Class 8

Students can Download Maths Chapter 9 Commercial Arithmetic Ex 9.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka Board Class 8 Maths Chapter 9 Commercial Arithmetic Ex 9.1

Question 1.
In a school 30% students play chess 60% play carrom and the rest play other games. If the total number of students in the school is 900 find the exact number of students who play each game.
Answer:
30% Students play chess. Total number of students is 900.
The number of players who play chess = 30% of 900.
=  × 900 = 270
The number of players who play carrom board = 60% of 900
=  × 900 = 540
The number of players who play other games = 10% of 900
=  × 900 = 90

Question 2.
In a school function Rs. 360 remained after spending 82% of the money. How much money was there in the beginning? verify your answer.
Answer:
Let the money, in the beginning, be ₹ 100 The money spent is ₹ 82.
∴ The amount remaining is 100 – 82 = ₹ 18.
If the remaining money is ₹ 18 the money, in the beginning, is ₹ 100.
If the remaining money is ₹ 360 the money in the beginning is
=  × 360 = 2000
Verification :
 × 2000 = 1640
Money remained = 2000 – 1640 = ₹ 360.

Question 3.
Akshay’s income is 20% less than that of Ajay what percent is Ajay’s income more than that of Akshay?
Answer:
Let Ajay’s income be Rs. 100.
Then Akshay’s income is (100 – 20) = Rs. 80.
If Akshay’s income is Rs. 80.
Ajay’s income is Rs. 100.
If Akshay’s income is Rs. 1 Ajay’s income is
\(\frac{100}{80}\) × 1 = Rs. \(\frac { 100 }{ 80 }\)
Changing the scale to 100.
Ajay’s income is Rs. 100
Akshay’s income is
= \(\frac { 100 }{ 80 }\) × 100 = 125
∴ Ajay’s income is 125 – 100 = 25% more than that of Akshay.

Question 4.
A daily wage employee spends 84% of his weekly earning. If he saves Rs. 384 find his weekly earning.
Answer:
Let his weekly earning be ₹ 100.
His expenditure is ₹ 84.
Therefore his savings is (100 – 84) = ₹ 16
If the savings is ₹ 16 then the income is ₹ 100.
If the savings is ₹ 384 then the income is
=  × 384 = ₹  2400
∴ His weekly earning is ₹ 2400.

Question 5.
A factory announces a bonus of 10% its employees. If an employee gets Rs. 10,780 find his actual salary.
Answer:
Let the salary of the employee be ₹ 100.
He gets a bonus of ₹ 10.
He gets 100 + 10 = ₹ 110 including bonus.
∴ If the employee gets ₹ 10.780 his salary is
=  × 10780 = 9800
∴ The salary of the employee is ₹ 9800

Students can Download Maths Chapter 1 Playing with Numbers Ex 1.3 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka Board Class 8 Maths Chapter 1 Playing with Numbers Ex 1.3

Question 1.
Find the quotient and remainder when each of the following number is divided by 13 : 8, 31, 44, 85, 1220.
Answer:

Question 2.
Find the quotient and the remainder when each of the following numbers are divided by 304.
128, 636, 785, 1038, 2236, 8858
Answer:

Question 3.
Find the least natural number larger than 100 which leaves the remainder 12 when divided by 19.
Answer:
The least number greater than 100 divisible by 19 is 114.
Adding 12 to this number we get 114 + 12 = 126.
126 leaves 12 as the remainder when divided by 19.

Question 4.
What is the least natural number you have to add to 1024 to get a multiple of 181?
Answer:
The Multiples of 181 are 181, 362, 543, 724, 905, 1086.
The nearest multiple of 181 to 1024 is 1.086.
Hence 62 (1086 – 1024) must be added to 1024 to get a multiple of 181.

