Class 7

Students can Download Kannada Lesson 1 Puttajji Puttajji Kathe Helu Questions and Answers, Summary, Notes Pdf, Siri Kannada Text Book Class 7 Solutions, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Siri Kannada Text Book Class 7 Solutions Gadya Bhaga Chapter 2 Sina Settaru Namma Teecharu

Sina Settaru Namma Teecharu Questions and Answers, Summary, Notes

Sina Settaru Namma Teecharu Summary in Kannada

Students can Download Chapter 1 Integers Ex 1.1, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 1 Integers Ex 1.1

Question 1.
Following number line shows the temperature in degree Celsius (°C) at different places on a particular day.

a) Observe this number line and write the temperature of the places marked on it.
Solution Places temperature
Solution:

b) What is the temperature difference between the hottest and the coldest places among the above ?
Solution:
Hottest temperature among the above places is Bengaluru } = 22°C

c) What is the temperature difference between Lahulspiti and Srinagar ?
Solution:

d) Can we say the temperature of Srinagar and Shimla taken together is less than the temperature at Shimla? Is it also the temperature at Srinagar?
Solution:
The temperature of Srinagar and Shimla took together = -2°C + 5°C = 3°C
Yes, it is also less than the temperature at Srinagar.

Question 2.
In a quiz, positive marks are given for correct answers, and negative marks are given for incorrect answers. If Jack’s scores in five successive rounds were 25, -5, -10, 15, and 10, what was his total at the end?
Solution:
The scores of five successive rounds = 25, -5, -10, 15, 10
Total scores of Jack at the end of the event
= 25 + 15+ 10+ = 50
= -5 -10 = -15
= 35

Question 3.
At Srinagar temperature was -5°C on Monday and then it dropped by 2°C on Tuesday. What was the temperature of Srinagar on Tuesday? On Wednesday, it rose by 4°C. What was the temperature on this day?
Solution:
The temperature on Monday at Srinagar = -5°C
The temperature on Tuesday at Srinagar = -5 – 2 = -7° C
The temperature on Wednesday at Srinagar = -7 + 4 = -3° C

Question 4.
A plane is flying at a height of 5000 m above the sea level. At a particular point, it is exactly above a submarine floating 1200m below the sea level. What is the vertical distance between them?
Solution:

Height of the plane flying above sea level = + 5000m
Level of the submarine below sea = – 1200m
Vertical distance between them = 5000m – (- 1200m) = 5000 + 1200 = 6200m.

Question 5.
Mohan deposits 2,000 in his bank account and withdraws 1,642 from it, the next day. If the withdrawal of amount from the account is represented by a negative integer, then how will you represent the amount deposited? Find the balance in Mohan’s account after the withdrawal.
Solution:

From the above number line
Distance from A to B = 20 kms
Distance from B to C = 30 kms
Final position is = C
Distance from A to C = – 10 kms
From the number line, the Balance amount represents the distance B to A.
B to A = Rs. 358.

Question 6.
Rita goes 20 km towards east from a point A to the point B. From B, she moves 30 km towards west along the same road. If the distance towards east is represented by a positive integer then, how will you represent the distance travelled towards west? By which integer will you represent her final position from A?
Solution:

From the above number line
Distance from A to B = 20 kms
Distance from B to C = 30 kms
Final position is = C
Distance from A to C = -10 kms
From the original starting point.

Question 7.
In a magic square each row, column and diagonal have the same sum. Check which of the following is a magic square.
Solution:

From the (i) square
1^st row 5 + (-1) + (-4) = 0
2^nd row -5 + -2 + 7 = 0
3^rd row 0 + 3 + -3 = 0
1^st column 5 + -5 + 0 = 0
2^nd column -1 + -2 + 3 = 0
3^rd column -4 + 7 – 3 = 0
1^st Diagonal 5 + -2 + -3 = 0
2^nd Diagonal 0 + -2 + -4 = -6
∴ It is not a magic square.

(ii) 1^st row 1 + -10 + 0 = -9
2^nd row – 4 + -3 – 2 = -9
3^rd row -6 + 4 – 7 = -9
1^st column 1 + – 4 – 6 = -9
2^nd column -10 + -3 + 4 = -9
3^rd column 0 + – 2 – 7 = -9
1^st Diagonal – 6 – 3 + 0 = -9
2^nd Diagonal 1 – 3 – 7 = -9
∴ It is a magic square.

