Category: Class 7

Students can Download Chapter 6 The Triangles and Its Properties Ex 6.2, 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 6 The Triangles and Its Properties Ex 6.2

Question 1.
Find the value of the unknown exterior angle x in the following diagrams :

i) ∠x = 50° + 70° = 120°
(∵ The exterior angle of the triangle is equal to the sum of its two interior opposite angles.)

ii) ∠x = 45° + 65° = 110°
(∵ The exterior angle of the triangle is equal to the sum of its two interior opposite angles.)

iii) ∠x = 30° + 40° = 70°
(∵ The exterior angle of the triangle is equal to the sum of its two interior opposite angles.)

iv) ∠x = 60° + 60° = 120°
(∵ The exterior angle of the triangle is equal to the sum of its two interior opposite angles.)

v) ∠x = 50° + 50° = 100°
(∵ The exterior angle of a triangle is equal to the sum of its two interior opposite angles.)

Question 2.
Find the value of the unknown interior angle x in the following figures:

Solution:
i) ∠x + 50° = 115°
∵ The exterior angle of a triangle is equal to the sum of its two interior opposite angles.
∵ ∠x = 115° – 50° = 65°

ii) ∠x + 70° = 100°
Solution:
(∵ The exterior angle of a triangle is equal to the sum of its two interior opposite angles)
∴ ∠x = 100° – 70°= 30°

iii) ∠x + 90° = 125°
∵ The exterior angle of a triangle is equal to the sum of its two interior opposite angles.
∴ ∠x = 125° – 90° = 35°

iv) ∠x + 60° = 120°
∵ The exterior angle of a triangle is equal to the sum of its two interior opposite angles.
∠x = 120° – 60° = 6o°

v) ∠x + 30° = 80°
∵ The exterior angle of a triangle is equal to the sum of its two interior opposite angles.
∵ ∠x = 80° – 30° = 50°

vi) ∠x + 35° = 75°
∵ The exterior angle of a triangle is equal to the sum of its two interior opposite angles.
∴ ∠x = 75° – 35° = 40°

Students can Download Chapter 11 Perimeter and Area Ex 11.3, 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 11 Perimeter and Area Ex 11.3

Question 1.
Find the circumference of the circles with the following radius :
(Take n = \(\frac{22}{7}\))
a) 14 cm
radius = r = 14 cm
∴ circumference of a circle = 2πr

b) 28 mm
radius = r = 28 cm
∴ circumference of a circle = 2πr

c) 21 cm
radius = r = 21 cm
circumference of a circle = 2πr

Question 2.
Find the area of the following circles, given that:

a) radius = 14 mm (Take π = \(\frac{22}{7}\))
radius = 14 mm
∴ Area of the circle = πr²

= 616 sq.cms

b) diameter = 49 m

Area of the circle = πr²

= 1,886.5 sq.metrs

c) radius = 5 cm
radius = r = 5 cm
∴ Area of the circle = πr²

Question 3.
If the circumference of a circular sheet is 154m, find its radius. Also find the area of the sheet. (Take π = \(\frac{22}{7}\))
Solution:
C = circumference of a circle = 154m = 2πr

Area of the sheet = πr²

= 1,886.5 sq.metrs

Question 4.
A gardener wants to fence a circular garden of diameter 21m. Find the length of the rope he needs to purchase, if he makes 2 rounds of fence. Also find the cost of the rope, if it costs ₹ 4 per meter.
(Take π = \(\frac{22}{7}\))
d = diameter = 21 m

∴ radius of the circular garden

circumference of the circular garden = 2πr

Length of the rope needed to make one round of fence } = 66 mts
Length of the rope needed to make 2 rounds =
66 × 2 = 132 mts
cost of rope per metre = Rs. 4
Total cost of rope = 132 × 4 = Rs. 528

Question 5.
From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet.
(Take π = 3.14)
Solution:
Radius of the outer circle = 4 cms.

∴ Area of the outer circle = πr²
= 3.14 × 4 × 4
= 28.26 sq.cms
= 50.24 sq.cms
π = 3.14
Radius of the inner circle = 3 cms
∴ Area of the inner circle = πr²
= 3.14 × 3 × 3 = 28.26 sq. cms
Area of the remaining part = 50.24 – 28.26 = 21.98 sq.cms

Question 6.
Saima wants to put lace on the edge of a circular table cover of diameter 1.5m. Find the length of the lace required and also find its cost if one meter of the lace
costs ₹ 15. (Take π = 3.14)
Solution:
Diameter of the table cover = 1.5m
Circumference of the table cover }
2πr – πd = 3.14 × 1.5 (∵ d = 2r) = 4.710 mts
∴ Length of the lace required = 4.71 mt
Cost of lace per meter = Rs. 15
∴ Total cost of the lace =4.71 × 15 = Rs 70.65

Question 7.
Find the perimeter of the adjoining figure, which is a semicircle including its diameter.

