ∆HGE.
\(\frac{\mathrm{CD}}{\mathrm{GH}}=\frac{\mathrm{BC}}{\mathrm{GF}}\)
∴ ∆DCB ~ ∆HGE

(iii) In ∆DCA and ∆HGE,
\(\frac{\mathrm{DC}}{\mathrm{GH}}=\frac{\mathrm{AD}}{\mathrm{EH}}=\frac{\mathrm{AC}}{\mathrm{EG}}\)
∴ ∆DCA ~ ∆HGE

Question 11.
In the following figure. E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ∆ABD ~ ∆ECF

Solution:
Data: E Is a point on side CB produced of an isosceles ∆ABC with AB = AC. AD ⊥ BC and EF ⊥ AC,
To Prove: ∆ABD ~ ∆ECF
In ∆ABD and ∆ECF.
∠ADB = ∠EFC = 90° (data)
∠ABD = ∠FCE (∵ ∠B = ∠C)
∠BAD = ∠FEC
∴ Equiangular triangles.
∴ Similarity criterion for triangles is A.A.A.
∴ ∆ABD ~ ∆ECF.

Question 12.
Sides AB and BC and median AD of a ¿ABC are respectively proportional to sides PQ and QR and median PM of ∆PQR. (Sec figure gIven below) Show that ∆ABC ~ ∆PQR

Solution:
Data: Sides AB and BC and median AD of a ∆ABC are respectively proportional to sides P9 and QR and medIan PM of ∆PQR.
To Prove: ∆ABC ~ ∆PQR
In ∆ABC and ∆PQR,
Side \(\frac{\mathrm{AB}}{\mathrm{PG}}=\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)
D is the mid-point of BC
∴ BD = DC
M is the mid-point of QR.
∴ QM = MR.
In ∆ADC and ∆PMR,
\(\frac{\mathrm{AD}}{\mathrm{PM}}=\frac{\mathrm{DC}}{\mathrm{MR}}=\frac{\mathrm{AC}}{\mathrm{PR}} \quad \left(\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\frac{1}{2} \mathrm{BC}}{\frac{1}{2} \mathrm{QR}}\right)\)
Similarity criterion is S.S.S.
∴ ∆AÐC ~ ∆PMR
Now, In ∆ABC and ∆PQR.
\(\frac{A B}{P Q}=\frac{B C}{Q R}=\frac{A C}{P R}\)
∴ SimilarIty criterion is S.S.S.
∴ ∆ABC ~ ∆PQR.

Question 13.
D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA² = CB.CD.

Solution:
Data: D is a point on the side BC of a triangle ∆ABC such that ∠ADC = ∠BAC.
To Prove: CA² = CB × CD
Let ∠ADC = ∠BAC = 100°
In ∆ABC, If ∠B = 50°,then ∠C = 30°
In ∆ADC, If ∠C = 30°. then ∠DAC = 50°
In ∆BCP, ∠A= 100°, ∠B= 50°, ∠C= 30°
In ∆ADC, ∠ADC = 100. ∠DAC = 50°.
∠ACD = 30°
Similarity criterion of ∆ is A.A.A.
∴ In ∆ABC and ∆ADC,
\(\frac{\mathrm{CA}}{\mathrm{BC}}=\frac{\mathrm{DC}}{\mathrm{CA}}\)
∴ CA × CA = BC × DC
∴ CA² = BC × DC.

Question 14.
Sides AB and AC and median AD of a triangle ∆ABC are respectively proportional to sides PQ and PR and mediam PM of another triangle ∆PQR. Show that ∆ABC ~ ∆PQR.

Solution:
Data: Sides AB and AC and median AD of a triangle ∆ABC are respectively proportional to sides PQ and PR and medIan PM of another triangle ∆PQR.
To Prove: ∆ABC ~ ∆PQR
In ∆ABC and ∆PQR
\(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)
Corresponding sides are proportional.
Similarity criterion for ∆ is S.S.S
∴ ∆ABC ~ ∆PQR.

Question 15.
A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Solution:
Height of PQ =?
In ∆ABC and ∆PQR,
∠B = ∠Q = 90°
∴ Hypotenuse AC is proportional to PQ
\(\quad \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BC}}{\mathrm{MR}}=\frac{\mathrm{AC}}{\mathrm{PR}}\)
\(\frac{6}{\mathrm{PQ}}=\frac{4}{28}\)
∴ 4 × PQ = 6 × 28
\(\quad \mathrm{PQ}=\frac{6 \times 28}{4} \quad \quad \mathrm{PQ}=42 \mathrm{m}\)

Question 16.
If AD and PM are medians of triangles ∆ABC and ∆PQR respectively, where ∆ABC ~ ∆PQR. prove that \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)

Solution:
Data : AD and PM are medians of triangles ∆ABC and ∆PQR respectively. ∆ABC ~ ∆PQR.
To Prove: \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)
∆ABC ~ ∆PQR.(data)
∴ we have \(\frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AC}}{\mathrm{PR}}=\frac{\mathrm{BC}}{\mathrm{QR}}\)
D is the mid-point of BC. BD = DC.
M Is the mid-point of QR. QM = MR
\(\frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\frac{1}{2} \mathrm{BC}}{\frac{1}{2} \mathrm{QR}}=\frac{\mathrm{BD}}{\mathrm{QM}}\)
\(\quad \frac{\mathrm{BC}}{\mathrm{QR}}=\frac{\mathrm{BD}}{\mathrm{QM}}\)
In ∆ABD and ∆PQM, we have
\(\quad \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{BD}}{\mathrm{QM}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)
\(\quad \frac{\mathrm{AB}}{\mathrm{PQ}}=\frac{\mathrm{AD}}{\mathrm{PM}}\)

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.3 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.3, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Exercise 5.2.

Karnataka SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Exercise 5.2

{Unless stated otherwise, use \(\pi=\frac{22}{7}\)}

Question 1.
Find the area of a sector of a circle with radius 6 cm if angle of the sector is 60°.
Solution:
Radius of a sector of a circle, r = 6 cm.
Angle of a sector of a circle, θ = 60°
Area of a sector of a circle = ?
Area of a sector of a circle \(=\frac{\theta}{360} \times \pi r^{2}\)

= 18.8 sq.cm.

Question 2.
Find the area of a quadrant of a circle whose circumference is 22 cm.
Solution:
Circumference of a circle, 2πr = 22 cm.
Area of a quadrant of a circle = ?
2πr = 22

∴ Area of a quadrant of a circle

\(=\frac{77}{8}\)
= 9.62 sq. cm.

Question 3.
The length of the minute hand of a clock is 14 cm. Find the area swept by the minute hand in 5 minutes.
Solution:
Length of the minute hand of a clock, r = 14 cm.
Area swept by the minute hand in 5 minutes \(=360 \times \frac{5}{60}\)
= 30°
Angle of segment of a circle, 0 = 30°
∴ Area swept by the minute hand in 5 minute,

Question 4.
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding:
(i) minor segment, (ii) major sector. (Use π = 3.14)
Solution:
Radius of the circle, r = 10 cm.
Angle at the centre of circle, θ = 90°
1) Area of minor segment = ?
2) Area of Major segment = ?
(i) Area of minor segment OAPB

ii) Area of Minor Segment, APB :
= Area of OAPB – Area of AOAB
\(=78.5 \mathrm{cm}^{2} \cdot-\frac{1}{2} \times \mathrm{OA} \times \mathrm{OB}\)
\(=78.5-\frac{1}{2} \times 10 \times 10\)
= 78.5 – 50
= 28.5 sq. cm.

(iii) Area of the Major Segment (AQBO):

Question 5.
In a circle of radius 21 cm, an arc subtends an angle of 60° at the Centre. Find:
(i) the length of the arc,
(ii) area of the sector formed by the arc,
(iii) area of the segment formed by the corresponding chord.
Solution:
Radius of the circle, r = 21 cm.
Angle at the centre of the circle, θ = 60°
1) the length of the arc?
2) Area of the sector formed by the arc?
3) Area of the segment.

1) The length of arc OAPB

= 22 cm

2) Area of the sector OAPB

= 231 sq. cm.

3) Area of the segment APB:
= Area of the sector OAPB – Area of the equilateral ∆OAB

= 231 – 190.7
= 40.3 sq. cm

Question 6.
A chord of a circle of radius 15 cm subtends an angle of 60° at the centre. Find the areas of the corresponding minor and major segments of the circle.
(Use π = 3.14 and \(\sqrt{3}=1.73\)).
Solution:
Radius of circle, r = 15 cm.
Angle at the centre, θ = 60°
1) Area of the minor segment =?
2) Area of the Major segment =?

(i) Area of Minor segment, APB :
= Area of sector OAPB – Area of ∆OAB

= 117.75 sq. cm. – 97.31
= 20.44 sq. cm.

ii) Area of the Major segment:
= Area of Circle – Area of a segment of circle OAPB

= 3.14 × 225 – 20.44
= 707.14 – 20.44
= 686.06 cm2.