Question 8.
Verify a – (-b) = a + b for the following values of a and b.
i) a = 21, b = 18

ii) a = 118, b = 125

iii) a = 75, b = 84

iv) a = 28, b = 11

Question 9.
Use the sign of >, < or = in the box to make the statements true.
Solution:

Question 10.
A water tank has stepped inside it. A monkey is sitting on the topmost step (i.e., the first step). The water level is at the ninth step.
Solution:
i) He jumps 3 steps down and then jumps back 2 steps up. In how many jumps will he reach the water level?
Solution:
Monkey jumps down = 3 steps
Monkey again jumps up = 2 steps
presenting this on number line,
Initial position = 1^st step
Final position = 9^th step
Representing this on number line,


First Jump = 1 + 3 = 4 steps
Second Jumb = 4 – 2 = 2 steps
Third Jump = 2 + 3 = 5 steps
Fourth Jump = 5 – 2 = 3 steps
Fifth Jump = 3 + 3 = 6 steps
Sixth Jump = 6 – 2 = 4 steps
Seventh Jump = 4 + 3 = 7 steps
Eight Jump = 7 – 2 = 5 steps
nineth Jump = 5 + 3 = 8 steps
tenth Jump = 8 – 2 = 6 steps
Eleventh Jump = 6 + 3 = 9 steps
∴ Monkey took 11 Jumps to reach 9 steps.

ii) After drinking water, he wants to go back. For this, he jumps 4 steps up and then jumps back 2 steps down in every move. In how many jumps will he reach back the top step?
Solution:
Initial position = 9^th step. Final position = 1^st step
Monkey Jumps up = 4 steps
Monkey Jumps down = 2 steps

First Jump = 9 – 4 = 5^th step
Second Jumb = 5 + 2 = 7^th step
Third Jump = 7 – 4 = 3^th step
Fourth Jump = 3 + 2 = 5^th step
Fifth Jump = 5 – 4 = 1^st step
∴ Monkey took 5 Jumps to reach back

iii) If the number of steps moved down is represented by negative integers and the number of steps moved up by positive integers, represent his moves in part (i) and (ii) by completing the following; In (a) the sum (-8) represents going down by eight steps. So, what will the sum 8 in (b) represent?
(a) -3 + 2 – ……. = -8
-3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 – 3 + 2 = -8

(b) 4 – 2 + …. = 8
Thus, Sum 8 in (b) represents going up by eight steps.

Students can Download Kannada Lesson 1 Puttajji Puttajji Kathe Helu Questions and Answers, Summary, Notes Pdf, Siri Kannada Text Book Class 7 Solutions, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Siri Kannada Text Book Class 7 Solutions Gadya Bhaga Chapter 1 Puttajji Puttajji Kathe Helu

Puttajji Puttajji Kathe Helu Questions and Answers, Summary, Notes

Puttajji Puttajji Kathe Helu Summary in Kannada

Students can Download Chapter 12 Algebraic Expressions Ex 12.4, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.4

Question 1.
Observe the patterns of digits made from line segments of equal length. You will find such segmented digits on the display of electronic watches or calculators.
If the number of digits framed is taken to be n, the number of segments required to form n digits is given by the algebraic expression appearing on the right of each pattern.
How many segments are required to form 5, 10, 100 digits of the kind 6, 4, 8.

Solution:
i) The number of segments required to form
5 digits of this kind
5n + 1 = 5 × 5 + 1 = 25 + 1 = 26

ii) The number of segments required to form 10 digits of this kind
5n + 1 = 5 × 10 + 1 = 50 + 1 = 51

iii) The number of segments required to form 100 digits of this kind
5n + 1 = 5 × 100 + 1 = 501

Let the number of digits formed be ‘n’.
Therefore the number of segments required to form ‘n’ digits is given by the algebraic expression 3n + 1.
i) The number of segments required to form 5 digits of this kind.
3n + 1 = 3 × 5 + 1 = 15 + 1 = 16

ii) The number of segments required to form 10 digits of this kind
3n + 1 = 3 × 10 + 1 = 30 + 1 = 31

iii) The number of segments required to form 100 digits of this kind
3n + 1 = 3 × 100 + 1 = 300 + 1 = 301