Students can Download Maths Chapter 7 Rational Numbers Ex 7.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka Board Class 8 Maths Chapter 7 Rational Numbers Ex 7.2

Question 1.
Write down ten rational numbers which are equivalent to \(\frac { 5 }{ 7 }\) and the denominator not exceeding 80.
Answer:
Multiply both numerator and denominator by 2, 3, 4………
\(\frac{10}{14}, \frac{15}{21}, \frac{20}{28}, \frac{35}{35}, \frac{30}{42}, \frac{35}{49}, \frac{40}{56}, \frac{45}{63}, \frac{50}{70}, \frac{55}{77}\)

Question 2.
Write down 15 rational numbers which are equivalent to \(\frac { 11 }{ 5 }\) and the numerator not exceeding 180.
Answer:
\(\begin{array}{l}{\frac{22}{10}, \frac{33}{15}, \frac{44}{20}, \frac{55}{25}, \frac{66}{30}, \frac{77}{35}, \frac{88}{40}, \frac{99}{45}} \\ {\frac{110}{50}, \frac{121}{55}, \frac{132}{60}, \frac{143}{65}, \frac{154}{70}, \frac{165}{75}, \frac{176}{80}}\end{array}\)

Question 3.
Write down 10 positive rational numbers such that the sum of the numerator and the denominator of each is 11. Write them in decreasing order.
Answer:

Question 4.
Write down ten positive rational numbers such that numerator – denominator for each of them is -2. Write to them in increasing order.
Answer:
Numerator – denominator = – 2
therefore the denominator is greater than the numerator by 2.
\(\frac{1}{3}, \frac{2}{4}, \frac{3}{5}, \frac{4}{6}, \frac{5}{7}, \frac{6}{9}, \frac{7}{9}, \frac{8}{10}, \frac{9}{11}, \frac{10}{12}\)

Question 5.
Is \(\frac { 3 }{ -2 }\) a rational number? If so, how do you write it in the form conforming to the definition of a rational number (that is, the denominator as positive integer)?
Answer:
\(\frac{3}{-2}\) is a rational number because the denominator is negative.
It can be written as \(\frac{3}{-2}\) since \(\frac{3}{-2}\) is same as \(\frac{3}{-2}\)

Question 6.
Earlier you have studied decimals 0.9, 0.8, can you’ write these as rational numbers?
Answer:
\(0.9=\frac{9}{10} \text { and } 0.8=\frac{8}{10}=\frac{4}{5}\)

Students can Download Maths Chapter 7 Rational Numbers Ex 7.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka Board Class 8 Maths Chapter 7 Rational Numbers Ex 7.1

1. Identify the property in the following statements.

Question (i)
2 + (3 + 4) = (2 + 3) + 4
Answer:
Associative property of addition.

Question (ii)
2 × 8 = 8 × 2
Answer:
Commutative property of multiplication

Question (iii)
8 × (6 + 5) = (8 × 6) + (8 × 5)
Answer:
Distributive property of multiplication over addition in integers.

Question 2.
Find the additive inverse of the following integers.
Answer:
Integer                    Additive inverse
6                                      -6
9                                      -9
123                                -123
-76                                   76
-85                                   85
1000                             -1000

3. Find the integer m in the following:

Question (i)
m + 6 = 8
Answer:
m = 8 – 6

Question (ii)
m + 25 = 15
Answer:
m =15 – 25
m = -10

Question (iii)
m – 40 = -26
Answer:
m = – 26 + 40
m = 14

Question (iv)
m + 28 = – 49
Answer:
m = – 49 – 28
m = – 77

Question 4.
Write the following in increasing order: 21, -8, -26, 85, 33, -333, -210, 0, 2011
Answer:
-333 < -210 < -26 < -8 < 0 < 21 < 33 < 85 < 2011.
-333, -210, -26, -8, 0, 21, 33, 85, 2011

Question 5.
Write the following in decreasing order: 85, 210, -58, 2011, -1024, 528, 364, -10000,12
Answer:
2011 > 528 > 364 > 210 > 85 > 12 > – 58 > -1024 < -10,000
2011, 528, 364, 210, 85, 12, -58, -1024, -10,000.

Students can Download Maths Chapter 1 Playing with Numbers Ex 1.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka Board Class 8 Maths Chapter 1 Playing with Numbers Ex 1.2

Question 1.
In the following find the digits represented by the letters.