Solution:
Diameter of the semi-circle = 10 cm
Circumference of semi-circle =

Perimeter of the semi-circle(including its diameter) } = 15.7 + 10 = 25.7 cms

Question 8.
Find the cost of polishing a circular table-top of diameter 1.6m, if the rate of polishing is ? 15/m². (Take π = 3.14)
Solution:
Diameter of the Table Top = 1.6 m
Area of the Table Top = πr²

Rate of polishing = Rs. 15/sq.m
Total cost of polishing = 2.0096 × 15
= Rs.30.14

Question 9.
Shazil took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area. If the same wire is bent into the shape of a square, what will be the length of each of its sides ? Which figure encloses more area, the circle or the square? (Take π = \(\frac{22}{7}\))
Solution:
Length of the wire = 44 cms
Total length of the circle = perimeter of the circle

∴ radius = 7 cms.

∴ Length of wire = perimeter of the square = 154 sq.cms
44 = 44

Area of the square = a² = 11 × 11 = 121 sq.cms.
∴ The area of the circle is more means the circle encloses more area.

Question 10.
From a circular card sheet of radius 14cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1 cm are removed, (as shown in the adjoining figure). Find the area of the remaining sheet. (Take π = \(\frac{22}{7}\))
Solution:
Radius of the circular Card = 14 cm
Area of the Card


Area of small 2 circles =

= 77 sq. cms.
Area of rectangle = l × b = 3 × 1 = 3 sq. cms
Area of the remaining sheet
= 616 – (77 + 3)
= 616 = 80 = 536 sq.cm

Question 11.
A circle of radius 2 cm is cut out from a square piece of an aluminium sheet of side 6 cm. What is the area of the leftover aluminium sheet? (Take π = 3.14)
Solution:
Area of square aluminium sheet = 6 × 6 = 36 sq. cms
Side = 6 cm radius = 2 cm
Area of circle =

Area of the left over Aluminium sheet = 36 – 12.56 = 23.44sq.cm

Question 12.
The circumference of a circle is 31.4. Find the radius and the area of the circle? (Take π = 3.14)
Solution:
The circumference of a circle = 2πr = 31.4 cm

Area of the cirlce = πr²
= 3.14 × 5 × 5
= 78.5 sq.cms

Question 13.
A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66m. What is the area of this path? (π = 3.14)
Solution:

Diameter of the flower bed = 66 m
Radius = 66/2 = 33 m
Area of the flower bed = πr²
= 3.14 × 33 × 33
= 3419.46 sq.cm
Radius of the flower bed with path = 33 + 4 = 37 ms.
Area of the flower bed with path =
= 3.14 × 37 × 37
= 4298.66 sq.cm
Area of the path = Area of the flower bed with path – Area of the flower bed
= 4298.66 – 3419.46
= 879.2 sq.mts

Question 14.
A circular flower garden has an area of 314 m². A sprinkler at the centre of the garden can cover an area that has a radius of 12 m. Will sprinkler water the
entire garden? (Take π = 3.14)
Solution:
Area of the circular flower garden = 314 sq.m
Radius of the sprinkler = 12m
Area of the sprinkler = πr² = 3.14 × 12 × 12
= 452.16 sq.mt
∴ 452.16 > 314
∴ Yes, the sprinkler water the entire garden.

Question 15.
Find the circumference of the inner and the outer circles, shown in the adjoining figure? (Take π = 3.14)
Solution:

Circumference of the outer circle = 2πr
radius = 19 m
= 2 × 3.14 × 19
= 119.32 mts
Circumference of the inner circle = 2πr
radius = 19 – 10 = 9 cm.
= 2 × 3.14 × 9
= 56.52 cm

Question 16.
How many times a wheel of radius 25cm must to go 352 m ? (Take π = \(\frac{22}{7}\))
Solution:
radius of the wheel = r = 28 cm
circumference of the wheel = 2πr

= 176 cms
To cover the distance of 352 mtrs (∵ 1 m = 100 cm)
No. of rotation

Question 17.
The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour? (Take π = 3.14)
Solution:
Length of the minute hand = 15 cm
1 hour = 1 round = circumference = 2πr
= 2 × 3.14 × 15
= 94.2 cms
∴ The tip of the minute hand in one hour moves 94.2 cms.

Students can Download Chapter 6 The Triangles and Its Properties Ex 6.1, 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 6 The Triangles and Its Properties Ex 6.1

Question 1.
In ∆PQR, D is the mid-point of \(\overline{\mathbf{Q R}}\) \(\overline{\mathbf{P M}}\) is ___
Solution:
\(\overline{\mathbf{P M}}\) is the altitude of the ∆PQR
∆ PQR

\(\overline{\mathbf{P D}}\) is ___.
\(\overline{\mathbf{P D}}\) is the median of the ∆ PQR
Is QM = MR?
No, QM ≠ MR ?

Question 2.
Draw rough sketches for the following:
Solution:
a) In ∆ ABC, BE is a median

b) In ∆ PQR, PQ and PR are altitudes of the triangle.

c) In ∆XYZ, YL is an altitude in the exterior of the triangle.

Question 3.
Verify by drawing a diagram if the median and altitude of an isosceles triangle can be same.
Solution:

In the ∆PQR
PQ = PR
PM is the Median (∵ QM = MR)
PM is the altitude. (∵ PMR = 90°)
∴ In this isosceles triangle median and altitude are the same. (proved / verified).