Let the number of digits formed be ‘n’. Therefore the number of segments required to form ‘n’ digits is given by the algebraic expression 5n + 2.
i) The number of segments required to form 5 digits of this kind.
5n + 2 = 5 × 5 + 2 = 25 + 2 = 27
ii) The number of segments required to form 10 digits of this kind
5n + 2 = 5 × 10 + 2 = 50 + 2 = 52

iii) The number of segments required to form 100 digits of this kind
5n + 2 = 5 × 100 + 2 = 500 + 2 = 502

Question 2.
Use the given algebraic expression to complete the table of number patterns.
Solution:

Students can Download Chapter 12 Algebraic Expressions Ex 12.3, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.3

Question 1.
If m = 2, find the value of:
i) m – 2
2 – 2 = 0

ii) 3m – 5
3 × 2 – 5 = 6 – 5 = 1

iii) 9 – 5m
9 – 5 × 2 ÷ 9 – 10 = -1

iv) 3m² – 2m – 7
= 3 × 2² – 2 × 2 – 7
= 3 × 4 – 4 – 7
= 12 – 11 = 1

v)

Solution:

Question 2.
If p = -2, find the value of :
Solution:
i) 4p + 7
= 4x – 2 + 7
= -8 + 7 = -1

ii) -3p² + 4p + 7
= -3 (-2)² + 4 (-2) + 7
= 3 × 4 – 8 + 7
= 12 – 8 + 7
= -13

iii) -2p³ – 3p² – 4p + 7
= -2 (-2)³ – 3(2)² + 4 (-2) + 7
= -2(-8) – 3(4) – 8 + 7
= 16 – 12 – 8 + 7
= 16 + 7 – 12 – 8
= 23 – 20 = 3

Question 3.
Find the value of the following expressions, when x = -1:

i) 2x – 7
= 2(-1) – 7
= -2 – 7 = -9

ii) -x + 2
= -1 + 2 = 1

iii) x² + 2x + 1
= 1 – 2 + 1 = 0

iv) 2x² – x – 2
= 2(-1)² – (-1) – 2
= 2 × 1 + 1 – 2
= 2 + 1 – 2 = 1

Question 4.
If a = 2, b = -2, find the value of:
i) a² + b²
= 2² + (-2)²
= 4 + 4 = 8

ii) a² + ab + b²
= 2² + 2 × -2 + (-2)
= 4 – 4 + 4 = 4

iii) a² – b²
= (2)² – (-2)² = 4 – 4 = 0

Question 5.
When a = 0, b = -1, find the value of the given expressions :
i) 2a + 2b
= 2 × 0 + 2 × -1
= o – 2 = -2

ii) 2a² + 2b² + 1
= 2(0)² + (-1)² + 1
= 0 + 1 + 1 = 2

iii) 2a²b + 2ab² + ab
= 2 × 0² × -1 + 2 × 0 × (-1)² + 0 × -1
= 0 + 0 + 0 = 0

iv) a² + ab + 2
= (0)² + 0 × -1 + 2
= 0 + 0 + 2 = 2

Question 6.
Simplify the expressions and find the value if x is equal to 2
Solution:
i) x + 7 + 4 (x – 5)
= x + 7 + 4x – 20 (re-arranging the terms)
= 5x – 13 = 5(2) – 12
= 10 – 13 = -3

ii) 3(x + 2) + 5x – 7
= 3x + 6 + 5x – 1
= 8x – 1 = 8(2) – 1
= 16 – 1 = 15

iii) 6x + 5 + 5(x – 2)
= 6x+ 5x – 10
= 11x – 10 = 11(2) – 10
= 22 – 10 = 12

iv) 4(2x – 1) + 3x + 11
= 8x – 4 + 3x + 11
= 8x + 3x – 4 + 11 (re-arranging the terms)
= 11x + 7 = 11(2) + 7 = 22 + 7 = 29

Question 7.
Simplify these expressions and find their values if x = 3, a = -1, b = -2.
i) 3x – 5 – x + 9
= 3x – x + 9 – 5
= 2x + 4 = 2(3) + 4 = 6 + 4 = 10

ii) 2 – 8x + 4x + 4
= 2 + 4 – 4x
= -4x + 6
= -4 × 3 + 6
= -12 + 6 = -6

iii) 3a + 5 – 8a + 1
= 3a – 8a + 5 + 1
= -5a + 6
= -5(-1) + 6
= 5 + 6 = 11

iv) 10 – 3b – 4 – 5b
= -3b – 5b + 10 – 4
= -8b + 6
= -8 (-2) + 6
= 16 + 16 = 22

v) 2a – 2b – 4 – 5 + a
= 2a – 2b – 9
= 3a – 2b – 9
= 3(-1) – 2 (-2) – 9
= -3 + 4 – 9
= -12 + 4 = -8