Answer:
Here : 3 + 4 = 7 ∴ B = 4

Answer:
Here 6 + A = 1 ∴ A = 5
6 + 5 = 11 is written in units place and 1 is carried.
∴ 1 + 1 + 2 = B
∴ B = 4

Answer:
Here A × A is A.
Hence A = 0, 1, 5, or 6 If A = 1 we get 21 × 1 = 21 but the product is 12A.
If A = 5 then 25 × 5 = 125.
∴ A = 5

Answer:
Here A + A = A
Hence A = 0

Answer: Here A × A = A
Hence A = 0, 1, 5 or 6
If A = 1, then 11 × 11 = 121
∴ A = 1 and B = 2

Answer:
3A × A
36 × 6
216
∴ A = 6 & B = 1

Question 2.
In the adjacent sum A, B, C are consecutive digits. In the third row A, B, C appear in some order. Find A, B, C.
Answer:

∴ A = 3, B = 4 and C = 5

Students can Download Maths Chapter 8 Linear Equations in One Variable Ex 8.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka Board Class 8 Maths Chapter 8 Linear Equations in One Variable Ex 8.1

1. Solve the following:

Question i.
x + 3 = 11
Answer:
x + 3 = 11
x = 11 – 3
x = 8

Question ii.
y – 9 = 21
Answer:
y – 9 = 21
y = 21 + 9
y = 30

Question iii.
10 = z + 3
10 = z + 3 10-3 =z
7 = z or z = 7

Question iv.
\(\frac{3}{11}+x=\frac{9}{11}\)
Answer:

Question v.
10x = 30
Answer:
10x = 30
x = \(\frac { 30 }{ 10 }\)
x = 3

Question vi.
\(\frac{s}{7}=4\)
Answer:
\(\frac{s}{7}=4\)
S = 4 × 7
S = 28

Question vii.
\(\frac{3 x}{6}=10\)
Answer:
\(\frac{3 x}{6}=10\)
3x = 10 × 6
3x = 60
x = \(\frac { 60 }{ 3 }\)
x = 20

Question viii.
\(1.6=\frac{x}{1.5}\)
Answer:
\(1.6=\frac{x}{1.5}\)
1.6 × 1.5 = x
2.40 = x
x = 2.4

Question ix.
8x – 8 = 48
Answer:
8x – 8 = 48
8x = 48 + 8
8x = 56
x = \(\frac { 56 }{ 8 }\)
x = 7

Question x.
\(\frac{x}{3}+1=\frac{7}{15}\)
Answer:

Question xi.
\(\frac{x}{5}=12\)
Answer:
\(\frac{x}{5}=12\)
x = 12 × 5
x = 60

Question xii.
\(\frac{3 x}{5}=15\)
Answer:

Question xiii.
3(x + 6) = 24
Answer:
3(x + 6) = 24
3x+ 18 = 24
3x = 24 – 18
3x = 6
x = \(\frac { 6 }{ 3 }\) = 2

Question xiv.
\(\frac{x}{4}-8=1\)
Answer:

Question xv.
3(x+2) – 2(x—1) = 7
Answer:
3(x + 2) – 2(x – 1) = 7
3x + 6 – 2x + 2 = 7
x + 8 = 7 .
x = 7 – 8
x = -1

2. Solve the equations :

Question i.
5x = 3x + 24
Answer:
5x = 3x + 24
2x = 24
x = \(\frac { 24 }{ 2 }\)
x = 12

Question ii.
8t + 5 = 2t – 31
Answer:
8t + 5 = 2t – 31
6t = -36
t = \(\frac { -36 }{ 6 }\)
t = -6

Question iii.
7x -10 = 4x + 11
Answer:
7x – 10 = 4x + 11
7x – 4x = 11 + 10
3x = 21
x = \(\frac { 21 }{ 3}\)
x = 7

Question iv.
4z + 3 = 6 + 2z
Answer:
4z + 3 = 6 + 2z
4z – 2z = 6 – 3
2z = 3

Question v.
2x – 1 = 14 – x
Answer:
2x – 1 = 14 – x
2x + x = 14 + 1
3x = 15
x = \(\frac { 15 }{ 3 }\)
x = 5

