Maths-

General

Easy

Question

# If the length of the median of an equilateral triangle is x cm, then what is the area of triangle.

Hint:

### In an equilateral triangle, median is the perpendicular of the base too.

## The correct answer is: Equilateral triangle

### It is given that length of the median = x cm

Let side of the triangle = a

Since median divides the base ⇒ BD =

Using Pythagoras in triangle ABD

AB^{2} = AD^{2} + BD^{2}

a^{2} = x^{2} + ( )^{2}

= x^{2 } ⇒ = x

a =

Area of equilateral triangle =

=

=

### Related Questions to study

Maths-

### Write an equation of the line with zero slope and passing through (3,7).Application

Hint:-

Equation of a line in slope point form-

(y-y1) = m × (x-x1)

tep-by-step solution:-

As per given information-

The given line passes through the point (3,7) and has a slope of 0.

∴ x1 = 3, y1 = 7 & m = 0

Using Slope point form of a line, we can find the equation of the given line as-

(y-y1) = m × (x-x1)

∴ y - 7 = 0 × (x - 3)

∴ y - 7 = 0

∴ y = 7

Equation of a line in slope point form-

(y-y1) = m × (x-x1)

tep-by-step solution:-

As per given information-

The given line passes through the point (3,7) and has a slope of 0.

∴ x1 = 3, y1 = 7 & m = 0

Using Slope point form of a line, we can find the equation of the given line as-

(y-y1) = m × (x-x1)

∴ y - 7 = 0 × (x - 3)

∴ y - 7 = 0

∴ y = 7

### Write an equation of the line with zero slope and passing through (3,7).Application

Maths-General

Hint:-

Equation of a line in slope point form-

(y-y1) = m × (x-x1)

tep-by-step solution:-

As per given information-

The given line passes through the point (3,7) and has a slope of 0.

∴ x1 = 3, y1 = 7 & m = 0

Using Slope point form of a line, we can find the equation of the given line as-

(y-y1) = m × (x-x1)

∴ y - 7 = 0 × (x - 3)

∴ y - 7 = 0

∴ y = 7

Equation of a line in slope point form-

(y-y1) = m × (x-x1)

tep-by-step solution:-

As per given information-

The given line passes through the point (3,7) and has a slope of 0.

∴ x1 = 3, y1 = 7 & m = 0

Using Slope point form of a line, we can find the equation of the given line as-

(y-y1) = m × (x-x1)

∴ y - 7 = 0 × (x - 3)

∴ y - 7 = 0

∴ y = 7

Maths-

### The edges of a triangular board are 6 cm, 8 cm and 10 cm. Find the area of triangle.

It is given that a = 6 cm , b = 8 cm and c = 10 cm

Using Pythagoras theorem

10

100 = 64 + 36

100 = 100 i.e. Pythagoras holds true

Hence given triangle is a right angles triangle with Base, b = 8 cm . height, h = 6 cm and hypotenuse = 10 cm

Area of a right angle triangle =

= = 24 cm

Using Pythagoras theorem

10

^{2}= 8^{2}+ 6^{2}100 = 64 + 36

100 = 100 i.e. Pythagoras holds true

Hence given triangle is a right angles triangle with Base, b = 8 cm . height, h = 6 cm and hypotenuse = 10 cm

Area of a right angle triangle =

= = 24 cm

^{2}### The edges of a triangular board are 6 cm, 8 cm and 10 cm. Find the area of triangle.

Maths-General

It is given that a = 6 cm , b = 8 cm and c = 10 cm

Using Pythagoras theorem

10

100 = 64 + 36

100 = 100 i.e. Pythagoras holds true

Hence given triangle is a right angles triangle with Base, b = 8 cm . height, h = 6 cm and hypotenuse = 10 cm

Area of a right angle triangle =

= = 24 cm

Using Pythagoras theorem

10

^{2}= 8^{2}+ 6^{2}100 = 64 + 36

100 = 100 i.e. Pythagoras holds true

Hence given triangle is a right angles triangle with Base, b = 8 cm . height, h = 6 cm and hypotenuse = 10 cm

Area of a right angle triangle =

= = 24 cm

^{2}Maths-

### An electrical geyser in cylindrical shape, having a diameter of 3.5 cm and height of 1.2 m neglecting the thickness of its walls. Calculate its capacity in litres?

Hint:

We simply calculate the volume of the geyser to find its capacity.