Question 8.
i) If z = 10, find the value of z³ – 3(z – 10)
= (10)³ – 3 (10 – 10)
= 1000 – 3(0)
= 1000 – 0 = 1000

ii) If p = -10, find the value of p² – 2p – 100
= (10)² – 2(-10) – 100
= 100 + 20 – 100
= 120 – 100
= 20

Question 9.
What should be the value of a if the value of 2x² + x – a equals to 5, when x = 0 ?
Solution:
= 2x² + x – a
= 2(0)² + 0 – a
= 5
= 0 + 0 – a = 5
∴ a = -5

Question 10.
Simplify the expression and find its value when a = 5 and b = -3, 2(a² + ab) + 3 – ab
= 2a² + 2ab + 3 – ab
= 2(5)² + 2 × 5 × -3 + 3 – 5 × -3
= 2 × 25 – 30 + 3 + 15
= 50 – 30 + 3 + 15
= 50 + 3 + 15 – 30
= 68 – 30 = 38

Students can Download Chapter 8 Comparing Quantities Ex 8.1, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 8 Comparing Quantities Ex 8.1

Question 1.
Find the ratio of :
a) 5 to 5o paise
Solution:
Rs. 5 : 50 paise
5 × 100 : 50 paise
500 paise : 50 paise (Dividing by 50)
10 : 1

b) 15 kg to 210 g
15 kg: 210 g
15 × 1000: 210 g
15000 g : 210 g (Dividing by 30) 500 : 7

c) 9 m to 27 cm
9 m : 27 cm
9 × 100 : 27 cm
900 cm : 27 cm (Dividing by 9)
100 : 3

d) 30 days to 36 hours
30 days : 36 hours
30 × 24 : 36 hours
720 hours : 36 hours (Dividing by 36)
20 : 1

Question 2.
In a computer lab, there are 3 computers for every 6 students. How many computers will be needed for 24 students?
Solution:
Let ‘x’ computers to be needed for 24 students. Then the ratio will be 3 : x = 6 : 24

∴ 24 students need 12 computers.

Question 3.
Population of Rajasthan = 570 lakhs and population of UP = 1660 lakhs. Area of Rajasthan = 3 lakh km² and area of UP= 2 lakh km².
Solution:
Population of Rajasthan = 570 lakhs
Population of UP = 1660 lakhs
Area of Rajasthan = 3 lakh sq. kms
Area of U.P = 2 lakh sq. kms

i) How many people are there per km² in both these States ?
Solution:

ii) Which State is less populated?
Solution:
Rajasthan State is less populated.

Students can Download Chapter 7 Congruence of Triangles Ex 7.2, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 7 Congruence of Triangles Ex 7.2

Question 1.
Which congruence criterion do you use in the following ?
a) Given : AC = D
AB = DE
BC = EF

So, ∆ ABC = ∆ DEF
SSS congruence criterion

b) Given : ZX = RP
RQ = ZY
∠PRQ = ∠XZY
So, ∆ PQR ≅ ∆ XYZ

SAS congruence criterion

c) Given: ∠MLN = ∠FGH
∠NML = ∠GFH
ML = FG
So, ∆LMN ≅ ∆GFH

ASA congruence criterion.

d) Given : EB = DB
AE = BC
∠A = ∠C = 90°
So, ∆ ABC ≅ ∆ CDB

RHS congruence criterion

Question 2.
You want to show that ∆ ART ≅ ∆ PEN,

a) If you have to use SSS criterion, then you need to show
i) AR =
AR = PE,
ii) RT =
RT = EN,
iii) AT =
AT = PN

b) If it is given that ∠T = ∠N and you are to use SAS criterion, you need to have

i) RT =
RT = EN and
ii) PN =
PN = AT
c) If it is given that AT = PN and you to use ASA criterion, you need to have
i) ?
∠ATR = ∠PNE and
ii) ?
∠TAR = NPE.