Question 7.
A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle.
(Use π = 3.14 and \(\sqrt{3}=1.73\)).
Solution:
Radius of the circle, r = 12 cm.
Angle at the centre, θ = 120°
Area of segment APB =?
(i) Area of sector OAPB:

(ii) Area of segment APB:

∴ Area of segment APB:
= Area of sector OAPB – Area of ∆AOB
= 150.72 – 62.68
= 88.44 sq. cm.

Question 8.
A horse is tied to a peg at one corner of a square-shaped grass field of side 15 m by means of a 5 m. long rope (see Figure). Find
i) the area of that part of the field in which the horse can graze.
ii) the increase in the grazing area if the rope were 10 m long instead of 5 m. (use π = 3.14)

Solution:
Length of the rope, r = 5 m, θ = 90°
Area of segment =?

(ii) Length of the rope, r = 10 cm, θ = 90°
Area of segment =?

∴ Area of the field which is increased in the grazing
= Equation (2) – Equation (1)
= 78.5 – 19.63
= 58.87 sq. cm.

Question 9.
A brooch is made with silver wire in the form of a circle with a diameter of 35 mm. The wire is also used in making 5 diameters which divide the circle into 10 equal sectors as shown in the Figure. Find
i) the total length of the silver wire required,
ii) the area of each sector of the brooch.

Solution:
Diameter, d = 35mm θ = \(\frac{360^{\circ}}{10}\) = 36°
Radius, r = \(\frac{35}{2}\) mm
(i) Length of the silver wire required :
Length of the wire used in making 5 diameter + Circumference of Silver, wire required.

(ii) Area of each sector of the brooch = ?
Diameter of the brooch, d = 35mm
Radius, r = \(\frac{35}{2}\) mm
Angle at the centre, θ = \(\frac{360^{\circ}}{10}\) = 36°
The Area of each sector of the Brooch:

Question 10.
An umbrella has 8 ribs which are equally spaced (see the figure). Assuming the umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.

Solution:
Radius of circle, r = 45 cm.
Angle at the centre, θ = \(\frac{360^{\circ}}{8}\) = 45°
Area between the two consecutive ribs of the umbrella =?
Area of each segment of the circle,

Question 11.
A car has two wipers which do not overlap. Each wiper has a blade of length 25 cm sweeping through an angle of 115°. Find the total area cleaned at each sweep of the blades.
Solution:
Radius of the wiper of the car, r = 25 cm.
Angle at the centre, θ = 115°
Area of the sector =?
Area of the blade when it is swept once

Area of two blades when swept:

Question 12.
To warn ships for underwater rocks, a lighthouse spreads a red coloured light over a sector of angle 80° to a distance of 16.5 km. Find the area of the sea over which the ships are warned. (use π = 3.14)
Solution:
Radius, r = 16.5 km. = \(\frac{33}{2}\) km.
Angle, θ = 80°
Area of the Sector =?
Area of the sea over which the ships are warned

Question 13.
A round table cover has six equal designs as shown in Figure. If the radius of the cover is 28 cm, find the cost of making the designs at the rate of Rs. 0.35 per cm².
(Use \(\sqrt{3}=\) = 1.7)

Solution:
Each angle at the centre,


Area of segment APB design 1 :
= Area of OAPB – Area of ∆AOB
= 410.6 – 333.2
= 77.46 sq.cm.
∴ Total are of SIX designs = 6 × 77.46
= 464.8 sq.cm.
Cost of designing for 1 sq.cm, is Rs. 0.35
∴ Cost of designing for 464.8 sq.cm ………….
= 464.8 × 0.35
= Rs. 162.68

Question 14.
Tick the correct answer in the following:
Area of a sector of angle p (in degrees) of a circle with radius R is

Solution:
D) \(\frac{p}{720}\) × 2πR²

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Ex 5.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 5 Areas Related to Circles Exercise 5.2, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.3.

Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.3

ex 3.3 class 10 Question 1.
Solve the following pair of linear equations by the substitution method.
(i) x + y = 14
x – y = 14
(ii) s – t = 3
\(\frac{s}{3}+\frac{t}{2}=6\)
(iii) 3x – y = 3
9x – 3y = 9
(iv) 0.2x + 0.3y = 1.3
0.4x + 0.5y = 2.3
(v) \(\sqrt{2} x+\sqrt{3} y=0\)
\(\sqrt{3} x-\sqrt{8} y=0\)
(vi) \(\frac{3 x}{2}-\frac{5 y}{2}=-2\)
\(\frac{x}{3}+\frac{y}{2}=\frac{13}{6}\)
Solution:
(i) x + y = 14 ………. (i)
x – y = 4 …………. (ii)
In eqn. (i), x + 4 = 14
∴ x = 14 – y
Substituting the value of ‘x’ in equation (ii), we have
x – y = 4
(14 – y) – y = 4
14 – y – y = 4
14- 2y = 4
-2y = 4 – 14
-2y = -10
2y = 10
∴ x = 5
∴ Substituting y = 5 in x = 14 – y,
x = 14 – y
= 14 – 5
x = 9
∴ x = 9, y = 5.

(ii) s – t = 3
\(\frac{s}{3}+\frac{t}{2}=6\)
s – t = 3 …………. (i)
\(\frac{s}{3}+\frac{t}{2}=6\) ………… (ii)
From eqn. (i), s – t = 3
-t = 3 – s
t = -3 + s
Substituting the value of ‘t’ in eqn. (ii),
\(\frac{s}{3}+\frac{-3+s}{2}=6\)
\(\frac{2 s-9+3 s}{6}=6\)
5s – 9 = 6 × 6
5s – 9 = 36
5s = 36 + 9
∴ 5s = 45
∴ s=\(\frac{45}{5}\)
∴ s = 9
Substituting the value of s = 9 in t = -3 + s
t = -3 + s
= -3 + 9
t = 6
∴ s = 9, t = 6

(iii) 3x – y = 3 …………. (i)
9x – 3y = 9 ………… (ii)
From eqn. (i), 3x – y = 3
-y = 3 – 3x
y = – 3 + 3x
Substituting the value of ‘y’ in eqn. (ii),
9x – 3y = 9
9x – 3(-3 + 3x) = 9
9x + 9 – 9x = 9
Here we can give any value for ‘x’ i.e., infinite solutions are there,
y = 3x – 3 Means x = 0 then y = -3
x = 1 then y = 0
x = 2 then y = 3
etc.

(iv) 0.2x + 0.3y = 1.3 …………. (i)
0.4x + 0.5y = 2.3 …………. (ii)
From eqn. (i)„
0.2x + 0.3y = 1.3
0.2x = 1.3 – 0.3y
\(x=\frac{1.3-0.37}{0.2}\)
Substituting the value of ‘x’ in eqn (ii)
0.4x + 0.5y = 2.3
\(0.4\left(\frac{1.3-0.37}{0.2}\right)+0.5 y=2.3\)
0.2 (1.3 – 0.3y) + 0.5y = 2.3
2.6 – 0.6y + 0.5y = 2.3
2.6 – 0.1y = 2.3
-0.1y = 2.3 – 2.6
-0.1y = -0.3
0.1y = 0.3
\(y=\frac{0.3}{0.1}=\frac{3}{1}=3\)
Substituting the vlue of ‘y’ in
\(x=\frac{1.3-0.37}{0.2}\)
\(=\frac{1.3-0.3(3)}{0.2}\)
\(=\frac{1.3-0.9}{0.2}\)
\(=\frac{0.4}{0.2}\)
\(=\frac{4}{2}\)
∴ x = 2
∴ x = 2, y = 3

(v) \(\sqrt{2} x+\sqrt{3} y=0\) ………….. (i)
\(\sqrt{3} x-\sqrt{8} y=0\) ………….. (ii)
From eqn (i)
\(\sqrt{2} x+\sqrt{3} y=0\)
\(\sqrt{2} x=-\sqrt{3} y\)
\(\mathbf{x}=-\frac{\sqrt{3}}{\sqrt{2}} \mathbf{y}\)
Substituting the value of ‘x’ in eqn (ii)

substituting the value of ‘y’ in
\(\mathrm{x}=-\frac{\sqrt{3}}{\sqrt{2}} \mathrm{y}\)
\(=\frac{-\sqrt{3}}{\sqrt{2}} \times 0\)
∴ x = 0
∴ x = 0, y = 0

(vi) \(\frac{3 x}{2}-\frac{5 y}{2}=-2\) ………… (i)
\(\frac{x}{3}+\frac{y}{2}=\frac{13}{6}\) …………….. (ii)
From eqn (ii)
\(\frac{x}{3}+\frac{y}{2}=\frac{13}{6}\)
\(\frac{x}{3}=\frac{13}{6}-\frac{y}{2}\)
\(\frac{x}{3}=\frac{13-3 y}{6}\)
\(x=\frac{3(13-3 y)}{6}\)
\(x=\frac{13-3 y}{2}\)
Substituting the value of ‘x’ in eqn (i)
\(\frac{3 x}{2}-\frac{5 y}{2}=-2\)
\(\frac{3}{2}\left(\frac{13-3 y}{2}\right)-\frac{5 y}{2}=-2\)
\(\frac{39-9 y}{4}-\frac{5 y}{2}=-2\)
\(\frac{39-9 y-10 y}{4}=-2\)
\(\frac{39-19 y}{4}=-2\)
39 – 19y = -2 × 4
39 – 19y = -8
-19y = -8 + 39
-19y = -47
Or 19y = 47
\(y=\frac{47}{19}\)
∴ y = 3
Substituting the value of ‘y’ in
\(x=\frac{13-3 y}{2}\)
\(=\frac{13-3(3)}{2}\)
\(=\frac{13-9}{2}=\frac{4}{2}\)
∴ x = 2
∴ x = 2, y = 3