Question vi.
6x + 1 = 3(x -1) + 7
Answer:
6x + 1 = 3(x —1) + 7
6x + l =3x-3 + 7
6x + l = 3x + 4
6x – 3x = 4 – 1
3x = 3
x = \(\frac { 3 }{ 3 }\)
x = 1

Question vii.
\(\frac{2 x}{5}-\frac{3}{2}=\frac{x}{2}+1\)
Answer:

Question viii.
\(\frac{x-3}{5}-2=\frac{2 x}{5}\)
Answer:

Question ix.
3(x + 1) = 12 + 4(x – 1)
Answer:
3(x + 1)= 12 + 4(x – 1)
3x + 3 = 12 + 4x – 4
3x + 3 = 4x + 8
3x – 4x = 8 – 3
-x = 5
x = -5

Question x.
2x – 5 = 3(x – 5)
Answer:
2x – 5 = 3(x – 5)
2x – 5 = 3x – 15
2x – 3x = -15 + 5
-x = -10
x= 10

Question xi.
6(1 – 4x) + 7(2 + 5x) = 53
Answer:
6( 1 – 4x) + 7(2 + 5x) = 53
6 – 24x+ 14 + 35x = 53
35x – 24x + 6 + 14 = 53
11x+ 20 = 53
11x = 53 – 20
x = \(\frac { 33 }{ 11 }\)
x = 3

Question xii.
3(x + 6) + 2(x + 3) = 64
Answer:
3(x + 6) + 2(x + 3) = 64
3x + 18 + 2x + 6 = 64
3x + 2x + 18 + 6 = 64
5x + 24 = 64
5x = 64 – 24
5x = 40
x = \(\frac { 40 }{ 5 }\)
∴ x = 8

Question xiii.
\(\frac{2 m}{3}+8=\frac{m}{2}-1\)
Answer:

Question xiv.
\(\frac{3}{4}(x-1)=(x-3)\)
Answer:
\(\frac{3}{4}(x-1)=(x-3)\)
3(x – 1) = 4(x – 3)
3x – 3 = 4x – 12
3x – 4x = -12 + 3
-x = -9
x = 9

Students can Download Maths Chapter 6 Theorems on Triangles Ex 6.3 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka Board Class 8 Maths Chapter 6 Theorems on Triangles Ex 6.3

Question 1.
The exterior angles obtained on producing the base of a triangle both ways are 104° and 136°. Find the angles of the triangle.
Answer:
In triangle ABC, BC is produced on either sides.
Let ∠ABC = 140° and ∠ACE = 136°
∠ABD +∠ABC = 180° [ Linear pair]
104 + ∠ABC = 180°
∠ABC = 180 – 104
∠ ABC = 76°

∠ACB + ∠ACE = 180° [Linearpair]
∠ACB + 136° = 180°
∠ACB = 180 – 136 ∠ACB = 44°
∠ABD + ∠ACB + ∠BAC = 180°
[Sum of the angles of a triangle 180° ]
76 + 44 + ∠BAC = 180°
120 + ∠BAC = 180°
∠BAC = 180 – 120 ∠BAC = 60°
Three angles are 76°, 44° & 60°

Question 2.
Sides BC, CA and AB of a triangle ABC are produced in order, forming exterior angles ∠ACD, ∠BAE, and∠CBF.
Show that ∠ACD + ∠BAE + ∠CBF = 360°.
Answer:

ABC + ∠CBF = 180° [Linear pair]………..(i)
∠ACB + ∠ACD = 180° [Linear pair]……(ii)
∠BAC + ∠BAE = 180° [Linear pair]…….(iii)
Byadding(i), (ii) and (iii)
∠ABC + ∠CBF + ∠ACB + ∠ACD+
∠BAC + ∠BAE = 180 + 180 + 180
∠ABC + ∠ACB + ∠BAC +
∠CBF + ∠ACD + ∠BAE = 540°
180 + ∠CBF + ∠ACD + ∠BAE = 540°
[Sum of the angles of a triangle 180°]
∠CBF + ∠ACD + ∠BAE = 360°

Question 3
Compute the value of x in each of the following figures
Answer:
i.