Explanations:

Step 1 of 1:

Given, base radius r = 3.5/2cm

Height h = 1.2m = 120cm

Capacity of the geyser = volume of the geyser

= 1155 cm

= 1.155 L (since 1 L = 1000 cm

Final Answer:

The capacity of the geyser is 1.155L

We simply calculate the volume of the geyser to find its capacity.

Explanations:

Step 1 of 1:

Given, base radius r = 3.5/2cm

Height h = 1.2m = 120cm

Capacity of the geyser = volume of the geyser

= 1155 cm

^{3}= 1.155 L (since 1 L = 1000 cm

^{3})Final Answer:

The capacity of the geyser is 1.155L

### An electrical geyser in cylindrical shape, having a diameter of 3.5 cm and height of 1.2 m neglecting the thickness of its walls. Calculate its capacity in litres?

Maths-General

Hint:

We simply calculate the volume of the geyser to find its capacity.

Explanations:

Step 1 of 1:

Given, base radius r = 3.5/2cm

Height h = 1.2m = 120cm

Capacity of the geyser = volume of the geyser

= 1155 cm

= 1.155 L (since 1 L = 1000 cm

Final Answer:

The capacity of the geyser is 1.155L

We simply calculate the volume of the geyser to find its capacity.

Explanations:

Step 1 of 1:

Given, base radius r = 3.5/2cm

Height h = 1.2m = 120cm

Capacity of the geyser = volume of the geyser

= 1155 cm

^{3}= 1.155 L (since 1 L = 1000 cm

^{3})Final Answer:

The capacity of the geyser is 1.155L

Maths-

### Write an equation of the line with undefined slope and passing through (5,11).Application

Hint:-

1. X-intercept is the point on a line at which the given line intersects the x-axis.

2. At this point y-coordinate = 0.

3. A verticle line/ line parallel to the y-axis/ Y-axis has an undefined slope.

Step-by-step solution:-

The given line has an undefined slope.

∴ When plotted on a graph, given line will be represented by a verticle line which is parallel to the y-axis.

We know that any line that is parallel to the Y-axis can be represented as-

x = a ....................................................................................... (Equation i)

where a = x-intercept.

Now, when a line is parallel to the y-axis, any point on the given line has its x-coordinate constant/ same.

Since the given line passes through point (5,11) where the x-coordinate = 5

The x-coordinate at any point on the line will be 5

i.e. for y= 0 also, x = 5

and x-intercept is the point at which y = 0

∴ x-intercept for the given line = a = 5 ................................ (Equation ii)

∴ Equation of the given line = x = 5 ..................................... (From Equation i & Equation ii)

1. X-intercept is the point on a line at which the given line intersects the x-axis.

2. At this point y-coordinate = 0.

3. A verticle line/ line parallel to the y-axis/ Y-axis has an undefined slope.

Step-by-step solution:-

The given line has an undefined slope.

∴ When plotted on a graph, given line will be represented by a verticle line which is parallel to the y-axis.

We know that any line that is parallel to the Y-axis can be represented as-

x = a ....................................................................................... (Equation i)

where a = x-intercept.

Now, when a line is parallel to the y-axis, any point on the given line has its x-coordinate constant/ same.

Since the given line passes through point (5,11) where the x-coordinate = 5

The x-coordinate at any point on the line will be 5

i.e. for y= 0 also, x = 5

and x-intercept is the point at which y = 0

∴ x-intercept for the given line = a = 5 ................................ (Equation ii)

∴ Equation of the given line = x = 5 ..................................... (From Equation i & Equation ii)

### Write an equation of the line with undefined slope and passing through (5,11).Application

Maths-General

Hint:-

1. X-intercept is the point on a line at which the given line intersects the x-axis.

2. At this point y-coordinate = 0.

3. A verticle line/ line parallel to the y-axis/ Y-axis has an undefined slope.

Step-by-step solution:-

The given line has an undefined slope.

∴ When plotted on a graph, given line will be represented by a verticle line which is parallel to the y-axis.

We know that any line that is parallel to the Y-axis can be represented as-

x = a ....................................................................................... (Equation i)

where a = x-intercept.

Now, when a line is parallel to the y-axis, any point on the given line has its x-coordinate constant/ same.