Question 2.
Solve 2x + 3y = 11 and 2x – 4y = -24 and hence find the value of ‘m’ for which y = mx + c
Solution:

ex 3.3 class 10 Question 2.
Solve 2x + 3y = 11 and 2x – 4y = -24 and hence find the value of ‘m’ for which y = mx + 3.
Solution:
2x + 3y = 11 …………… (i)
2x – 4y = -24 ………….. (ii)
From eqn. (i),
2x = 3y = 11
2x = 11 – 3y
\(x=\frac{11-3 y}{2}\)
Substituting the value of ‘x’ in eqn. (ii),
2x – 4y = -24
\(2\left(\frac{11-3 y}{2}\right)-4 y=-24\)
11 – 3y – 4y = -24 – 11
-7y = – 35
7y = 35
\(y=\frac{35}{7}\)
∴ y = 5.
Substituting the value of ‘y’ in
\(x=\frac{11-3 y}{2}\)
\(=\frac{11-3(5)}{2}\)
\(=\frac{11-15}{2}\)
\(=\frac{-4}{2}\)
∴ x = -2
∴ x = -2, y = 5
Now, y = mx + 3
5 = m(-2) + 3
5 = -2m + 3
5 – 3 = -2m
-2m = 2
\(\mathrm{m}=\frac{2}{-2}\)
∴ m= -1.

ex 3.3 class 10 Question 3.
Form the pair of linear equations for the following problems and find their solution by substitution method :
(i) The difference between the two numbers is 26 and one number is three times the other. Find them.
Solution:
Let one number be ‘x’.
another number be ‘y’.
Their difference is 26.
∴ x – y = 26 ………. (i)
One number is three times the other,
∴ x = 3y …………. (ii)
Substituting x = 3y in eqn. (i),
x – y = 26
3y – y = 26
2y = 26
\(y=\frac{26}{2}\)
∴ y = 13
Substituting the value of ‘y’ in eqn. (ii),
x = 3y
∴ x = 3 × 13
∴ x = 39, y = 13

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.
Solution:
In two supplementary angles, let one angle be ‘x’.
another angle be ‘y’
Sum of these = 180°
∴ x + y = 180 ………… (i)
The larger of two supplementary angles exceeds the smaller by 18 degrees.
∴ x-y = 18 ………… (ii)
From eqn. (ii),
x – y = 18
x = 18 + y
Substituting the value of ‘x’ in eqn. (i),
x + y = 180 .
18 +y + y = 180
18 + 2y = 180
2y = 180 – 18
2y = 162
\(\mathrm{y}=\frac{162}{2}\)
∴ y = 81
Substituting the value of y’ in x = 18 + y
x = 18 + 81
x = 99
∴ x = 99°, y = 81°.

(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs. 3800. Later, she buys 3 bats and 5 balls for Rs. 1750. Find the cost of each bat and each ball.
Solution:
Let the cost of each bat b Rs. x.
Let the Cost of each ball be Rs. y.
∴ 7x + 6y = 3800
3x + 5y= 1750
From eqn. (ii),
3x + 5y = 1750
3x = 1750 – 5y
\(x=\frac{1750-5 y}{3}\)
Substituting the value of ‘x’ in eqn. (i),
\(7\left(\frac{1750-5 y}{3}\right)+6 y=3800\)
\(\frac{12250-35 y}{3}+\frac{6 y}{1}=3800\)
\(\frac{12250-35 y+18 y}{3}=3800\)
12250 – 17y = 3800 × 3
12250 – 17y = 11400
-17y = 11400 – 12250
-17y = -850
\(y=\frac{850}{17}\)
∴ y = Rs. 50
Substituting the value of ‘y’ in
\(x=\frac{1750-5 y}{3}\)
\(=\frac{1750-5 \times 50}{3}\)
\(=\frac{1750-250}{3}\)
\(=\frac{1500}{3}\)
∴ x = Rs. 500
∴ x = Rs. 500, Rs. y = 50
∴ Cost of each bat is Rs. 500,
Cost of each ball is Rs. 50.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs. 105 and for a journey of 15 km, the charge paid is Rs. 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
Solution:
Let the fixed charge of the taxi be Rs. x.
If the charge for each km is Rs. y, then
Fixed charge + Charge for 10 km
= Rs. 105
∴ x + 10y = 105 ………. (i)
Fixed charge + Charge for 15 km travelled
= Rs. 155
∴ x + 15y = 155 ……….. (ii)
From eqn. (i),
x + 10 y = 105
x = 105 – 10y
Substituting the value of ’x’ in eqn. (ii),
x + 15y = 155
105 – 10y + 15y = 155
105 + 5y = 155
5y = 155 – 105
5y = 50
∴ \( \mathrm{y}=\frac{50}{5}\)
∴ Rs. y = 10.
Substituting the value of y in
x = 105 – 10y,
= 105 – 10 × 10
= 105 – 100
∴ x = Rs. 5
∴ Fixed Charge of Taxi is Rs. 5.
Charge for each km is Rs. 10
Charge for 1 km is Rs. 10
Charge for 25 km = 25 × 10 = Rs. 250.

(v) A fraction becomes \(\frac{9}{11}\). if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator it becomes \(\frac{5}{6}\). Find the fraction.
Solution:
If the Numerator is ‘x’ and denominator is ‘y’, then the Fraction is \(\frac{x}{y}\).
Adding 2 to both Numerator and denominator, then fraction is \(\frac{9}{11}\).
\(\frac{x+2}{y+2}=\frac{9}{11}\)
11 (x + 2) = 9(y + 2)
11x + 22 = 9y + 18
11x – 9y + 22 – 18 = 0
11x – 96 + 4 = 0 ……….. (i)
If 3 is added to both Numerator and denominator, then the fraction is \(\frac{5}{6}\).
\(\frac{x+3}{y+3}=\frac{5}{6}\)
6(x + 3) = 5 (y + 3)
6x + 18 = 5y + 15
6x – 5y + 18 – 15 = 0
6x – 5y + 3 = 0 ………… (ii)
From eqn. (i),
11x – 9y + 4 = 0
11x = 9y – 4
\(x=\frac{9 y-4}{11}\)
Substituting the value of ‘x’ in eqn. (ii),
6x – 5y + 3 = 0
\(6\left(\frac{9 y-4}{11}\right)-5 y+3=0\)
\(\frac{54 y-24}{11}-\frac{5 y}{1}+\frac{3}{1}=0\)
\(\frac{54 y-24-55 y+33}{11}=0\)
-y + 9 = 0
-y = -9
∴ y = 9
Substituting the value of ‘y’ in
\(x=\frac{9 y-4}{11}\)
\(=\frac{9 \times 9-4}{11}\)
\(=\frac{81-4}{11}\)
\(=\frac{77}{11}\)
∴ x = 7.
∴ Fraction is \(\frac{x}{y}=\frac{7}{9}\)

(vi) Let the present age of Jacob = x years
and the present age of his son = y years
∴ 5 years hence: Age of Jacob = (x + 5) years
Age of his son = (y + 5) years
According to given condition,
[Age of Jacob] = 3[Age of his son]
x + 5 = 3(y + 5) ⇒ x + 5 = 3y + 15
⇒ x – 3y -10 = 0 … (1)
5years ago : Age of Jacob = (x – 5) years,
Age of his son = (y – 5) years
According to given condition,
[Age of Jacob] = 7[Age of his son]
∴ (x – 5) = 7(y – 5) ⇒ x – 5 = 7y – 35
⇒ x – 7y + 30 = 0 … (2)
From (1) , x = [10 + 3y] … (3)
Substituting this value of x in (2) , we get
(10 + 3y) – 7y + 30 = 0
⇒ -4y = -40 ⇒ y = 10
Now, substituting y = 10 in (3) ,
we get x = 10 + 3(10)
⇒ x = 10 + 30 = 40
Thus, x = 40 and y = 10
⇒ Present age of Jacob = 40 years and present age of his son = 10 years

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.3, drop a comment below and we will get back to you at the earliest

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Karnataka State Syllabus Class 10 Social Science History Chapter 10 The Political Developments of 20th Century

Class 10 Social Science The Political Developments of 20th Century Textual Questions and Answers

I. Fill in the blanks with suitable words:

Question 1.
The Communist Government in Russia was founded by ……………
Answer:
Lenin.

Question 2.
The First World War ended in …………
Answer:
1918.

Question 3.
The Fascist dictator was ………….
Answer:
Mussolini.

Question 4.
The Second World War started in ………..
Answer:
1939.

Question 5.
Japan attacked …………. which was the Naval base of USA.
Answer:
Pearl Harbour.