Question 3.
You have to show that ∆ AMP ≅ ∆ AMQ.
In the following proof, supply the missing reasons

Question 4.
In ∆ ABC, ∠A = 30°, ∠B = 40° and ∠C = 110°
In ∆ PQR,

A student says that ∆ ABC ≅ ∆ PQR by AAA congruence criterion. Is he justified? Why or why not?
Solution:
No, he is not justified because AAA is not a criterion for the congruence of triangles.

Question 5.
In the figure, the two triangles are congruent. The corresponding parts are marked. We can write ∆ RAT ≅ ?
Solution:

Question 6.
Complete the congruence statement:
Solution:

∆ BCA ≅ ? ∆ QRS ≅ ?
∆ BCA ≅ ? ∆ BTA ≅ ?
∆ QRS ≅ ? ∆ TPQ

Question 7.
In a squared sheet, draw two triangles of equal areas such that

i) The triangles are congruent.
Solution:
Consider the ∆^S PQS and SQR

In ∆ PQS and ∆ SQR
PS = QR = 6 cms
∠SPQ = ∠QRS = 90°
QS = QS = common
By RHS congruence criterion
∆PQS ≅ ∆SQR.
Perimeter of the ∆ PQS = PQ + QS + PS
Perimeter of the ∆ SQR = SR + QS + QR
∴ Perimeter of the ∆ PQS = Perimeter of the ∆ SQR (∵ PQ = SR & PS = QR)

ii) the triangles are not congruent. What can you say about their perimeters?
Solution:

∴ area of ∆ PQS = Area of ∆ PQM.
By seeing the figure the ∆ PQS = PQ + PS + SQ = 8 + 6 + 10 = 24 cms.
Perimeter of the ∆ PQM = PQ + PM + QM
= 8 + 7.2 + 7.2
= 22.4 cms.
∴ Their perimeters are not equal.
PM = QM
(PM)² = PN² + MN²
= 4² + 6²
= 16 + 36 = 52
PM = QM = \(\sqrt{52}\) = 7.2

Question 8.
Draw a rough sketch of two triangles such that they have five pairs of congruent parts but still the triangles are not congruent.
Solution:

Considering the two triangles PQR and XYZ
In ∆ PQR and XYZ
PQ = XZ = 7 cms
PR = YZ = 6 cms
RQ = XY = 5 cms
∠PRQ = ∠XYZ
∠PQR = ∠XZY
In the above 2 triangles 5 pairs of the congruent present. Still, they are not congruent.

Question 9.
If ∆ ABC and ∆ PQR are to be congruent, name one additional pair of corresponding parts. What criterion did you use?

The additional corresponding part is BC = RQ by ASA congruence rules

Question 10.
Explain, why ∆ ABC ≅ ∆ FED

Solution:
∠ABC = ∠DEF = 90°
BC = DE ∠ACD = ∠EDF
( ∵ The sum of the measures of three angles of a triangle is 180°.)
∴ ∆ ABC ≅ ∆ DEF
(By ASA congruence criterion.)

Students can Download Chapter 12 Algebraic Expressions Ex 12.2, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 12 Algebraic Expressions Ex 12.2

Question 1.
Simplify combining like terms :
i) 21b – 32 + 7b – 20b
(re-arranging the terms)
= 21b + 7b – 20b – 32
= 28b – 20b – 32
= 8b – 32

ii) z² + 13z² – 5z + 7Z² – 15z
(re-arranging the terms)
= -z² + 13z² – 5z – 15z + 7z³
= 12z² – 20z + 7z³

iii) p – (p – q) – q – (q – p)
= p – p + q – q – q + p
(re-arranging the terms)

= p – q

iv) 3a – 2b – ab – (a – b + ab) + 3ab + b – a
= 3a – 2b – ab – a + b – ab + 3ab + b – a
= 3a – a – a – 2b + b + b – ab – ab + 3ab

= a + ab

v) 5x²y – 5x² + 3yx² – 3y² + x² – y² + 8xy² – 3y²
(re-arranging the terms)
= 5x²y + 3yx² – 5x² + x² – 3y² – y² – 3y² + y²
= 8x²y – 4x² – 7y² + 8xy²

vi) (3y² + 5y – 4) – (8y – y² – 4) .
3y² + 5y – 4 – 8y – y² – 4
(re-arranging the terms)
3y² + y² + 5y – 8y – 4 + 4
= 14y² – 3y