Since the given line passes through point (5,11) where the x-coordinate = 5

The x-coordinate at any point on the line will be 5

i.e. for y= 0 also, x = 5

and x-intercept is the point at which y = 0

∴ x-intercept for the given line = a = 5 ................................ (Equation ii)

∴ Equation of the given line = x = 5 ..................................... (From Equation i & Equation ii)

1. X-intercept is the point on a line at which the given line intersects the x-axis.

2. At this point y-coordinate = 0.

3. A verticle line/ line parallel to the y-axis/ Y-axis has an undefined slope.

Step-by-step solution:-

The given line has an undefined slope.

∴ When plotted on a graph, given line will be represented by a verticle line which is parallel to the y-axis.

We know that any line that is parallel to the Y-axis can be represented as-

x = a ....................................................................................... (Equation i)

where a = x-intercept.

Now, when a line is parallel to the y-axis, any point on the given line has its x-coordinate constant/ same.

Since the given line passes through point (5,11) where the x-coordinate = 5

The x-coordinate at any point on the line will be 5

i.e. for y= 0 also, x = 5

and x-intercept is the point at which y = 0

∴ x-intercept for the given line = a = 5 ................................ (Equation ii)

∴ Equation of the given line = x = 5 ..................................... (From Equation i & Equation ii)

Maths-

### The difference between the sides of the right angles triangle containing the right angle is 7 cm, and its area is 60 cm2 . Find the perimeter of the triangle.

It is given that difference between perpendicular and base = 7

i.e. b – h = 7 ⇒ b = 7 + h

Now, Area of triangle = 60 cm

= 60

= 120

h

h

h(h + 15)-8(h + 15) = 0

(h + 15)(h – 8) = 0

h = 8 , - 15

Since, perpendicular is always positive so h = 8

Base, B = 7 + 8 = 15

Using Pythagoras theorem ,

H

H

H = = 17 cm

Perimeter of triangle = Sum of all sides

= 17 + 8 + 15 = 40 cm

i.e. b – h = 7 ⇒ b = 7 + h

Now, Area of triangle = 60 cm

^{2}= 60

= 120

h

^{2}+ 7h – 120 = 0h

^{2}+15h – 8h – 120 = 0h(h + 15)-8(h + 15) = 0

(h + 15)(h – 8) = 0

h = 8 , - 15

Since, perpendicular is always positive so h = 8

Base, B = 7 + 8 = 15

Using Pythagoras theorem ,

H

^{2}= B^{2}+ P^{2}H

^{2}= 15^{2}+ 8^{2}= 225 + 64 = 289H = = 17 cm

Perimeter of triangle = Sum of all sides

= 17 + 8 + 15 = 40 cm

### The difference between the sides of the right angles triangle containing the right angle is 7 cm, and its area is 60 cm2 . Find the perimeter of the triangle.

Maths-General

It is given that difference between perpendicular and base = 7

i.e. b – h = 7 ⇒ b = 7 + h

Now, Area of triangle = 60 cm

= 60

= 120

h

h

h(h + 15)-8(h + 15) = 0

(h + 15)(h – 8) = 0

h = 8 , - 15

Since, perpendicular is always positive so h = 8

Base, B = 7 + 8 = 15

Using Pythagoras theorem ,

H

H

H = = 17 cm

Perimeter of triangle = Sum of all sides

= 17 + 8 + 15 = 40 cm

i.e. b – h = 7 ⇒ b = 7 + h

Now, Area of triangle = 60 cm

^{2}= 60

= 120

h

^{2}+ 7h – 120 = 0h

^{2}+15h – 8h – 120 = 0h(h + 15)-8(h + 15) = 0

(h + 15)(h – 8) = 0

h = 8 , - 15

Since, perpendicular is always positive so h = 8

Base, B = 7 + 8 = 15

Using Pythagoras theorem ,

H

^{2}= B^{2}+ P^{2}H

^{2}= 15^{2}+ 8^{2}= 225 + 64 = 289H = = 17 cm

Perimeter of triangle = Sum of all sides

= 17 + 8 + 15 = 40 cm

Maths-

### An artist charges $20 constant fee plus $10 per extra hour he does. Write the linear model and find the charges taken by him if he works for 4 and a half hours.

Hint:-

1. Cost of extra hours = number of hours performed × rate per extra hour

2. Total cost = Constant fees + cost of extra hours performed.

Step-by-step solution:-

We are given that-

constant fees = $20

Cost of extra hours performed = $10 / hour

Let x be the number of extra hours performed by the artist and y be the total fees charged by him for the entire performance.