AB = AC
∴ ∠ABC = ∠ACB = 50°
∠ACB + ∠ACD = 180° [Linear pair]
50 + x = 180°
x = 180 – 50 = 130°

ii.

∠ABC + ∠CBF = 180° [Linear pair].
106°+∠ABC = 180°
∠ABC = 180 – 106 = 74°
∠EAC = ∠ABC + ∠ACB
130° =74 +x
130 – 74 = x
∴x = 56°

iii.

∠QPR = ∠TPU = 65°
[Vertically opposite angles]
∠PRS = ∠PQR + ∠QPR [Exterior angle=Sum of interior opposire angle]
100 = 65 + x
100 – 65 = x
35 = x
∴ x = 35°.
iv.

∠BAE + ∠BAC = 180° [Linear pair]
120° + ∠BAC = 180°
∠BAC = 180 – 120
∠BAC = 60°
∠ACD = ∠BAC + ∠ABC
[Exterior angle = Sum of interior opposuite angle]
112° = 60+x
112 – 60 = x
52 = x
x = 52°
v.

In Δ ABC, BA = BC (data)
∴ ∠BAC = ∠BCA = 20°
[Base angles of an isosceles traingle]
∠ABD = ∠BAC + ∠BCA
[Exterior angle=Sum of interior opposite angle]
x = 20 + 20
x=40°

Question 4.
In figure QT ⊥ PR, ∠TOR = 40° and ∠SPR = 30° find ∠TRS and ∠PSQ.

Answer:
In ΔTQR,
∠TQR + ∠QTR + ∠TRQ = 180°
40 + 90 + ∠TRQ = 180°
130 + ∠TRQ = 180
∠ TRQ = 180 – 130
∠ TRQ = 50°
∠ TRS = 50° [∠PRS is same as ∠TRS ]
In Δ PRS, RS is produced to Q
∴Exterior∠PSQ = ∠SPR + ∠PRS = 30 + 50
∠PSQ = 80°

Question 5.
An exterior angle of a triangle is 120° and one of the interior opposite angle is 30 °. Find the other angles of the triangle.
Answer:

In Δ ABC, BC is produced to D. Let
∠ACD = 120° and ∠ABC = 30°
Exterior ∠ACD = ∠BAC + ∠ABC
120 = ∠BAC + 30
120 – 30 = ∠BAC
90° = ∠BAC
∴∠BAC = 90
∠ACB +∠ACD = 180° [Linear pair]
∠ACB + 120 = 180°
∠ACB = 180 – 120
∠ACB = 60°
Other two angles are 90° & 60°

Students can Download Maths Chapter 16 Mensuration Ex 16.2 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka Board Class 8 Maths Chapter 16 Mensuration Ex 16.2

Question 1.
Find the total surface area and volume of a cube whose length is 12 cm.
Answer:
l = 12 cm
T.S.A of cube = 6 l² = 6 x 12² = 6 x 144 = 864 cm²
Volume of cube = P = (12)³ = 1728 cm³

Question 2.
Find the volume of a cube whose surface area is 486 cm².
Answer:
T.S.A of a cube = 486 cm²
6¹² = 486
l² = \(\frac{486}{6}\)
l² = 81
∴ l = √81 = 9 cm
Volume = l³ = 9³= 729 cm³

Question 3.
A tank, which is cuboidal in shape has a volume 6.4m³. The length and breadth of the base are 2m and 1.6m respectively. Find the depth of the tank.
Answer:
V = 6.4 m³
l = 2 m
b = 1.6 m
h = ?
Volume of cuboid = 6.4 m³
l × b × h = 6.4
2 × 1.6 × h = 6.4
3.2h = 6.4
h = \(\frac{6.4}{3.2}\)
h = 2m
h = 2m .
∴ The depth of the tank is 2 m.