II. Discuss in groups and answer the following questions :

Question 1.
Explain the role of Lenin in Russian Revolution.
Answer:
Lenin played an important role in the Russian revolution. When Japan defeated Russia in 1905, it was a great insult to the people of Russia. It was due to the worst administration of the Tsars. The people started rebellions against the Tsars. Thousands of factory workers and peasants fought with weapons. At this juncture, Lenin, the leader of the Bolsheviks Labour Party guided the workers and peasant in the revolution against the Tsars. The working class engaged in more and more agitations. The last Russian Tsar Nicholas II was unable to control the agitations and deserted the country. As a result, the Mensheviks party gained the power of Russia. Lenin who was in exile came back to his country. He assured the people a good government. The people responded wholeheartedly to this. He declared Russia the Socialist Republic in 1917. He became the President of Russia.

Question 2.
Discuss the immediate causes of the First World War.
Answer:
World War I took place between 1914 to 1918. The immediate cause of the war was the assassination of the Austrian prince. Archduke Franz Ferdinand on 28th July 1914. This incident, caused an immediate war between the two groups Triple Entente and Triple Alliance.

Question 3.
How did Nazism destroy Germany? Explain.
Answer:

• Hitler declared that the Nazi party as the only party of Germany
• He instilled fear of Jewish people among Germans by spreading false rumors that they are going to dominate the world.
• He put forward the supremacy of German race theory and nurtured ultra-nationalism carefully.
• Hitler utilized Nazism to inculcate these values among the GermAnswer:
• Nazism advocated that Germans are the superior race of the world and they are the only fit to rule the world.
• Gobbels’ was appointed to spread the theory of Nazism A cruel army named ‘Brown Shirts’ was formed for this purpose.
• In 1935. Hitler implemented the ‘Nuremberg Laws’.
• According to one estimate, six million Jews were killed and one million others were also killed.

Question 4.
What were the reasons for the Second World War?
Answer:
The causes for World War II were :

1. The ambition of Hitler was to win the entire world and popularize the Aryan race.
2. The Versailles Treaty imposed a heavy loss on Germany. Germany lost most of its areas to allied nations. Economically, Germany was destroyed.
3. Among the defeated countries in World War I feelings of shame and humiliation gave rise to aggressive nationalism.
4. The compensation for the loss in the war and other decisions that were heaped on Germany affected the people adversely. Unemployment, poverty led to the widespread Germans.
5.  At this juncture, Hitler took the power of Germany. His main ambition was to take revenge on the allied nations.
6. He employed his dictatorial power for mass massacres in Germany.
7. The immediate cause for World War II was Hitler’s attack on Poland, which was an ally of Britain.

Question 5.
What is Cold War?
Answer:
After World War II, the constant fear, hatred, jealousy and anxiety in political, economic, military and other affairs between the two global powers is called ‘Cold War’.

Question 6.
What were the effects of Chinese Revolution?
Answer:
The effects of Chinese Revolution were:

1. Mao-Tse-Tung became the President of China. During his rule community farming was adopted.
2. People enjoyed free education, health and sports facilities.
3. A lot of importance was given to science and technology in order to achieve industrial development.
4. The “Leap Forward” project was adopted.
5. Private property was converted to the property of society.

Question 7.
How did the USA come out of its Great Economic Depression?
Answer:
In 1929 economic development in America stagnated. As a result industrial and agricultural production collapsed. Besides mining, shipbuilding, production of consumer goods suffered a lot. The economic crisis was witness to political changes. Under such economic conditions, Roosevelt, the President of America encouraged women to work in the social sector. He brought many reforms to change the economic structure of America. It was due to the efforts of Roosevelt, America overcame its economic depression.

Class 10 Social Science The Political Developments of 20th Century Additional Questions and Answers

I. Fill In The Blanks With Suitable Answers:

Question 1.
The October 1917 revolution in Russia was created by the __________ party of workers.
Answer:
Bolsheviks

Question 2.
The situations emerging after the first world war brought into power in Italy _______________________
Answer:
Mussolini

Question 3.
The communist party was formed in china in _____________
Answer:
1925

Question 4.
The first world war ended by treaty of __________________
Answer:
Versailles

Question 5.
In 19th century Russia was ruled by _____________________
Answer:
Tsars

Question 6.
Tsar Nicholas II ran away from the country this was called as ______________
Answer:
February 1917 revolution

Question 7.
Lenin became president of Russia on _____________________
Answer:
October 7, 1917

Question 8.
Hitlers title was ________________________
Answer:
Fnlirer

Question 9.
Hitlers formed cruel army named as _________________
Answer:
Brownshirts

Question 10.
Mussolini founded party ______________________
Answer:
Fascist

Question 11.
Russia entered “a Nonwar pact” with Germany on ________________
Answer:
August 24, 1939

Question 12.
The private property was converted into public property by revolution, in China ________________
Answer:
Cultural

Question 13.
The Suez canal crisis solved by UNO in __________________
Answer:
1956

II. Multiple Choice Questions:

Question 1.
‘League of Nation’ came into existence in 1919 because
a. In order to prevent future occurrences of war.
b. In order to step the first world war.
c. To follow terms of Versailles treaty
d. To provide peace and security
Answer:
a. In order to prevent future occurrence of war

Question 2.
Lenin became the president of China it was the results of
a. Support of Menhiviles
b. October revolution
c. The help of Bolsheviks
d. February Revolution
Answer:
b. October revolution

Question 3.
The founder of the Fascist party was
a. Hindenburg
b. Victor Emanual
c. Hitler
d. Mussolini
Answer:
d. Mussolini

Question 4.
The leader of komingtong party was
a. Sun yat sen
b. Chiang – Kai – Sliek
c. Mao Tse Tung
d. Karl Mark
Answer:
a. Sun yat-sen

Question 5.
America faced the war with Japan because
a. Japan attacked Pearl Harbour of U.S.A
b. It was the main trade centre
c. In 1929 USA faeed economic crisis
d. Japan was the main trade centre of Asia.
Answer:
a. Japan attacked the pearl Harbour of USA

III. One Marks Questions:

Question 1.
More competition had created among the European countries?
Answer:
Industrialization and the invention of new’ technologies.

Question 2.
Why did Russia withdraw from first world war?
Answer:
With occurrence of revolution in Russia in November of 1917.

Question 3.
The rule of Tsars was “a nationalistic jail” Why?
Answer:
Tsars exploited the landlords in turn were exploiting laborers and small farmers.

Question 4.
Which were the watchwords of Lenin in the Russia revolution?
Answer:
‘’Peace, Food and Land”.

Question 5.
Who became the first Astronaut of the world?
Answer:
Yurigarin of Russia

Question 6.
Why ‘Brown Shirt’ army was formed by Hitler?
Answer:
To create violence

Question 7.
Why Hitler committed suicide?
Answer:
Germany surrendered due to the March of Red army and Allied forces into Berlin.

Question 8.
Why Chiang Kai Shek runaway to Faiwan?
Answer:
By 1949 most of the cities came under the control of communist party.

Question 9.
Why China implemented‘The Cultural Revolution’?
Answer:
The private property was converted into public property.

IV. Two Marks Questions:

Question 1.
What were the reasons for first world war?
Answer:

• Extreme form of nationalism
• Rivelary Alliance
• Triple Alliance
• Triple Entente
• Arms race
• Assassination of Archduke Francis Ferdinand

Question 2.
Write the importance of February revolution .
Answer:

• Lenin guided the formers and workers
• Lenin was declared as a traitor
• Workers and farmers became mere intense
• The Tsar Nicholar II was ran away from the country.

Question 3.
What were achievements of Lenin?
Answer:

• Lenin declared that land belonged to the farmers.
• He implemented political and economical policies.
• Ensured free health schooling and residence to all t
• Implemented Karl Marx’s scientific communism in practice.

Question 4.
How Gorbachev had main role to stop the cold war?
Answer:

• Garbachev introduced many reformations.
• He introduced Glasnost in 1985 it means ‘Liberal’.
• He implemented perestroika. It means ‘Re-Organising’.

Question 5.
Which were the features of pasist party?
Answer:

• Mussolini started pasist party in Italy.
• Ultra Nationalism
• Patronizing violence.
• Racial superiority
• Expansions of national boundary.
• Support to human executions

V. Three Marks Questions

Question 1.
What was the role of Sun vat Sen in Chinese revolution?
Answer:

• In 1911 under the leadership of Sun yat Sen anti-imperialistic democratic revolution [ took place.
• The communist party started in China in the year 1925
• It groomed movements in rural areas and city areas.
• It supported farmers movements
• In order to build a unified China
• The communist party and Kuomintang party worked together.

We hope the KSEEB SSLC Class 10 History Solutions Chapter 10 The Political Developments of 20th Century help you. If you have any query regarding Karnataka SSLC Class 10 History Solutions Chapter 10 The Political Developments of 20th Century, drop a comment below and we will get back to you at the earliest.

Students can Download History Chapter 9 Post Independent India Questions and Answers, Notes Pdf, KSEEB SSLC Class 10 Social Science Solutions helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 10 Social Science History Chapter 9 Post Independent India

Class 10 Social Science Post Independent India Textual Questions and Answers

I. Fill in the blanks with appropriate answers :

Question 1.
The British government’s last Governor General was …………….
Answer:
Lord Mountbatten.