Cost of extra hours = Number of extra hours performed × rate per extra hour

∴ Cost of extra hours = x × 10

∴ Cost of extra hours = 10x .................................................................................................. (Equation i)

Now, we know that-

Total fees charged by the artist = y = constant fees + cost of extra hours performed

∴ y = 20 + 10x ........ (From given information & Equation i) ....................... (Equation ii)

Now, we need to find the total fees charged by the artist if he works 4 and a half extra hours.

∴ x = 4

We substitute x = 4 in Equation ii-

y = 20 + 10x

∴ y = 20 + 10 (4)

∴ y = 20 + 40

∴ y = 60

Final Answer:-

∴ Linear model that represents the fees charged by the artist is y = 20 + 10x and the charges taken by him if he works for 4 and a half hours $60.

1. Cost of extra hours = number of hours performed × rate per extra hour

2. Total cost = Constant fees + cost of extra hours performed.

Step-by-step solution:-

We are given that-

constant fees = $20

Cost of extra hours performed = $10 / hour

Let x be the number of extra hours performed by the artist and y be the total fees charged by him for the entire performance.

Cost of extra hours = Number of extra hours performed × rate per extra hour

∴ Cost of extra hours = x × 10

∴ Cost of extra hours = 10x .................................................................................................. (Equation i)

Now, we know that-

Total fees charged by the artist = y = constant fees + cost of extra hours performed

∴ y = 20 + 10x ........ (From given information & Equation i) ....................... (Equation ii)

Now, we need to find the total fees charged by the artist if he works 4 and a half extra hours.

∴ x = 4

We substitute x = 4 in Equation ii-

y = 20 + 10x

∴ y = 20 + 10 (4)

∴ y = 20 + 40

∴ y = 60

Final Answer:-

∴ Linear model that represents the fees charged by the artist is y = 20 + 10x and the charges taken by him if he works for 4 and a half hours $60.

### An artist charges $20 constant fee plus $10 per extra hour he does. Write the linear model and find the charges taken by him if he works for 4 and a half hours.

Maths-General

Hint:-

1. Cost of extra hours = number of hours performed × rate per extra hour

2. Total cost = Constant fees + cost of extra hours performed.

Step-by-step solution:-

We are given that-

constant fees = $20

Cost of extra hours performed = $10 / hour

Let x be the number of extra hours performed by the artist and y be the total fees charged by him for the entire performance.

Cost of extra hours = Number of extra hours performed × rate per extra hour

∴ Cost of extra hours = x × 10

∴ Cost of extra hours = 10x .................................................................................................. (Equation i)

Now, we know that-

Total fees charged by the artist = y = constant fees + cost of extra hours performed

∴ y = 20 + 10x ........ (From given information & Equation i) ....................... (Equation ii)

Now, we need to find the total fees charged by the artist if he works 4 and a half extra hours.

∴ x = 4

We substitute x = 4 in Equation ii-

y = 20 + 10x

∴ y = 20 + 10 (4)

∴ y = 20 + 40

∴ y = 60

Final Answer:-

∴ Linear model that represents the fees charged by the artist is y = 20 + 10x and the charges taken by him if he works for 4 and a half hours $60.

1. Cost of extra hours = number of hours performed × rate per extra hour

2. Total cost = Constant fees + cost of extra hours performed.

Step-by-step solution:-

We are given that-

constant fees = $20

Cost of extra hours performed = $10 / hour

Let x be the number of extra hours performed by the artist and y be the total fees charged by him for the entire performance.

Cost of extra hours = Number of extra hours performed × rate per extra hour

∴ Cost of extra hours = x × 10

∴ Cost of extra hours = 10x .................................................................................................. (Equation i)

Now, we know that-

Total fees charged by the artist = y = constant fees + cost of extra hours performed

∴ y = 20 + 10x ........ (From given information & Equation i) ....................... (Equation ii)

Now, we need to find the total fees charged by the artist if he works 4 and a half extra hours.

∴ x = 4

We substitute x = 4 in Equation ii-

y = 20 + 10x

∴ y = 20 + 10 (4)

∴ y = 20 + 40

∴ y = 60

Final Answer:-

∴ Linear model that represents the fees charged by the artist is y = 20 + 10x and the charges taken by him if he works for 4 and a half hours $60.