Question 2.
Add:
i) 3mn, -5mn, 8mn, -4mn
3mn + 8nm – 5mn – 4mn
(re-arranging the terms)
= 11mn – 9mn = 2mn

ii) t – 8tz, 3tz – z, z – t

(re-arranging the terms)
= -8tz + 3tz = -5tz

iii) -7mn + 5, 12mn + 2, 9mn – 8, – 2mn – 3
-7mn + 12 mn + 9mn – 2mn + 5 + 2 – 8 – 3
(re-arranging the terms)
= -7mn – 2mn + 12mn + 9mn + 7 – 11
= -9mn + 21 mn -4
= 12mn – 4

iv) a + b -3, b – a +3, a – b + 3

(re-arranging the terms)
= (2a – a + 2b – b + 6 – 3)
= a + b + 3

v) 14x + 10y – 12xy – 13, 18 – 7x – 10y + 8xy, 4xy
(re-arranging the terms)

= 17x + 51

vi) 5m – 7n, 3n – 4m + 2, 2m – 3mn – 5
(re-arranging the terms)
= 5m – 4m + 2m – 7n + 3n – 3mn + 2 – 5
= 7m – 4m – 7n + 3n – 3mn – 3
= 13m – 4n – 3mn – 3

vii) 4x²y, – 3xy², 5y², 5x²y
(Re-arranging the terms)
4x²y + 5x²y – 3xy² – 5xy²
= 9x²y – 8xy²

viii) 3p²q² – 4pq + 5, 10p²q², 15 + 9pq + 7p²q²
3p²q² – 10p²q² + 7p²q² – 4pq + 9pq + 5+ 15
(Re-arranging the terms)
= 3p²q² + 7p²q² – 10p²q² – 4pq + 9pq + 5 + 15

= 5pq + 20

ix) ab – 4a, 4b – ab, 4a – 4b

(Re-arranging the terms)
(All terms are cancelling)
= 0

x) x² – y² – 1, y² – 1 – x², 1 – x² – y²
(Re arranging the terms)

= -x² – y² – 1

Question 3.
Subtract :
i) -5y² from y²
= y² – (-5y²)
= y² + 5y² = 6y²

ii) 6xy from – 12xy
= -12xy – (6xy)
= -12xy – 6xy = -18xy

iii) (a – b) from (a + b)
= a + b – (a – b)

= 2b

iv) a(b – 5) from b(5 – a)
= ab – 5a from 5b – ab
= (5b – ab) – (ab – 5a)
= 5b – ab – ab + 5a
= 5a – 5b – 2ab

v) -m² + 5mn from 4m² – 3mn + 8
= (4m² – 3mn + 8) – (-m² + 5mn)
= 4m² – 3mn + 8 + m² – 5mn
= 4m² + m² – 3mn – 5mn + 8
= 5m² – 8mn + 8

vi) -x² + 10x – 5 from 5x – 10
= (5x – 10) – (-x² + 10x – 5)
= 5x – 10 + x² – 10x + 5
= 5x – 10x + x² – 10 + 5
= x² – 5x – 5

vii) 5a² – 7 ab + 5b² from 3ab – 2a² – 2b²
= (3ab – 2a² – 2b²) – (5a² – 7ab + 5b²)
= 3ab – 2a² – 2b² – 5a² + 7ab – 5b
= -2a² – 5a² – 2b² – 5b² + 7ab + 3ab
= -7a² – 7b² + 10ab

viii) 4pq – 5q² – 3p² from 5p² + 3q² – pq
= (5p² + 3q² – pq) – (4pq – 5q² – 3p²)
= 5p² + 3q² – pq² – 4pq + 5q² + 3p²
= 5p² + 3p² + 3q² + 5q² – 4pq – pq
= 8p² + 8q² – 5pq

Question 4.
a) What should be added to x² + xy + y² to obtain 2x² + 3xy ?
Solution:
= 2x² + 3xy – (x² + xy + y²)
= 2x² + 3xy – x² – xy – y² – 2x² – x² + 3xy – xy – y²
(re-arranging the terms)
∴ x² + 2xy – y² should be added to x² + xy + y² to get 2x² + 3xy

b) What should be subtracted from 2a + 8b + 10 to get -3a + 7b + 16 ?
Solution:
= (2a + 8b + 10) – (-3a + 7b + 16)
= 2a + 8b+ 10 + 3a – 7b – 16
= 2a + 3a + 8b – 7b + 10 – 16
(re-arranging the terms)
= 5a + b – 6
∴ (5a + b – 6) should be subtracted from 2a + 8b +10 to get -3a + 7b + 16.