Question 2.
India’s first Home Minister was ……………..
Answer:
Sardar Vallabhbhai Patel.

Question 3.
Inia’s first President was ……………
Answer:
Dr. Babu Rajendra Prasad.

Question 4.
Pondicherry became a union territory in the year …………..
Answer:
1963.

Question 5.
State Reorganization law was implemented in ………… year.
Answer:
1953.

II. Discuss in groups and answer the following :

Question 1.
What were the problems faced in independent India?
Answer:

1. The problems faced by India after Independence were the Problem of refugees.
2. Problem of communaliriots.
3. Formation of the government.
4. Integration of various provinces.
5. Linguistic formation of states.
6. Production of food and many others.

Question 2.
How did the nation face the refugee problem?
Answer:

1. In 1951 most of the refugees from west Pakistan were taken care of.
2. After the liberation of Bangladesh 10 lakh, people were arriving to India as refugees. Our government provided resettle them in Tripura, Meghalaya and Assam States.
3. Many Bangladesh refugees came to Bengal. India did not leave its humanitarian concern and tried to provide better facilities.
4. More than one lakh and twenty thousand Tibetan .refugees came to India. In 1960 our Karnataka government-sanctioned 3000 Acres of land to them.

Question 3.
How was Pondicherry liberated from the French? Explain.
Answer:
Even after the independence of India, the French had treated Pondicherry as their colony. The Congress, the communists, and other organizations urged that Pondicherry should be the part of India. After a long effort, Pondicherry was liberated from the French and it was declared a Union territory in 1963.

Question 4.
How was Goa liberated from the Portuguese?
Answer:
Even after independence, Goa was under the rule of Portuguese. Though they were ordered to give up Goa, the Portuguese strengthened their hold on Goa by suppressing the movements. In 1955, Satyagrahis from different ports of India gathered at Goa and staged dharnas and liberation movements to quit Portuguese from Goa. In 1961 the Indian army intervened and took it under its control. Thus Goa was liberated in 1961 and became a Union Territoiy of India. Now it has become one of the states of India.

Question 5.
Explain the process of State Reorganization based on language.
Answer:
After India achieved Independence, it had to face the linguistic formation problem. Many organizations and the people demanded to mark the boundaries based on the language of the areas. The desire of the people for linguistic formation of states was intense. In 1953 the government formed Justice Fazal Ali Commission in which Fazal Ali was the Chairman, K.M. Panickker and H.N. Kunjru were the members. Andhra Pradesh became the first state on the base of language. As per the report of the Commission, the State Reorganisation Act came into force in 1956. Accordingly, 14 states and 6 Union Territories were formed in the country. On November 1st, 1956, Mysore State came into existence on the base of Kannada speaking areas. In 1973 Mysore was renamed as Karnataka State.

Class 10 Social Science Post Independent India Additional Questions and Answers

I. Multiple Choice Questions:

Question 1.
The Chairman of the constitutional drafting committee was
a. Dr. B.R.Ambedkar
b. Gandhiji
c. Dr. Babti Rajendra Prasad
d. Nehru
Answer:
a. Dr. B.R.Ambedkar

Question 2.
Karnataka sanctioned the land to Tibetan refugees at
a. Mysore
b. Cliamaraj Nagar
c. Bylukuppe
d. Mandagadde
Answer:
c. Bylukuppe

Question 3.
Sardar Vallabhabai patcFwho is known as Iron Man of India because
a. Successful in integrating the princely states.
b. He appointed as Home minister of India
c. He maintained India’s stability
d. He could find a solution and firm decision.
Answer:
a. Successful in integrating the princely states.

Question 4.
Integrated Hyderabad into India in
a. 1946
b. 1949
c. 1948
d. 1954
Answer:
c. 1948

Question 5.
India was strengthening its military forces during post-independent India because of
a. Pakistan occupied Kashmir
b. To control partition of India
c. India had to protect its freedom
d. To avoid communal clashes
Answer:
c. India had to protect its freedom

Question 6.
Junagadh joined the Indian federation in
a. 1948
b. 1949
c.1947
d. 1946
Answer:
b. 1949

Two Marks Questions:

Question 1.
How did Junagadh province join to Indian federation?
Answer:

• The Nawab of this princely state had signed the agreement to join the state of Pakistan
• His citizens revolted against him and flooded the streets.
• The king fled from the kingdom
• The dewan requested the Indian government to the military to Junagadh to maintain law and order
• Later Junagadh joined the Indian federation in 1949.

III. Three Marks Questions:

Question 1.
Explain briefly about communal violence problems and solutions.
Answer:

• It is said that people have lost more lives in communal violence than in wars.
• The clash that takes place on the grounds of religion creates panic in all the societies.
• The British to create divide and rule policy with this religious suspicions entered public life later leading communal violence.
• Like this Hindu Muslim clashes took a nasty turn and resulted in many communal clashes.
• Asa result, when India was partitioned in 1947, the north India had to face a lot of communal violence
• When India was celebrating its independence in New Delhi. Gandhiji was in Naukali and other places in meeting the victims consoling them.
• The evil violence that emerged during the communal violence made Nehru to think about forming a secular nation.
• The constitution of India has accepted religion as the personal choice of the individual.
• It foresees a situation where India remains secular country.The Indians need to understand this intention of the constitution live accordingly.

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Students can Download History Chapter 6 The First War of Indian Independence (1857) Questions and Answers, Notes Pdf, KSEEB SSLC Class 10 Social Science Solutions helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 10 Social Science History Chapter 6 The First War of Indian Independence (1857)

Class 10 Social Science The First War of Indian Independence (1857) Textual Questions and Answers

I. Fill in the blanks with suitable answers :

Question 1.
The Mutiny of 1857 was called as …………. by the British historians.
Answer:
Sepoy mutiny.

Question 2.
The policy implemented by Dalhousie is …………
Answer:
Doctrine of policy.

Question 3.
During the mutiny of 1857, …………. killed a British officer.
Answer:
Mangal Pandey.

Question 4.
Jhansi Rani took over ……….. from the British during her war against them.
Answer:
Gwalior.

Discuss in groups and answer the following questions :

Question 1.
What were the results of the ‘Doctrine of Lapse? Discuss.
Answer:
Lord Dalhousie introduced Doctrine of Lapse. The effects of this policy were:

1. Many kingdoms lost their rights due to this policy. 2. Satara, Jaipur, Jhansi, Udaipur and other kingdoms came under the British control.
2. Dalhousie canceled the princely titles of the Nawab of Tanjore and Carnatic kingdoms. 3. The Moghul Sultan, the Nawab of Oudh and other kings were thrown out of their kingly status. 4. Thousands of soldiers dependent on these kings became unemployed.

Question 2.
How did the economic policies result in the mutiny of 1857? Explain.
Answer:

• Due to the development of industrialization in England, the Indian handicrafts and industries diminished.
• The artisans of India became unemployed
• The w eavers became the first victims as w’ool and cloth-making industries suffered a lot
• The Indian handicrafts became financially pathetic.
• Due Zamindaru system, the farmers were exploited by Zamindars.
• Land tax collection rights awarded to Talukadars were withdrawn.
• By forming Inam commissions, Inam lands were also withdrawn.
• As result, the farmers had to undergo lot of economic hardships and felt insulted.
• Hence, the farmers also protested all these.

Question 3.
What were the issues that outraged the religious feelings of the Soldiers?
Answer:
The factors that disturbed the religious sentiment of soldiers were:

1. The soldiers in the British army were being given new rifles called Royal Enfield. The cartridge used in the rifles were greased with the fat of cow and pigs. The soldiers had to pull the safety catch with the help of their teeth. But, it was impossible for both the Hindu and Muslim soldiers, because cow was sacred for the Hindus, while pig was blasphemous for Muslims. This factor disturbed the religious sentiments for both religions.
2. Another factor that disturbed the Indian religious sentiment, was that the soldiers were forced to cross the oceans to serve on foreign land.

Question 4.
What were the immediate causes for the First War of Indian Independence?
Answer:
The immediate causes for the First War of Independence were:

1. A rumour spread among the soldiers that the cartridges used in the rifles were greased with the fat of cow and pigs. Cows were sacred to the Hindus, whereas pigs were blasphemous for Muslims. This incident became the immediate for the war.
2. The rumour spread to Barackpur soldiers, which resulted in severe dissatisfaction.
3. When the British officers ordered the soldiers to use the rifles by biting the cartridges, they refused to do so.
4. An Indian soldier named as Mangal Pandey killed a British Officer. Soon he was arrested by the British, tried and hanged to death. This was also an immediate cause for the First War of Independence.

Question 5.
List out the reasons that led to the failure of the mutiny.
Answer:
The reasons for revolt were:

• It did not cover every part of India.
• It was mainly concentrated on the issues of the rights of kings and Queens rather than liberation of the country.
• The unity among the British and the disunity among the Indian soldiers resulted in its failure,
• The mutiny lacked direction and leadership.
• The soldiers also lacked discipline and organizing skills.
• The Indian soldiers lacked military strategies.planning capabilities and soldering skills. –
• The freedom fighters lacked a definite aim.
• Many of the Indian kings extended their loyalties to the British and did not support the freedom fighters.
• The plundering and other crimes committed by the Sepoys made them to lose the faith of common people.