Students can Download Chapter 8 Comparing Quantities Ex 8.3, 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 8 Comparing Quantities Ex 8.3

Question 1.
Tell what is the profit or loss in the following transactions. Also find profit per cent or loss per cent in each case.
a) Gardening shears bought for ₹ 250 and sold for ₹ 325.
Solution:
C. P = Rs 250
S. P = Rs. 325
S.P > C.P
∴ It is profit.
Total profit = S.P – C.P
= 325 – 250 = Rs 75

b) A refrigerator bought for ₹ 12,000 and sold at ₹ 13,500.
Solution:
C. P = Rs. 12,000
S.P = Rs. 13,500
S.P > C.P
It is profit
Total profit = S. P – C. P
= 13,500 – 12,000
= Rs.1,500/-

c) A cupboard bought for ₹ 2,500 and sold at ₹ 3,000.
Solution:
C.P = Rs. 2,500
S. P = Rs. 3,000
S.P > C.P
It is profit
Total profit = S. P – C. P
= 3000 – 2500
= Rs. 500

d) A skirt bought for 250 and sold at ₹ 150.
C.P = Rs. 250/-
S.P = Rs. 150/-
S.P < C.P
∴ It is loss
Total loss = C.P – S.P
= 250 – 150 = Rs. 100

Question 2.
Convert each part of the ratio to percentage :
a) 3 : 1
Solution:
Total parts = 3 + 1 = 4
Percentage of First part of the ratio =

Percentage of 2^nd part of the ration =

b) 2 : 3 : 5
Solution:
Total parts = 2 + 3 + 5 = 10

% of second part of the ratio =

% of third part of the ratio =

c) 1 : 4
Total parts = 1 + 4 = 5

d) 1 : 2 : 5
Total parts. = 1 + 2 + 5 = 8

% of second part of the ratio =

Question 3.
The population of a city decreased from 25,000 to 24,500. Find the percentage decrease.
Solution:
Decreased population = 25,000 – 24,500
= 500
Percentage of decrease =

Question 4.
Arun bought a car for 3, 50, 000. The next year, the price went upto ₹ 3,70,000. What was the percentage of price increase ?
Solution:
Price of a car = Rs. 3,50,000
Increased price of a car = Rs. 3,70,000
Total increase = Rs. 20,000
Percentage of price increase =

Question 5.
I buy a T. V. for ₹ 10,000 and sell it at a profit of 20%. How much money do I get for it ?
Solution:
C.P of a T.V = Rs. 10,000
Profit = 20%
20% of Rs. 10,000

∴ Amount I get = C. P + profit
= 10,000 + 2,000
= Rs. 12,000

Question 6.
Juhi sells a washing machine for ₹ 13,500. She loses 20% in the bargain. What was the price at which she bought it ?
Solution:
S. P = C. P – Loss

∴ She bought that washing machine for Rs. 16,875.

Question 7.
i) Chalk contains calcium, carbon and oxygen in the ratio 10 : 3 : 12. Find the percentage of carbon in chalk.
Solution:
Total parts = 10 + 3 + 12 = 25

% of carbon in chalk =

ii) If in a stick of chalk. carbon is 3g, what is the chalk stick?
Solution:
Let the weight of chalk stick be ‘M’
Then 12% of M = 3

∴ The weight of the chalk stick is 25 grams.

Question 8.
Amina buys a book for ₹ 275 and sells it at a loss of 15%. How much does she sell it for ?
Solution:
C. P of a book = Rs. 275
loss = 15% of C. P

S.P = C.P – Loss
= 275 – 41.25
= Rs. 233.751
∴ She sells it for Rs. 233.75.

Question 9.
Find the amount to be paid at the end of 3 years in each case:
a) Principal = ₹ 1,200 at 12 % p.a
P = Rs. 1200/-
R = 12%
T = 3 years

A = P + I = 1200 + 432 = Rs. 1,632
∴ The amount to paid at the end of 3 years is Rs. 1,632

b) Principal = ₹ 7,500 at 5% p.a
P = Rs. 7,500
R = 5%
T = 3 years

A = P + I
= 7500 + 1125
= Rs. 8,625
∴ The amount to be paid at the end of 3 years is Rs. 8,625.

Question 10.
What rate gives ₹ 280 as interest on a sum of ₹ 56,000 in 2 years ?
Solution:
P = Rs. 56,000/-
R = ?
T = 2 years
I = Rs. 280

∴ The rate of interest is 0.25% per annum.

Question 11.
If Meena gives an interest of ? 45 for one year at 9% rate p.a. What is the sum she has borrowed ?
Solution:
P = ?
R = 9%
T = 1 years
I = Rs. 45

∴ She has borrowed the sum of Rs. 500

Students can Download Chapter 13 Exponents and Powers Ex 13.1, 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 13 Exponents and Powers Ex 13.1

Question 1.
Find the value of :
(i) 2⁶
(ii) 9³
(iii) 11²
(iv) 5⁴
Solution:
i) 2⁶
2⁶ = 2 × 2 × 2 × 2 × 2 × 2 = 64

ii) 9³
9³ = 9 × 9 × 9 = 729

iii) 11²
11² = 11 × 11 = 121

iv) 5⁴
5⁴ = 5 × 5 × 5 × 5 = 625.