Question 4.
How many m³ of soil has to be excavated from a rectangular well 28cm deep and whose base dimensions are 10cm and 8m. Also, find the cost of plastering its vertical walls at the rate of Rs. 15/m2.
Answer:
l = 10 m
b = 8
h = 28 m
Volume of soil = volume of cuboid
= l × b × h = 10 × 8 × 28 = 2240 m³
2240 m³ of soil has to be excavated.
Area to be plastered = L.S.A of cuboid
= 2h(l + b)
= 2 × 28(10 + 8)
= 56 × 18
= 1008 m²
The cost of plastering 1m² = Rs. 15
∴ The cost of plastering = 15 × 1008 = Rs. 15,120

Question 5.
A solid cubical box of fine wood costs Rs. 256 at the rate of Rs. 500/m3. Find its volume and length of each side.
Answer:
The cost of Rs. 500 is for 1m³.
The cost of Rs. 256 is for
\(\frac{256}{500}=\frac{128}{250}=\frac{64}{125} \mathrm{m}^{3}\)
∴ Volume of the wood = \(\frac{64}{125} \mathrm{m}^{3}\) = 0.512 m³
Volume = 0.512m
l³ = 0.512
\(l=\sqrt[3]{0.512}\)
l = 0.8 m²
= 0.8 × 100
Length of the side = 80cm, l= 80cm

Students can Download Maths Chapter 11 Congruency of Triangles Ex 11.3 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka Board Class 8 Maths Chapter 11 Congruency of Triangles Ex 11.3

Question 1.
In a Δ ABC, AB = AC andlA.= 50° find ∠B and ∠C.
Answer:
∠A+ ∠B + ∠C = 180°
(Sum of the angles of a triangle is 180° )
50 +∠[B + ∠B = 180°
∠B = ∠C Base angles of an isosceles triangle
50 + 2∠B= 180°
2∠B = 180 – 50
2∠B = 130°
∠B = \(\frac{130^{\circ}}{2}\)
∠B = 65°
∠B = ∠C = 65°

Question 2.
In AABC,AB = BCand|B = 64°find|£,
Answer:
AB = BC [data]
∴ ∠C = ∠A [Theorem 1]
∠A + ∠B + ∠C = 180°
(Sum of the angles of a triangle is 180°)
∠C + 64 + ∠C = 180° [∠A = ∠C]
64 + 2∠C = 180°
2∠C = 180 – 64
2∠C = 116
∠C = \(\frac { 116 }{ 2 }\) = 58°

Question 3.
In each of the following figure find the value of x :
i.

Answer:
In ∆ABC, AB = AC
∴ ∠ABC = ∠ACB
∴ ∠BAC + ∠ABC + ∠ACB = 180°
(Sum of the angles of a triangle is 180°)
40 + ∠ABC + ∠ABC = 180°
(∠ABC =∠ACB)
40 + 2∠ABC = 180°
2∠ABC = 180 – 40
2∠ABC = 140°
∠ABC = 70°
∠ACB = ∠ ABC = 70°
∠ACB + ∠ACD = 180°
70 + x = 180°
x = 180 – 70
x = 110°

ii

AC = CD
∠CAD = ∠CDA
∠CAD = ∠CAD = 30°
∠ACD + ∠CAD + ∠CDA = 180°
(Sum of the angles of a triangle is 180°)
∠ACD + 30 + 30=180°
∠ACD + 60=180°
∠ACD = 180-60
∠ACD = 120°
∠ACD = ∠BAC + ∠ABC
120 = 65° + x
120 – 65 = x
55 = x
x=55°

iiii

Answer:
AB = AC
∠ABC = ∠ACB = 55° [Theorem l]
Exterior ∠APB = ∠DAC +∠ACD
75 = x + 55
75 – 55 = x
20 = x
x = 20°

iv.

BD = DC = Ad
BD = DC = AD & ∠ABD = 50 °
∠ABD = ∠BAD(Th. l)
∠BAD = 50°
∠ABD + ∠BAD + ∠ADB = 180°
50 + 50 +∠ADB = 180°
∠ADB = 180 – 100 = 80°
∠APB = 80°
∴ ∠APB +∠ADC = 180°
80 +∠ADC = 180°
∠ADC = 100°
Now AD = DC
∴∠DAC = ∠DCA = x°
∴ x + x +∠ADC = 180°
2x + 100 = 180°
2x = 180-100=80°
x=40°