Question 5.
What should be taken away from – 4y² + 5xy + 20 to obtain -x² – y² + 6xy + 20?
Solution:
= (3x² – 4y² + 5xy + 20) – (-x² – y² + 6xy + 20)
= 3x² – 4y² + 5xy + 20 + x² + y² – 6xy – 20
= 3x² + x² – 4y² + y² + 5xy – 6xy + 20 – 20
(re – arranging the terms)
= 4x² – 3y – xy
∴ 4x² + 3y² – xy should be taken away from 3x² – 4y² + 5xy + 20 to obtain -x² – y² + 6xy + 20

Question 6.
From the sum of 3x – y + 11 and -y – 11, subtract 3x – y – 11.
Solution:

b) From the sum of 4 + 3x and 5 – 4x + 2x², subtract the sum of 3x² – 5x and -x² + 2x + 5.
Solution:

or
[(4 + 3x) + (5 – 4x + 2x²)] – [(3x² – 5x) + (x² + 2x + 5)]
= 4 + 3x + 5 – 4x + 2x² – 3x² + 5x + x² – 2x – 5 = 4 + 3x – 4x + 2x² + 5 – 3x² + x² + 5x – 2x – 5
(re-arranging the terms)

= 2x + 4

Students can Download Chapter 10 Practical Geometry Ex 10.5, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 10 Practical Geometry Ex 10.5

Question 1.
Construct the right angled ∆ PQR? where m∠Q = 90°, QR = 8 cm and PR = 10 cm.
Solution:

Solution:
Steps of Construction

1. Draw a line segment of length 8 cm and named as QR. At Q draw QM ⊥ QR.
2. With R as centre, draw an arc of radius 10 cm and cut the ⊥^le line at P.
3. Join PR. Now we get the required PQR ∆

Question 2.
Construct a right-angled triangle whose hypotenuse is 6 cm long and one of the legs is 4
cm long.
Solution:

Steps of Construction

1. Draw a line segment of length 4 cm and named it AB. At A draw a ⊥ line AM.
2. With B as centre, draw an arc of 6 cms cut the x line at ‘C ’. Now we obtained the required triangle ABC.

Question 3.
Construct an isosceles right-angled triangle ABC, where m∠ACB = 90° and AC = 6 cm.
Solution:

Steps of Construction

1. Draw a line segment of 6 cm, ie., AC. At C draw CM ⊥ CA.
2. With ‘C’ as the centre draw an arc of radius 6 cm to intersect CM at ‘B’
3. Join AB. Now we get the required ∆ACB.

Students can Download Chapter 10 Practical Geometry Ex 10.4, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 7 Maths Chapter 10 Practical Geometry Ex 10.4

Question 1.
Construct ∆ ABC, given m∠A= 30° and m∠B = 30° and AB = 5.8 cm.
Solution:

Steps of Construction.

1. Draw a line segment AB of length 5.8 cm, at A draw a ray AM making an angle of 60° with AB.
2. At ‘B’ draw a ray ABN making an angle of 30° with BA.
3. Mark the point of intersection of two rays as ‘C’. Now we get the A ABC which is required.

Question 2.
Construct ∆ PQR if PQ = 5 cm, m∠PQR = 105° and m∠QRP = 40°. (Hint Recall angle-sum property of a triangle).
Solution:

By angle-sum property of a triangle ∠RPQ + ∠PQR + ∠QRP = 180°
∠RPQ + 105° + 40° = 180°
∠RPQ + 145° = 180°
∴ ∠RPQ = 180° – 145° = 35°,

Steps of Construction

1. Draw a line segment PQ of length 5 cm. At P draw a ray PM making an angle of 35° with PQ.
2. At Q draw a ray QN making an angle of 105° with QP.
3. Mark the point of intersection of two rays as R. We get the required P Q R triangle.

Question 3.
Examine whether you can construct ∆ DEF such that E F = 7.2 cm. m∠E = 110° and m∠F = 80°. Justify your answer.
Solution:
∠E + ∠F = 110°+ 80°= 190°
This triangle is not possible to construct because the measures of two angles exceed 180°.
The sum of three angles of a triangle is always equal to 180°.
So it is not possible.