Question 2.
Express the following in exponential form :
(i) 6 × 6 × 6 × 6
(ii) t × t
(iii) b × b × b × b
(iv) 5 × 5 × 7 × 7 × 7
(v) 2 × 2 × a × a
(vi) a × a × a × c × c × c × c × d
Solution:
i) 6 × 6 × 6 × 6 = 6⁴

ii) t × t = t²

iii) b × b × b × b = b⁴

iv) 5 × 5 × 7 × 7 × 7 = 5² × 7³

v) 2 × 2 × a × a = 2² × a²

vi) a × a × a × c × c × c × c × d = a³ × c⁴ × d

Question 3.
Express each of the following numbers using exponential notation :
(i) 512
(ii) 343
(iii) 729
(iv) 3125
Solution:
i) 512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 2⁹
The exponential notation of 512 is 2⁹

ii) 343 = 7 × 7 × 7 = 7³
∴ The exponential notation of 343 is 7³

iii) 729 = 3 × 3 × 3 × 3 × 3 × 3 = 3⁶
∴ The exponential notation of 729 is 3⁶

iv) 3125 = 5 × 5 × 5 × 5 × 5 = 5⁵
∴ The exponential notation of 3125 is 5⁵

Question 4.
Identify the grater number, wherever possible, in each of the following ?
Solution:
i) 4³ or 3⁴
4³ = 4 × 4 × 4 = 64
3⁴ = 3 × 3 × 3 × 3 = 81
∴ 3⁴ > 4³

ii) 5³ or 3⁵
5³ = 5 × 5 × 5 = 125
3⁵ = 3 × 3 × 3 × 3 × 3 = 243
∴ 3⁵ >5³

iii) 2⁸ or 8²
2⁸ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 256
8² = 8 × 8 = 64
∴ 2⁸ > 8²

iv) 100² or 2¹⁰⁰
100² = 100 × 100= 10,000
2⁸ = (2¹⁰)¹⁰ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
× 2 × 2
= (1024)¹⁰ = [(1024)²]⁵ = [10,48,576]
= (10,48,476)⁵ > 10,000
= 2¹⁰⁰ > 100²

v) 2¹⁰ or 10²
2¹⁰ = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024
10² = 10 × 10 = 100
∴ 2¹⁰ > 10²

Question 5.
Express each of the following as product of powers of their prime factors :
Solution:
i) 648
648 = 2 × 2 × 2 × 3 × 3 × 3 × 3
= 2³ × 3⁴

ii) 405
405 = 5 × 3 × 3 × 3 × 3
= 5 × 3⁴

iii) 540
540 = 2 × 2 × 3 × 3 × 3 × 5
= 2² × 3³ × 5

iv) 3,600
3,600 = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 5
= 2⁴ × 3² × 5²

Question 6.
Simplify :
(i) 2 × 10³
(ii) 7² × 2²
(iii) 2³ × 5
(iv) 3 × 4⁴
(v) 0 × 10²
(vi) 5² × 3³
(vii) 2⁴ × 3²
(viii) 3² × 10⁴
Solution:
i) 2 × 10³
= 2 × 10 × 10 × 10 = 2,000

ii) 7² × 2²
= 7 × 7 × 2 × 2
= 49 × 4 = 196

iii) 2³ × 5
= 2 × 2 × 2 × 5
= 8 × 5 = 40

iv) 3 × 4⁴
= 3 × 4 × 4 × 4 × 4
= 12 × 64 = 768

v) 0 × 10²
= 0 × 10 × 10 = 0

vi) 5² × 3³
= 5 × 5 × 3 × 3 × 3
= 25 × 27 = 675

vii) 24 × 32
=2 × 2 × 2 × 2 × 3 × 3
= 16 × 9 = 144

viii) 3^{2 × 104
= 3 × 3 × 10 × 10 × 10 × 10
= 90 × 1000 = 90,000}

Question 7.
Simplify :
Solution:
i) (-4)³
(-4)³ = (-4) × (-4) × (-4) = -64

ii) (-3) × (-2)³
= (-3) × (-2) × (-2) × (-2) = (-3) × (-8) = 24

iii) (-3)² × (-5)²
= (-3) × (-3) × (-5) × (-5)
= 9 × 25 = 225

iv) (-2)³ × (-10)³
= (-2) × (-2) × (-2) × (-10) × (-10) × (-10)
= +4 × -2 × 100 × -10
= -8 × -1000 = 8000

Question 8.
Compare the following numbers :
Solution:
i) 2 : 7 × 10¹² ; 1.5 × 10⁸
= 2.7 × 10¹²
= 2.7 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10
= 2.70000000000000
= 27 × 10¹¹

1.5 × 10⁷
= 1.5 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10
= 1.5 × 10⁷
∴ 2.7 × 10¹² >1.5 × 10⁸

ii) 4 × 10¹⁴ ; 3 × 10¹⁷
=4 × 10¹⁴ = 4 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10
= 4 × 10¹⁴
3 × 10⁷ = 3 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10 × 10
∴ 3 × 10¹⁷ > 4 × 10¹⁴

Students can Download Chapter 11 Perimeter and Area Ex 11.2, 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 11 Perimeter and Area Ex 11.2

Question 1.
Find the area of each of the following parallelograms :
a)
Solution:
Base = B = 7cm
Height = h = 4 cm
∴ Area of the parallelogram = B × h
= 7 × 4 = 28 sq.cms

b) Base = B = 5 cm
Height = h = 3 cm.