Maths-

### A square and equilateral triangle have equal perimeters. If the diagonal of the square is 12 cm, find the area of triangle.

It is given that diagonal of the square = 12 cm

Using Pythagoras theorem in triangle ACD

AD

(12)

288 = 2x

144 = x

12 = x

So, side of the square = 12 cm

Let side of an equilateral triangle = a

Perimeter of triangle = Perimeter of square

⇒ 3a = 4(side) = 4(12) = 48 cm

⇒ a = 16 cm

Now, Area of an equilateral triangle = a

= 16

( = 1.73)

= 110.85 cm

Using Pythagoras theorem in triangle ACD

AD

^{2}= AC^{2}+ CD^{2}(12)

^{2}= x^{2}+ x^{2}= 2x^{2}288 = 2x

^{2}144 = x

^{2}12 = x

So, side of the square = 12 cm

Let side of an equilateral triangle = a

Perimeter of triangle = Perimeter of square

⇒ 3a = 4(side) = 4(12) = 48 cm

⇒ a = 16 cm

Now, Area of an equilateral triangle = a

^{2}= 16

^{2}= 64( = 1.73)

= 110.85 cm

^{2}### A square and equilateral triangle have equal perimeters. If the diagonal of the square is 12 cm, find the area of triangle.

Maths-General

It is given that diagonal of the square = 12 cm

Using Pythagoras theorem in triangle ACD

AD

(12)

288 = 2x

144 = x

12 = x

So, side of the square = 12 cm

Let side of an equilateral triangle = a

Perimeter of triangle = Perimeter of square

⇒ 3a = 4(side) = 4(12) = 48 cm

⇒ a = 16 cm

Now, Area of an equilateral triangle = a

= 16

( = 1.73)

= 110.85 cm

Using Pythagoras theorem in triangle ACD

AD

^{2}= AC^{2}+ CD^{2}(12)

^{2}= x^{2}+ x^{2}= 2x^{2}288 = 2x

^{2}144 = x

^{2}12 = x

So, side of the square = 12 cm

Let side of an equilateral triangle = a

Perimeter of triangle = Perimeter of square

⇒ 3a = 4(side) = 4(12) = 48 cm

⇒ a = 16 cm

Now, Area of an equilateral triangle = a

^{2}= 16

^{2}= 64( = 1.73)

= 110.85 cm

^{2}Maths-

### Identify a line with an undefined slope .Application

a. y = x

b. y+ x =0

c. y = 10

d. x = 1

Hint:-

Verticle lines / lines parallel to y-axis have undefined slope.

Step-by-step solution:-

We know that when a line, plotted on a graph is represented by a verticle line, its slope is undefined.

i.e. slopes of lines that are parallel to y-axis are undefined.

Equations of lines that are parallel to y-axis are in the form x = a where a is the value of x-intercept.

From the given options, option d is the only option with equation of line in this format i.e. x = a i.e. x = 1.

∴ option d is the correct option.

Verticle lines / lines parallel to y-axis have undefined slope.

Step-by-step solution:-

We know that when a line, plotted on a graph is represented by a verticle line, its slope is undefined.

i.e. slopes of lines that are parallel to y-axis are undefined.

Equations of lines that are parallel to y-axis are in the form x = a where a is the value of x-intercept.

From the given options, option d is the only option with equation of line in this format i.e. x = a i.e. x = 1.

∴ option d is the correct option.

### Identify a line with an undefined slope .Application

a. y = x

b. y+ x =0

c. y = 10

d. x = 1

Maths-General

Hint:-

Verticle lines / lines parallel to y-axis have undefined slope.

Step-by-step solution:-

We know that when a line, plotted on a graph is represented by a verticle line, its slope is undefined.

i.e. slopes of lines that are parallel to y-axis are undefined.

Equations of lines that are parallel to y-axis are in the form x = a where a is the value of x-intercept.

From the given options, option d is the only option with equation of line in this format i.e. x = a i.e. x = 1.

∴ option d is the correct option.

Verticle lines / lines parallel to y-axis have undefined slope.

Step-by-step solution:-

We know that when a line, plotted on a graph is represented by a verticle line, its slope is undefined.

i.e. slopes of lines that are parallel to y-axis are undefined.

Equations of lines that are parallel to y-axis are in the form x = a where a is the value of x-intercept.

From the given options, option d is the only option with equation of line in this format i.e. x = a i.e. x = 1.

∴ option d is the correct option.