Question 6.
What was the main aspect of the declaration of the British Queen?
Answer:
The features of the declaration of the British Queen in 1857 are

1. Accepting the agreements made by the East India Company.
2. Giving up the ambitious expansion plans made by the British.
3. Providing a stable government for Indians.
4. Treating everyone equally before law.
5. Exhibiting religious tolerance and not interfering in the religious matters of the country.
   These are the features of the declaration of the British Queen in 1858.

Class 10 Social Science The First War of Indian Independence (1857) Additional Questions and Answers

II. Fill in the blanks with suitable words:

Question 1.
Due to the development of ……….. in England, the Indian handicrafts and industries diminished.
Answer:
Industrialization.

Question 2.
………….guns were provided to soldiers by the British.
Answer:
Royal Enfield.

Question 3.
In the Freedom Struggle, Indian soldiers declared the Mughal King …………. as the emperor of India.
Answer:
Bahadur Shah II

Question 4.
…………… provided leadership to the revolt.
Answer:
Jhansi, Rani Laxmibai.

Question 5.
In the year 1858, Britain queen declared …………..
Answer:
Magna Carta.

Question 6.
In 1857 freedom struggle, ……………. helped Nana Saheb.
Answer:
TatyaTope.

III. Two Marks Questions

Question 1.
Who was Laxrni Bai? What was her role in the 1857 revolt?
Answer:
Laxmi Bai was Queen of Jhansi

• Rani Laxmi Bai who was angered by the Doctrine of lapse declared war on the British.
• She captured Gwalior
• She died a heroic death while battling the British in another battle.

Question 2.
Which were the administrative reasons for the Sepoy revolt?
Answer:

• The British brought in many civil and criminal laws.
• There were lot of partialities
• English became the language of the court.
• English judges gave judgment in favor of the English.

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Karnataka State Syllabus Class 10 Social Science History Chapter 5 Social and Religious Reformation Movements

Class 10 Social Science Social and Religious Reformation Movements Textual Questions and Answers

I. Fill in the blanks with suitable answers :

Question 1.
The nineteenth century is called the period of ………..
Answer:
“Indian Renaissance”.

Question 2.
Raj Ram Mohan Roy started …………. periodical.
Answer:
Samvada Kaumudi.

Question 3.
The founder of Prarthana Samaj is …………
Answer:
Atmaram Pandurang.

Question 4.
The Young Bengal Movement was started by ………..
Answer:
Henry Louis Vivian Derozio

Question 5.
The Guru of Swami Vivekananda was …………
Answer:
Ramakrishna Parmahamsa.

Question 6.
The Anglo-Oriental College was established at ………….
Answer:
Aligarh.

II. Discuss in groups and answer the following questions :

Question 1.
What are the preaching of the Brahma Samaj?
Answer:
The preachings of Brahmo Samaj are :

1. Opposing idol worship.
2. Opposing exploitation of women.
3. Supporting the prohibition of Sati.
4. encouraging monotheism.
5. opposing priestly class.
6. Condemning performance.
7. Condemning performance of Yagas and rituals etc.

Question 2.
Discuss the declaration of Dayananda Saraswathi’s ‘Back to Vedas’.
Answer:

• He realized that the remedies to the various maladies of India are present in Vedas.
• One should study Vedas and other ancient sacred texts and preach them.
• The existence of caste and superstitions had pushed people towards other religions.
• Hence, he declared ‘Back to Vedas’. He was more of a renaissance person than a reformation person.

Question 3.
Discuss the reformation advocated by Sathyashodak Samaj.
Answer:
The reforms propagated by Satya Shodak Samaj are:

1. The Samaj propagated that freedom was the basic necessity of every individual and no individual could express his ideas if he did not have freedom.
2. It urged for the prohibition of liquor.
3. It opposed gender inequality denial of human rights.
4. It opposed the exploitation of people.
5. It opposed the practice of untouchability.
6. It started the movement of social justice.
7. It advocated free and compulsory education in order to bring about reform in the social system.
8. Phule condemned slavery.
9. He gave stress on girls’ education. Therefore he started a school for girls.

Question 4.
Discuss the aims of Aligarh Movement.
Answer:
The Aligarh Movement was started under the leadership of Sir Syed Ahmed Khan. The objectives of the Movement are:

• Syed Ahmad Khan said without an open mind, any social and intellectual development is impossible.
• He did not support the purdah system for Muslim woman.
• He did not accept polygamy. All these aspects were part of his struggles. In order to implement his ideals into practise.
• He founded Mohammadan Anglo-Oriental College in Aligarh in 1875.
• He utilized this institution to spread western scientific and cultural ideas.
• This institution later became Aligarh Muslim university.
• Syed Ahmad Khan preached religion tolerance.
• He thrived for unity among Hindus and Muslims.

Question 5.
Explain the views of Ramakrishna Mission.
Answer:
Ramakrishna Mission was founded by Swami Vivekananda. The vision of the mission is the loving life and the significance of the individual, his presence and ability. Social service is necessaiy for achievement of salvation (moksha). The vision of mission is also nurturing culture through education and social service. Tolerance of all religions and maintaining all religions as truth is the vision of this mission.

Question 6.
Explain how Swami Vivekananda was a source of inspiration for youngsters?
Answer:
Swami Vivekananda was a revolutionary monk. He opened the eyes of the Indians towards the importance of loving life. He stressed the significance of the individual, his presence, and his ability. He gave importance to social service, which is necessary for the achievement of salvation (moksha). Gandhiji and many national leaders were inspired by his philosophy. He emphasized nation and religion like the two faces of a coin.

In 1893 at the Chicago Conference of World Religions, Vivekananda upheld the greatness of Indian by advocating tolerance of all religions and maintaining that all religions are true. He introduced Indian culture to the world. Really such a great personality, the revolutionary monk is a great inspiration to the youth of India.

Question 7.
What were the reformation activities of Annie Besant?
Answer:
The reformations brought by Annie Besant are:

1. She aroused pride in Indian culture through her lectures.
2. She attempted to establish equality, universal brotherhood and harmony in society.
3. She gave full support to Indian struggle for freedom. As a token of this, she started Home Rule Movement in 1916.
4. She contributed immensely to Indian philosophy and freedom struggle.

Question 8.
Explain the contributions of Sri Narayana Dharma Paripalana Yogam.
Answer:
Sri Narayana Guru (1854-1928) started the reformation movement in 1903. This started Elavu of Kerala. This movement aimed at strengthening the backward and exploited communities. The caste differences were too much in Kerala and there were many prohibitions in place. During that period, all were not allowed to use facilities like tanks and roads. They were barred from wearing footwear. Restrictions were there on women’s dress also. These communities had to live without basic human rights. In order to address this issue, Sri Narayana Guru started the Dharma Paripalana Yogam movement. One Caste, One Religion and One God for human beings was the basic idea of Sri Narayana Guru. Education is the only path to achieve this he declared. He built temples for the backward communities as they were denied entry into temples.

Question 9.
Make a list of main aspects of the Periyar Movement.
Answer:
The non-Brahmin movement took a new shape under the Justice Party. This became more radical. ‘Self-Respect Movement’ led by E.V. Ramaswami Naicker became a new force. Ramaswamy who came out of Congress in 1925, started Self Respect League in 1926. He was called Periyar (Senior Person) out of love by people.

Periyar was born in Erode to a rich family. He believed that Congress is in favour of Varna system and started a new movement-based Dravida Racial identity. He rejected the racial supremacy in the name Aiya and Brahmin.

He said Tamil is language of Dravidians. He opposed Sanskrit language and literature. He also rejected Rama as the Vedic leader and accepted Ravana as the Dravidian leader. Angered by a bad experience at Banaras, he converted Sanskrit, Rama and Ravana as cultural ideals and patronized Ravana.

He championed equality and criticized caste and gender based discrimination. He participated in temple entry movement held at Vaikom of Kerala. He became the president of Justice Party in 1939.

He started an association called “Dravida Kalagam’. He also started Justice Periodical. The ideological Non-Brahminical movement started by Ayonthidas and T.M.Nayar was turned into a cultural movement by Periyar. His life goal was to establish an equal society where there will be no inequality based on caste, religion and gender.

Class 10 Social Science Social and Religious Reformation Movements Additional Questions and Answers

Question 1.
Raja Ram Mohan Roy even learned ………. and ………. to read Bible in its original form.
Answer:
Hebrew, Greek

Question 2.
……….were started to protect cows.
Answer:
Cow Protection Associations

Question 3.
……….. was one of the important programmes of Arya Samaj.
Answer:
Shuddi Movement

Question 4.
The wife’s name of Jyothiba Phule was …………
Ans:
Savithribai Phule

Question 5.
Ramakrishna Paramahamsa was a worshipper of ……………
Answer:
Kali

Question 6.
Swami Vivekananda attracted the attention of the listeners while speaking at held at ……….. Paris.
Answer:
Congress Religions

Question 7.
One Caste, One Religion and One God for human beings was the basic idea of ………..
Answer:
Sri Narayana Guru

Question 8.
Annie Besant translated ……….. to English.
Answer:
Bagavath Geetha

Question 9.
Periyar started ………….
Answer:
Justice Party

Question 10.
Dayananda Saraswathi declared ………….
Answer:
‘Back to Vedas’.