Area of the parallelogram = Base × height
= B × h = 5 × 3
= 15 sq.cms

c) Base = B = 2.5 cms
Height = h = 3.5 cms

Area of the parallelogram
Base × height = 2.5 × 3.5 = 8.75 sq.cms

d) Base = B = 5 cms
Height = h = 4.8 cms

Area of the parallelogram = Base × height
= 5 × 4.8 = 24 sq.cms

e) Base = B = 2 cms

Height = h = 4.4 cms
∴ Area of the parallelogram = Base × height
2 × 4.4 = 8.8 sq.cms

Question 2.
Find the area of each of the following triangles :
Solution:
a) Base = B = 4 cms
Height = h = 3 cms

b) Base = B = 5cms
Height = h = 3.2 cms

∴ Area of the triangle

c) Base = B = 3 cms
Height = h = 4 cms
Area of the triangle

d) Base = B = 3 cms
Height = h = 2 cms
Area of the parallelogram

Question 3.
Find the missing values :

Solution:
a) Area of the parallelogram = Base × height
Base = 20 cm
Area = 246 sq cm
Height = ?
246 = 20 × h

Height = 12.3 cms

b) Height = 15 cm
Area = 154.5 cm²
Base = ?
Area of the parallelogram = Base × height
154.5 = B × 15

∴ Base = 10.3 cms

c) Height = 8.4 cm
Area = 48.72 cm²
Base = ?
Area of the parallelogram = Base × height
48.72 = B × 8.4

∴ Base = 5.8 cms

d) Base = 15.6 cm
Area = 16.38 sq cm
Height = ?
Area of the parallelogram = Base × height
16.38 = 15.6 × h

∴ Height = 1.05 cms

Question 4.
Find the missing values :

Solution:
i) Base = 15 cm Area = 87 cm² Height = ?

ii) Height = 31.4 mm
Area = 1256 mm²
Base = ?

iii) Base = 22 cm Area = 170.5 cm²
Height = ?

Question 5.
PQRS is a parallelogram (Fig.23). QM is the height from Q to SR and QN is the height from Q to PS. If SR = 12 cm and QM = 7.6 cm. Find :
Solution:

a) the area of the parallelogram PQRS
Area of the parallelogram PQRS = Base × height
Base = SR =12 cm
Height = QM = 7.6 cm
= 12 × 7.6 = 91.2 sq.cms
∴ Area = 91.2 cm²

b) QN, if PS = 8 cm
Base = PS = 8 cm Height = QN = ?
Area = 91.2 cm²
Area of the parallelogram = Base × height
91.2 = 8 × QN

∴ QN = 11.4 cms.

Question 6.
DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD (Fig 11.24). If the area of the parallelogram is 1470 cm², AB = 35 cm and AD = 49 cm, find the length of BM and DL.
Solution:

Area of the parallelogram = Base × height
i) Base = AB = 35 cm
Height = DL = ?
Area = 1470 cm²
1470 = 35 × DL

ii) Base = AD = 49 cm
Height = BM = ?
Area = 1470 cm²
Area of the parallelogram = Base × height
1470 = 49 × BM

∴ BM = 30 cms.

Question 7.
∆ ABC is right-angled at A (Fig 11.25). AD Is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, Find the area of ∆ ABC. Also, find the length of AD.
Solution:

Question 8.
∆ ABC is isosceles with AB AC = 7.5 cm and BC = 9 cm (Fig 11.26). The height AD from A to BC is 6 cm. Find the area is 6 cm. Find the area of ∆ABC. What will be the height from C to AB i.e., CE?
Solution:



∴ The height from C to AB = CE = 7.2 cms

Students can Download Chapter 11 Perimeter and Area Ex 11.1, 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 11 Perimeter and Area Ex 11.1

Question 1.
The length and the breadth of a rectangular piece of land are 500 m and 300 m respectively. Find
i) its area
Solution:

Length of the land = l = 500 m
Breadth of the land = b = 300 m
∴ The Area of rectangular piece of land
l × b = 500 × 300
= 150,000 Sq. m.

ii) The cost of the. land, if 1 m² of the land costs 10,000.
Solution:
The cost of rectangular piece of land per
1 Sq m = ₹ 10,000
∴ Total cost of that land = 150,000 × 10,000 = ₹ 1,500,000,000

Question 2.
Find the area of a square park whose perimeter is 320 m.
Solution:

Perimeter of the sq park = 320 m
∴ One side of the sq part = \(\frac{320}{4}\) = 80 m
∴ Area of the square park = a² = side² = 80²
80 × 80 = 6400 Sq.m

Question 3.
Find the breadth of a rectangular plot of land, if its area is 440 m² and the length is 22 m. Also find its perimeter.
Solution:
Area of a rectangular filed = l × b = 440 sq m.