Question 4.
Suppose ABC is an equilateral triangle. Its base BC is produced to D such that BC = CD.
Calculate: l.∠ACD
2.∠ADC
Answer:
∠ABC = ∠ACB = ∠BAC = 60°
(ABC is an equilateral triangle)
∠ACB +∠ACD = 180° (Linearpoint)
60 + ∠ACD = 180°
∠ACD = 180 – 60
∠ACD = 120°

In ∆ACD,AC=CD
∠ CAD = ∠CPA (Theorem l)
∠ACB +∠ACD = 180° [linear pair]
60° +∠ACD = 180°
∠ACD = 180° – 60° = 120°
∠ACA +∠CAD + ∠CDA = 180°
2∠CDA = 180 – 120°
2∠CDA = 60°
∠CDA = \(\frac { 60 }{ 2 }\) = 30°
∠CDA = 30

Question 5.
Show that the perpendicular drawn from the vertices of the base of an isosceles triangle to the opposite sides are equal. ,
Answer:
Data : In ∆ABC, AB = AC,
BD ⊥AC & CE⊥ AB
To prove : BD = CE

Proof:
In ∆ABC, AB = AC [data]
∠ABC = ∠ACB [Theorem l]
In ∆EBC and ∆DCB
∠EBC =∠DCB( Base angles )
∠BEC = ∠CDB [= 90° ]
BC = BC (Common side)
∆EBC = ∆DCB [ASA postulate]
BD = CE [Corresponding sides]

Question 6.
Prove that an ∆ABC is an isosceles triangle if the altitude AD from A on BC bisects BC.
Answer:
In ∆ADB and ∆ADC
AD = AD [Common side]
∠APB =∠ ADC [90° ]

BD = DC [AD bisects BC]
∴∆ADB ≅ ∆ADC [SAS postulate]
∴AB = AC [Correspondingsides]
∴∆ ABC is an isosceles triangle

Question 7.
Suppose a triangle is equilateral, prove that it is equiangular.
Answer:

To prove: ∠A = ∠B = ∠C
Proof: In ∆ABC, AB = BC
∠C = ∠B [Theorem l]….(i)
BC = AC
∠A =∠B [Theorem l]…(ii)
From (i) and (ii)
∠A =∠B = ∠C
∆ABC is equiangular

Students can Download Maths Chapter 1 Playing with Numbers Ex 1.1 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 8 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka Board Class 8 Maths Chapter 1 Playing with Numbers Ex 1.1

Question 1.
Write the following numbers in generalized form 39, 52, 106, 359, 628, 3458, 9502, 7000.
Answer:
39 = 30 + 9 = (3 × 10) + (9 × 1)
52 = 50 +2 = (5 × 10) + (2 × 1)
106 = 100 + 6 = (1 × 100) + (0 × 10)+(6 × 1) 359 = 300 + 50 + 9 = (3 × 100) + (5 × 10) + (9 × 1)
628 = 600 + 20 + 8 = (6 × 100) + (2 × 10) + (8 × 1)
3458 = 3000 + 400 + 50 + 8 = (3 × 1000) + (4 × 100) + (5 × 10) + (8 × 1)
9502 = 9000 + 500 + 2 = (9 × 1000) + (5 × 100) + (0 × 10) + (2 × 1)
7000 = 7000 + 0 + 0 + 0 = (7 × 1000) + (1 × 100) + (0 × 10) + (0 × 1)

Question 2.
Write the following in the decimal form.
(i) (5 × 10) + (6 × 1);
(ii) (7 × 100)+ (5 × 10)+ (8 × 1);
(iii) (6 × 1000) + (5 × 10) + (8 × 1);
(iv) (7 × 1000) + (6 × 1);
(v) (1 × 1000) + (1 × 10);
Answer:
(i) (5 × 10) + (6 × 1);
50 + 6 = 56

(ii) (7 × 100) + (5 × 10)+ (8 × 1);
700 + 50 + 8 = 758

(iii) (6 × 1000) + (5 × 10) + (8 × 1);
6000 + 50 + 8 = 6058

(iv) (7 × 1000) + (6 × 1);
7000 + 6 = 7006

(v) (1 × 1000) + (1 × 10);
1000 + 10 = 1010