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Students can Download History Chapter 7 Freedom Movement Questions and Answers, Notes Pdf, KSEEB SSLC Class 10 Social Science Solutions helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 10 Social Science History Chapter 7 Freedom Movement

Class 10 Social Science Freedom Movement Textual Questions and Answers

I. Fill in the blanks with suitable answers:

Question 1.
The Indian National Congress was found in the year ………….
Answer:
1885.

Question 2.
The Drain Theory was forwarded by ………….
Answer:
Dadabai Navoroji.

Question 3.
Swarajya is my birth right was declared by ……………..
Answer:
Bal Gangadhar Tilak.

Question 4.
Bala Gangadhar Tilak published ……………… newspaper in Marathi.
Answer:
Kesari.

Question 5.
A secret organization by name ‘Abhinava Bharathi’ belonged ……………….
Answer:
Revolutionaries.

II. Choose the right option and fill in the blanks:

Question 1.
The founder of Indian National Congress is ………………
a) Mahatma Gandhiji
b) A.O. Hume
c) Balagandhar Tilak
d) Gopala Krishna Gokhale
Answer:
(b) A. O. Hume

Question 2.
‘Maratha’ paper was published by ………………
a) Jawahara Lai Nehru
b) Ras Bihari Bose
c) Balagandhara Tilak
d) V.D.Saavarkar
Answer:
(c) Bal Gangadhar Tilak

Question 3.
Muslim League was founded in ……………….
a)1924
b) 1922
c) 1929
d) 1906
Answer:
(d) 1906

Question 4.
The Viceroy who implemented the Bengal division was ………………….
a) Lord Cornwallis
b)Dalhousie
c) Lord Curzon
d)Robert Clive
Answer:
(c) Lord Curzon

III. Discuss in a group and answer the following:

Question 1.
Which were the organizations that were present before the founding of the Indian National Congress?
Answer:
The Hindu Mela, the East Indian Association, Poona Public Sabha, and the Indian Association.

Question 2.
What were the demands of Moderates placed in front of the British?
Answer:
The demands put forward before the British by the moderates were:

1. Development of Industries in India,
2. Reduction of military expenses in the British army.
3. Improvement in the educational standard.
4. Forcing the British Government to take up studies about poverty in the country.

Question 3.
Explain the Drain Theory.
Answer:
Moderates were the first to study the ill effects of British rule on India. They explained the drain of resources of India into England through scientific statistics and called it ‘Drain Theory’. By increasing the import and reducing the export, the British facilitated the draining out of precious Indian resources into India. Just like Dadabai Navoroji, R.C. Datta too published books explaining the draining of Indian resources into England.

Question 4.
Name the revolutionaries of the Indian Independence Movement.
Answer:
Aurobindo Gosh, V. D. Saavarakar, Ashwini Kumar Datta, Rajanarayana Bose, Rajguru, Chakikar brothers, Vishnu Shastri, Champukar, Shyamaji Krishnaverma, Ras Bihari Gosh, Madam Cama, Kudiram Bose, Ramprasad Bismil, Ashvakulla Khan, Bagath Singh, Chandrashekar Azad, Jatin Das are more prominent among the revolutionaries of Indian Independence Movement.

Question 5.
Discuss the role of Balagangadhar Tilak in the Indian Independence Movement.
Answer:
Bala Gangadhar Tilak was one of the members of the Radical group. The aim of the Radical group was an Independent India. Bala Gangadhar Tilak started preparing the common people for freedom struggle. He declared “Swaraj is My Birth Right and I will get it back”. Through religious functions like Shivaji Jayanti, Ganesh festival he started organizing people for the freedom movement. He published ‘Kesari’ in Marathi and ‘Maratha’ in English news paper and used them as weapons to criticize the British administration. He called the people for the active participation in the freedom struggle. Thus Tilak played an important role in the freedom struggle.

Question 6.
What were the reasons for the withdrawal of the Bengal Division?
Answer:
The division of Bengal in 1905 was opposed by the Indian National Congress. Still, the Bengali language could unite the Hindu and Muslim communities. Raksha Bandhan, a Cultural festival was held to bring in unity among Hindus and Muslims. The division of Bengal resulted in widespread protection across the country. The radicals took the issue to the doorsteps of common people. They called for boycotting of foreign goods and the institutions that encourage it. Indians were encouraged to use local goods. The British government withdrew the Bengal division order in 1911.

Class 10 Social Science Chapter 7 Freedom Movement Additional Questions and Answers

I. Fill In The Blanks With Suitable Answers:

Question 1.
Due to their differences in ideology, beliefs and execution styles they are identified as ……….. and ………..
Answer:
Moderates, Radicals.

Question 2.
The period between …………. is called as the Age of Moderates.
Answer:
CE 1885 and 1905.

Question 3.
Bengal had more concentration of ………….. and ………… people.
Answer:
Hindu, Muslim.

Question 4.
The British divided Bengal in ……………
Answer:
1905.

Question 5.
Bal Gangadhar Tilak published …………… paper in the English language.
Answer:
‘Maratha’.

II. Multiple Choice Questions:

Question 1.
The participation of Indians in the legislature process was
a. 1833
b. 1861
c. 1909
d. 1919
Answer:
b. 1861

Question 3.
Vernacular press Act was implemented by
a. Lord Lytton
b. A.O. Hume
c. Warren Mattings
d. Lord Rippon
Answer:
a. Lord Lytton

Question 5.
Moderates had faith in the rule of
a. British rule and Judiciary
b. British admisnistration
c. Self rule and Hema pulse
d. Own government
Answer:
a. British rule and Judiciary

Question 7.
‘Abhianava Bharatha’ and ‘Anushecla Sarniti’ were the Iwo important secret organizations of_______
a. Revolutionaries
b. Radicals
e. Moderates
d. Gandh Ian era
Answer:
Revuluthnuiries

Question 9.
“Swaraj is my Birth Right. I would definitely get it back”, declared by
a. Gandhiji
b. Nehru
c. Tilak
d. C.R.Das
Answer:
c. Tilak

Four Marks Questions:

Question 1.
Who founded INC? Which were its aims?
Answer:
A.O.Hume plays ed an important role in the formation of Indian national congress in 1885. Hume was a retired British civil servant and met political leaders in cities like Madras, Bombay, and Calcutta and discussed the various issues of public importance.

Aims of INC:

• The congress declared that achieving national unity as its primary aim during its first national convention.
• It thrived to achieve unity among the diverse cultural and social paths of India.
• The leaders of this period also had the commitment to achieve it.

Question 4.
What was role of Revolutionaries in freedom struggle?
Answer:

• Revolutionaries dreamed ofattaining complete freedom.
• They believed that they can drive away the British by employing violent methods. Aurobindo Gosh.
• D. Saavarkar, Ashwini Kumar Datta. Bagath Singh, Chandrashekar Azad are important revolutionaries.
• They established secret associations across the country
• Started collecting weapons and money for an armed struggle against the British.
• A secret organization named ‘Lotus and Dragger was founded in England.
• “Gadha’ in USA can be recalled here
• Abhinava Bharatha’ and ‘Anusheela Samiti’ were two important secret

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Karnataka State Syllabus Class 10 Social Science History Chapter 4 Opposition to British Rule in Karnataka

Class 10 Social Science Opposition to British Rule in Karnataka Textual Questions and Answers

I. Fill in the blanks with suitable answers :

Question 1.
The First Anglo-Mysore war took place between ……….. and ………
Answer:
Hyder ali and British.

Question 2.
The Second Anglo-Mysore war ended with …….. treaty.
Answer:
Salbhai.

Question 3.
Kittur Chenamma adopted a boy named ………….
Answer:
Shivalingappa.

Question 4.
Rayanna of Kittur state belonged to ………. village.
Answer:
Sangolli.

Question 5.
Surapura’ is in the present district of ………..
Answer:
Yadagiri.

Question 6.
The Bedas of ……….. village of Belgaumdistrict rebelled against the British.
Answer:
Hulagali.

Question 7.
The Amara Sulya rebellion was basically a ………. rebellion.
Answer:
Farmers.

II. Discuss in groups and answer later :

Question 1.
How did Hyder Ali come to power?
Answer:
After the death of Aurangzeb, Moghul empire was weakened. As a result the Moghul lost political control over South India. A lot of political struggles took place in Carnatic region. Before this, the death of Chikkadevaraj wodeyar in 1704 created the various political challenges in Mysuru state. His death created the problems of succession and administration.

All these developments Cloded the politics of Mysuru. Hyder Ali gained prominence in this scenario of uncertainty that clouded over the Mysuru and carnatic region. He joined Mysore Army as an ordinary soldier, but was known for his shrewd political moves. He came into prominence during the siege of Devanahalli and militaiy action against Nizam of Arcot.

Question 2.
What are the effects of Second Anglo-Mysore war?
Answer:
The Second Anglo-Mysore was started in 1780. Hyder Ali was defeated in a battle held in Port Novae by the British. This increased the confidence of the British and also changed the direction of the battle. But they suffered financial setbacks in Pulieat and Soliungur.