Length of a rectangular field =

perimeter of the rectangular filed = 2(l + b)
2 (20 + 22)
2(42) = 84 m

Question 4.
The perimeter of a rectangular sheet is 100 cm. If the length is 35 cm, find its breadth. Also find the area.
Solution:
Perimeter of the rectangular sheet = 100 cm
length of the rectangular sheet = 35 cm
Breadth of the rectangular sheet = 2(l + b) = 100
= 2(35 + b)= 100
= 70 + 2b = 100
= 2b = 100 – 70

Breadth = 15 cms
∴ Area = l × b = 35 × 15 = 525 sq.cms

Question 5.
The area of a square park is the same as of a rectangular park. If the side of the square park is 60 m and the length of the rectangular park is 90 m, find the breadth of the rectangular park.
Solution:

Area of the sq park = Area of rectangular park
a = side = 60 m
a² = l × b
60² = 90 × b
3600 = 90 × b

∴ The breadth of a rectangular park = 40 mts.

Question 6.
A wire is the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the same wire is rebent in the shape of a square, what will be the measure of each side? Also, find which shape encloses more area?
Solution:
The shape of the wire rectangle.
The perimeter of this rectangle
Total or length of wire =2 (l + b)
length = 40 cm breadth = 22 cm
= 2(40 + 22)
= 2(62) = 124 cm.
If it is reshaped in the form of sq, its side}

∴ Its square = side × side = 31 × 31 = 961 sq. cms
Rectangle’s Area = l × b = 40 × 22 = 880 sq.cms.
The area of the square encloses more area than the area of the rectangle.

Question 7.
The perimeter of a rectangle is 130 cm. If the breadth of the rectangle 30 cm, find its length. Also find the area of the rectangle.
Solution:
p = perimeter of a rectangle = 130 cm
l = length of a rectangle = l = ?
b = breadth of a rectangle = b = 30 cm
p = 2 (l + b)
130 = 2l + 2 × 30
2l = 130 – 60 = 70

∴ Area of a rectangle = l × b = 30 × 35
= 1050 sq.cms

Question 8.
A door of length 2 m and breadth 1 m is fitted in a wall. The length of the wall is 4.5 m and the breadth is 3.6 m (Fig 11.6). Find the cost of whitewashing the wall, if the rate of whitewashing the wall is ₹ 20 per m²
Solution:

Area of the wall = l × b
length of the wall = 4.5 m
the breadth of the wall = 3.6 m
Area = 4.5 × 3.6 = 16.2 sq.m
length of the door = 2 m
the breadth of the door = 1 m
Area of the door = 2 × 1 = 2 sq. m
Area of the wall excluding (without) door = 16.20 – 2 = 14.20 sq.m
Rate of whitewashing the wall = Rs. 20/ sq.m
∴ The total cost of whitewashing the wall = 14.20 × 20 = Rs. 284.00

Students can Download Chapter 3 Data Handling Ex 3.1, 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 3 Data Handling Ex 3.1

Question 1.
Find the range of heights of any ten students of your class.
Solution:
Students can try themselves.
(This is practical question)

Question 2.
Organise the following marks ion a class assessment, in a tabular form.
4, 6, 7, 5, 3, 5, 4, 5, 2, 6, 2, 5, 1, 9, 6, 5, 8, 4, 6, 7

i) Which number is the highest ?
Solution:
Highest number is 9

ii) Which number is the lowest ?
Solution:
Lowest number is 1

iii) What is the range of the data ?
Solution:
Range of the given data = Highest number – Lowest number = 9 – 1 = 8

iv) Find the arithmetic mean.
Solution:
Arithmetic Mean =

Question 3.
Find the mean of the first five whole numbers.
Solution:
The first whole numbers are 0, 1, 2, 3, 4
Arithemetic mean of five whole numbers =

Question 4.
A cricketer scores the following runs in eight innings :
58, 76, 40, 35, 46, 45, 0, 100. Find the mean score.
Solution:
The scores of 8 innings = 58, 76, 40, 35, 46, 45, 0, 100

Question 5.
Following table shows the points of each player scored in four games :

i) Find the mean to determine as average number of points scored per game.
Solution:
A’s average number of points scored

ii) To find the mean number of points per game for C, would you divide the total points by 3 or 4 ? Why ?
Solution:
To find the mean number of points per game for C, we shall divide the total points by 3 only because he did not play the 3rd game.

iii) B played in all the four games. How would you find the mean ?
Solution:
‘B’ played all the four games. So we find

iv) Who is the best performer ?
Solution:
To find out the best performer, we should know mean of all the 3 players. So we should calculate ‘C’s mean.