Meanwhile, by entering ‘Salbai Agreement’, the British were successful in winning over the Marathas and Nizam of Hyderabad to their side. Hyder Ali died due to illness during the war. The British tried to take advantage of Hyder Ali’s death by invading Mangalore and Bidanoor. They also tried to instigate the rulers of Calicut and Malabar regions against Tippu Sultan. Tippu Sultan defeated the British. The “Treaty of Mangalore” ended the Second Anglo-Mysore War in 1784.

Question 3.
What were the conditions of Srirangapatanam treaty?
Answer:
Following are the conditions of Srirangapatanam treaty:

1. Tippu was forced to. part with half of his Kingdom, was forced pay three crore rupees as war damage fee.
2. He had to pledge two of his children as a guarantee against the payment.
3. He was also forced to release the Prisoners of War.

Question 4.
The Fourth Anglo Mysore strengthened the position of British in Mysore. Discuss.
Answer:
Fourth Anglo-Mysore War took place between Tippu Sultan and British in 1799. The British were able to destroy the strong fort. Tippu died while fighting the British in 1799. With the death of Tippu Sultan, the British were happy as if the whole of India came under their rule. Most of the territories under Tippu’s rule were shared among the British, Marathas, and Hyderabad Nizam. A small territory was handed were to the royal representative of Mysore Wodeyars. This region came to be known as Mysore Princely State.

Question 5.
Explain the method of resisting British power by Dondiya Wagh.
Answer:
Dondiya started his career as a cavalry soldier in Hyder Ali’s army and grew to the position of military general. He built his own private army and fought along with Tippu Sultan. Due to differences with Tippu, he was imprisoned. The British released him from prison after the Fourth Anglo-Mysore War. He built a small army and started his operations. He organized the army with the unhappy soldiers of Tippu’s army and the feudatory rulers who had lost power. He captured Bidanoor and Shivamogga forts and made an unsuccessful attempt to capture Chitradurga fort. Lord Wellesley tried check this rebellion.

An attack was organized on Shivamogga, Honalli, Harihara and other places under the control of Dondiya.
Dondiya lost his base. After the capture of Shikaripura, Dondiya ran away towards Gutti, which was under the control of Nizam of Hyderabad. When Nizam’s army attacked Gutti, Dondiya had to run towards the regions of Maratha. The Maratha army attacked him and captured most of his horses, camels and arms. In spite of these, he continued his war fare. When he was caught in between Maratha Army and Nizam’s army, the British attacked him near Yelaparavi and killed him at Konagal.

Question 6.
Explain the method adopted by Rayanna to fight the British.
Answer:
Sangoli Rayanna developed a sense of nationalism and went on organizing an army. He went on organizing secret meetings at sensitive geographical locations. He aimed at looting the treasury and taluk offices of the British. He had an army of five hundred men. He became furious with the villagers who were assisting the British army. The British devised a cunning strategy to capture Rayanna. They encouraged Desais who yere opposing Rani Chenamma. An Amaldhar named Krishnaraya joined hand with them. Thus Rayanna was cunningly captured and brought down to Dharwad. He was declared as an offender and was hanged till death.

Question 7.
Explain the contribution of Puttabasappa of Kodagu in freedom struggle.
Answer:
Puttabasappa took over the leadership of the rebellion. The rebellion started in the hilly region. Puttabasappa organized the rebels and calmed down the people. He declared that tax on tobacco and saflt will be with drawn, if the rebel government assumes power. The rich farmers, land owners and local chieftains were assured of this move. Puttabasappa killed an Amaldhar who was known for his brutality further increased the popularity of Puttabasappa.

The rebels marched towards Mangalore to capture it. The British were engaged in fortifying their fort in Mangalore. The rebels marched towards Mangalore through Panimangalore and Bantwal. They looted the treasury and prison of Bantwal.

The British sought the army of Thalacheri, Kannur and Bombay to quell this uprising. On hearing this development, Puttabasappa and his associated fled towards Sulya. The British captured them with the help of people in Kodagu. Puttabasappa, Lakshmappa, Bangarasa, Kedambadi Ramayaih Gowda and Guddemane Appaih were hanged till death.

Question 8.
Discuss the Surapura rebellion in brief.
Answer:
Surapura is at fifty kilometers from the present day Yadgir. Venkatappa Nayaka came to throne at an early age. His ascendance to throne was opposed by Krishna Nayaka’s brother Peddanayaka. This resulted in internal struggles. The British interfered in the affairs of Surapura. In 1842, they appointed Medes Taylor as their political agent and gained proxy power over Surapura.

In 1857, it came to the notice of the government that the representatives of Nana Saheba were present in Surapura. The British appointed an officer named Campbell to report on the various activities of the King. The officer submitted a report to the resident of Hyderabad that the King is involved in misadministration. The British army captured Surapura in 1858. The war continued, there are confusions regarding Venkatappa Nayaka’s end.

Class 10 Social Science Opposition to British Rule in Karnataka Additional Questions and Answers

Question 1.
British entered an agreement with Hyder Ali through ……….. treaty.
Answer:
Madras.

Question 2.
The treaty of ……….. ended the second Anglo-Mysore War.
Answer:
Mangalore.

Question 3.
Tippu Sultan signed the Treaty of ……….. and with this Third Anglo-Mysore War came to an end.
Answer:
Srirangapatna.

Question 4.
The British attacked Kittur under the leadership of ………….
Answer:
Colonel Deak.

Question 5.
In the year ………. Rayanna was hanged till death.
Answer:
1831.

Question 6.
The rebellion of Hulagali is called ………… rebellion.
Answer:
Bedas.

We hope the KSEEB SSLC Class 10 History Solutions Chapter 4 Opposition to British Rule in Karnataka help you. If you have any query regarding Karnataka SSLC Class 10 History Solutions Chapter 4 Opposition to British Rule in Karnataka, drop a comment below and we will get back to you at the earliest.

Expert Teachers at KSEEBSolutions.com has created KSEEB Karnataka SSLC Model Question Papers 2019-2020 with Answers Pdf Download of KSEEB Previous Year Model Question Papers, Sample Papers for Class 10 all subjects in Kannada Medium and English Medium are part of KSEEB Solutions.

Here we have given Karnataka Secondary Education Examination Board KSEEB SSLC Model Question Papers 2019-20 with Answers for Class 10th Std. Students can view or download the Karnataka Sample Papers for Class 10 for their upcoming Karnataka SSLC board examinations. Students can also read Karnataka SSLC KSEEB Solutions for Class 10.

These Karnataka SSLC Model Question Papers for 10th all subjects in Kannada Medium and English Medium are useful to understand the pattern of questions asked in the board exam. Know about the important concepts to be prepared for Karnataka Board Exams and Score More marks.

KSEEB SSLC 10th Model Question Papers 2019-20 Karnataka Pdf with Answers

These model question papers are designed according to the latest exam pattern, so it will help students to know the exact difficulty level of the question papers.

Karnataka Board KSEEB Model Papers for Class 10 all subjects in both Kannada and English medium are part of Karnataka Board Model Papers. Here we have given Karnataka SSLC Sample Papers for Class 10 all subjects.

Karnataka SSLC Model Question Papers in English and Kannada Medium

• Karnataka SSLC Maths Model Question Papers
• Karnataka SSLC Science Model Question Papers
• Karnataka SSLC Social Science Model Question Papers
• Karnataka SSLC Kannada Model Question Papers
• Karnataka SSLC English Model Question Papers
• Karnataka SSLC Hindi Model Question Papers

To enhance the academic quality of students, in order to measure their learning quality, and to control mass copying as well rote memorization, Karnataka Secondary Education Examination Board has brought a change in the design of the S.S.L.C Examination Model Question Papers from 2019-2020. The board hopes to make some significant changes will bring in students’ attitudes towards periodic growth of social conditions.

In the previous question paper design (KSEEB Question Papers 2019 with answers), the unit vice marks were allotted in all the subjects. Students may ignore certain units because some teachers do not teach those units with less mark, as a result, they can lose certain concepts/valuable aspects which may play a significant role in their future. Keeping this in view theme-based marks are distributed in question paper format from 2019-20 i.e KSEEB Question Papers 2020 with answers.

In the above, we have provided the sample papers of all the subjects like Maths and Science etc. You can download the papers very easily. We will work on Karnataka board. And very soon we will provide Solved Papers for Karnataka board all classes and subjects as per the revised syllabus of 2020.

Practice each and every concept of subjects with great sincerity and dedication. Try to solve questions related to each and every topic, and try to make yourself clear with each and every concept. Work hard, you will get good marks.

It is necessary that students will understand the new pattern and style of question paper according to the latest exam pattern. We hope these model papers will help to prepare for Karnataka SSLC Board Exams 2020.

We hope Karnataka State Board Syllabus KSEEB SSLC 10th Model Question Papers 2019-2020 with Key Answers Pdf Download of KSEEB Previous Year Model Question Papers, Sample Papers for Class 10 all subjects in Kannada Medium and English Medium will help you.

If you have any query regarding Karnataka State NCERT Syllabus KSEEB 10th Standard Model Question Papers 2019-20 with Answers Pdf, drop a comment below and we will get back to you at the earliest.