∴ The best performer is ‘A’

Question 6.
The marks (out of 100) obtained by a group of students in a science test are 85, 76, 90, 85, 39, 48, 56, 95, 81 and 75. Find the:
Solution:
Marks obtained by a group (in ascending order) = 39, 48, 56, 75, 76, 81, 85, 85, 90, 95
i) Highest and the lowest marks obtained by the students.
Solution:
Highest and the lowest marks obtained by the student 95 and 39 respectively

ii) Range of the marks obtained.
Range = Highest score – Lowest score
= 95 -39 = 56

iii) Mean marks obtained by the group.

= 39 + 48 + 56 + 75 + 76 + 81 + 85 + 85 + 90 + 95

Question 7.
The enrollment in a school during six consecutive years was as follows :
1555, 1670, 1750, 2013, 2540, 2820. Find the mean enrollment of the school for this period.
Solution:
The enrollment in a school for 6 successive years } = 1555, 1670, 1750, 2013, 2540, 2820.

∴ Mean enrollment of the school for 6 years is 2,058

Question 8.
The rainfall (in mm) in a city on 7 days of a certain week was recorded as follows:

i) Find the range of rainfall in the above data.
Solution:
Range of the rainfall = Highest Rainfall – Lowest Rainfall
= 20.5 – 0.0 = 20.5

ii) Find the mean rainfall for the week
Solution:

iii) On how many days was the rainfall less than the mean rainfall.
Solution:
No. of days of rainfall which was less than Mean rainfall } = 5 days

Question 9.
The heights of 10 girls were measured in cms and the results are as follows: 135, 150, 139, 128, 151, 132, 146, 149, 143, 141.
Solution:
The height of girls are arranged in ascending order} = 128, 132, 135, 139, 141, 143, 146, 149, 150, 151.

i) What is the height of the tallest girl?
Solution:
The height of the tallest girl is 151 cms

ii) What is the height of the shortest girl?
Solution:
The height of the shortest girl is 128cms

iii) What is the range of the data ?
Solution:
The range of the data = Highest height – lowest height
= 151 – 128 = 23

iv) What is the mean height of the girls?
Solution:

= 128 + 132 + 135 + 139 + 141 + 143 + 146 + 149 + 150 + 151

v) How many girls have heights more than the mean height.
Solution:
No. of girls who were more than mean height = 5

Students can Download Chapter 2 Fractions and Decimals Ex 2.1, 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 2 Fractions and Decimals Ex 2.1

Question 1.
Solve:
i)

Solution:

ii)

Solution:

iii)

Solution:

L.C.M = 5 × 7 × 1 × 1 = 35

iv)

Solution:

Online Subtracting Fractions Calculator subtracts the fractions 9/11 and 4/5 i.e. 1/55

The given fractions are 9/11 and 4/5

Firstly the L.C.M should be done for the denominators of the two fractions 9/11 and 4/5

9/11 – 4/5

The LCM of 11 and 5 (denominators of the fractions) is 55

Given numbers has no common factors except 1. So, there LCM is their product i.e 55

 

= 9 x 5 – 4 x 11/55

= 45 – 44/55

= 1/55

Result: 1/55

v)

Solution:

vi)

⇒ convert mixed fractions to improper fraction

vii)

convert mixed fractions to improper fraction

Question 2.
Arrange the following in descending order:
i)

= We need to arrange these in descending order,
To find which number is greater or smaller, we make their denominators equal.

ii)

⇒ We make their denominators equal, to find the descending order.

Question 3.
In a “magic square”, the sum of the numbers in each row, in each column and along the diagonals is the same. Is this a magic square ?

Solution:
For Row,


For diagonals,

Since Sum of air rows, columns and diagonals are equal.

Question 4.
A rectangular sheet of paper is 12\(\frac{1}{2}\) cm long and 10\(\frac{2}{3}\) cm wide. Find its perimeter.
Solution:
Length of rectangular sheet of paper = 12\(\frac{1}{2}\) cm
(breadth) width of rectangular sheet paper = 10\(\frac{2}{3}\) cm

Perimeter of rectangle = 2 (length + breadth)

converting the above fraction to mixed fraction,

Question 5.
Find the perimeters of
(i) ∆ ABE
(ii) the rectangle BCDE in this figure. Whose perimeter is greater ?

Solution:
(i) ∆ ABE
perimeter of ∆ ABE
perimeter = AB + AE + BE

ii) The rectangle BCDE in this figure

Perimeter of rectangle BCDE,
As it is a rectangle, opposite sides are equal
BC = DE CD = BE

perimeter of rectangle = 2(l + b)


Also,
We have to find which perimeter is greater

To find which fraction is greater, we make its denominator equal.


Perimeter of ∆ ABE > Perimeter of BCDE
(Thus, Perimeter of ∆ ABE is greater)

Question 6.
Salil wants to put a picture in a frame. The picture is 7\(\frac{3}{10}\) cm wide. How much should the picture be trimmed ?
Solution:
There are two things here – picture, and frame

so, picture has to be trimmed

\(=\frac{3}{10}\)
∴ Picture has to trimmed by = \(=\frac{3}{10}\) cm

Question 7.
Ritu ate \(\frac{3}{5}\) part of an apple and the remaining apple was eaten by her brother Somu. How much part of the apple did Somu eat ? Who had the larger share ? By how much ?
Solution:
Total part = 1
Part eaten by Ritu = \(\frac{3}{5}\)
Part eaten by Somu = Total part – part eaten by Ritu

Now,
We have to tell who ate the larger share so, we have to compare,

∴ Ritu ate the larger share.

Ritu ate large share by \(\frac{1}{5}\)

Question 8.
Michael finished colouring in \(\frac{7}{12}\) hour. Vaibhav finished colouring the same picture in \(\frac{3}{4}\) hour. Who worked longer ? By what fraction was it longer ?
Solution:
Michael finished work in = \(\frac{7}{12}\) hour
Vaibhav finished work in = \(\frac{3}{4}\)
We need to find who worked longer.
i.e., we need to find greater of = \(\frac{7}{12}\) & \(\frac{3}{4}\)
We make their denominators equal

∴ Vaibhav worked longer.
We also need to find by how much

Vaibhav worked longer by \(\frac{1}{6}\) hours.

Students can Download Chapter 4 Simple Equations Ex 4.4, 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 4 Simple Equations Ex 4.4

Question 1.
Set up equations and solve them to find the unknown numbers in the following cases :
a) Add 4 to eight times a number; you get 60.
Solution:
Let the unknown number be’x’
8 times a number be 8x
According to the question
8x + 4 = 60
8x = 60 – 4
8x = 56

∴ The unknown number x be 7

b) One-fifth of a number minus 4 gives 3.
Solution:
Let the unknown number be x
Then according to the question

∴ That unknown number be 35.

c) If I take three-fourths of a number and add 3 to it, I get 21.
Solution:
Let the unknown number be x
According to the question

∴ That unknown number be 24.

d) When I subtracted 11 from twice a number, the result was 15.
Solution:
Let the unknown number be x According to the question
2x – 11 = 15
2x = 15 + 11
2x = 26

∴ That unknown number be 13.

e) Munna subtracts thrice the number of notebooks he has from 50, he finds the result to be 8.
Solution:
Let the unknown number be x
According to the question
50 – 3x = 8
– 3x = 8 – 50

Therefore that unknown number be 14.

f) Ibenhal thinks of a number. If she adds 19 to it and divides the sum by 5, she will get 8.
Solution:
Let the unknown number be x
According to the question

∴ That unknown number be 21

g) Anwar thinks of a number. If he takes away 7 from \(\frac{5}{2}\) of the number, the result is 23.
Solution:
Let the unknown number be x
According to the question

That unknown number be 12

Question 2.
Solve the following :
a) The teacher tells the class that the highest marks obtained by a student in her class is twice the lowest marks plus 7. The highest score is 87. What is the lowest score?
Solution:
Let the lowest score be x
According to the question
2x + 7 = 87
2x = 87 – 7 = 80
x = 80/2 = 40 x = 40
∴ The lowest score be 40.

b) In an isosceles triangle, the base angles are equal. The vertex angle is 40°. What are the base angles of the triangle? (Remember, the sum of three angles of a triangle is 180°).
Solution:
Let the base angles of an isosceles triangle be x°
According to the question
2x° + 40° = 180°
2x° = 180° – 40°
2x°= 140

The base angles are 70° and 70°

c) Sachin scored twice as many runs as Rahul. Together, their runs fell two short of a double century. How many runs did each one score?
Solution:
Let the score of Rahul be ‘x’ runs
According to the question
x + 2x = 198
3x = 198

∴ Rahul’s score = x = 66 runs
Sachins score = 2x = 2 × 66 = 132 runs

Question 3.
Solve the following :

i) Irfan says that he has 7 marbles more than five times the marbles Parmit has. Irfan has 37 marbles. How many marbles does Parmit have?
Solution:
Let Parmits marble be x
According to the question
5x + 7 = 37
5x = 37 – 7 = 30

∴ Parmits marble be = x = 6

ii) Laxmi’s father is 49 years old. He is 4 years older than three times Laxmi’s age. What is Laxmi’s age?
Solution:
Let the age of Laxmi be ‘x’ years
According to the question
3x + 4 = 49
3x = 49 – 4 = 45
3x = 45
x = 45/3 = 15
∴ Lakshmi’s age is x = 15 years.

iii) People of Sundargram planted trees in the village garden. Some of the trees were fruit trees. The number of non-fruit trees were two more than three times the number of fruit trees. What was the number of fruit trees planted if the number of non-fruit trees planted was 77?
Solution:
Let the number of fruit trees be ‘x’
According to the question
3x + 2 = 77
3x = 77 – 2 = 75

∴ No. of fruit trees = x = 25

Question 4.
Solve the following riddle :
I am a number, Tell my identity!
Take me seven times over And add a fifty!
To reach a triple century You still need forty!
Solution:
Let the number be ‘x’
According to the question
7x + 50 = 300 – 40
7x + 50 = 260
7x = 260 – 50 = 210

∴ That number be x = 30