Maths-

### The lengths of the two sides of right triangle containing the right angle differ by 2 cm. If the area of triangle is 24 cm^{2}. Find the perimeter of triangle

It is given that difference between perpendicular and base = 2 cm

i.e. b – h = 2 ⇒ b = 2 + h

Now, Area of triangle = 24 cm

= 24

= 48

h

h =

h =

h = 6 , - 8

Since, perpendicular is always positive so h = 6

Base, B = 2 + 6 = 8

Using Pythagoras theorem ,

H

H

H = = 10 cm

Perimeter of triangle = Sum of all sides

= 8 + 10 + 6 = 24 cm

i.e. b – h = 2 ⇒ b = 2 + h

Now, Area of triangle = 24 cm

^{2}= 24

= 48

h

^{2}+ 2h – 48 = 0h =

h =

h = 6 , - 8

Since, perpendicular is always positive so h = 6

Base, B = 2 + 6 = 8

Using Pythagoras theorem ,

H

^{2}= B^{2}+ P^{2}H

^{2}= 8^{2}+ 6^{2}= 64 + 36 = 100H = = 10 cm

Perimeter of triangle = Sum of all sides

= 8 + 10 + 6 = 24 cm

### The lengths of the two sides of right triangle containing the right angle differ by 2 cm. If the area of triangle is 24 cm^{2}. Find the perimeter of triangle

Maths-General

It is given that difference between perpendicular and base = 2 cm

i.e. b – h = 2 ⇒ b = 2 + h

Now, Area of triangle = 24 cm

= 24

= 48

h

h =

h =

h = 6 , - 8

Since, perpendicular is always positive so h = 6

Base, B = 2 + 6 = 8

Using Pythagoras theorem ,

H

H

H = = 10 cm

Perimeter of triangle = Sum of all sides

= 8 + 10 + 6 = 24 cm

i.e. b – h = 2 ⇒ b = 2 + h

Now, Area of triangle = 24 cm

^{2}= 24

= 48

h

^{2}+ 2h – 48 = 0h =

h =

h = 6 , - 8

Since, perpendicular is always positive so h = 6

Base, B = 2 + 6 = 8

Using Pythagoras theorem ,

H

^{2}= B^{2}+ P^{2}H

^{2}= 8^{2}+ 6^{2}= 64 + 36 = 100H = = 10 cm

Perimeter of triangle = Sum of all sides

= 8 + 10 + 6 = 24 cm

Maths-

### Identify a line having zero slope. Application

a. x = y

b. x = 100

c. y = 100

d. x = 0

Hint:-

Horizontal lines / lines parallel to x-axis have slope = 0.

Step-by-step solution:-

We know that when a line, plotted on a graph is represented by a Horizontal line, its slope is 0.

i.e. slopes of lines that are parallel to x-axis are 0.

Equations of lines that are parallel to x-axis are in the form y = b where b is the value of y-intercept.

Option C i.e. y = 100 is the only option that satisfies this condition.

Hence, y = 100 is the line having zero slope.

Horizontal lines / lines parallel to x-axis have slope = 0.

Step-by-step solution:-

We know that when a line, plotted on a graph is represented by a Horizontal line, its slope is 0.

i.e. slopes of lines that are parallel to x-axis are 0.

Equations of lines that are parallel to x-axis are in the form y = b where b is the value of y-intercept.

Option C i.e. y = 100 is the only option that satisfies this condition.

Hence, y = 100 is the line having zero slope.

### Identify a line having zero slope. Application

a. x = y

b. x = 100

c. y = 100

d. x = 0

Maths-General

Hint:-

Horizontal lines / lines parallel to x-axis have slope = 0.

Step-by-step solution:-

We know that when a line, plotted on a graph is represented by a Horizontal line, its slope is 0.

i.e. slopes of lines that are parallel to x-axis are 0.

Equations of lines that are parallel to x-axis are in the form y = b where b is the value of y-intercept.

Option C i.e. y = 100 is the only option that satisfies this condition.

Hence, y = 100 is the line having zero slope.

Horizontal lines / lines parallel to x-axis have slope = 0.

Step-by-step solution:-

We know that when a line, plotted on a graph is represented by a Horizontal line, its slope is 0.

i.e. slopes of lines that are parallel to x-axis are 0.

Equations of lines that are parallel to x-axis are in the form y = b where b is the value of y-intercept.

Option C i.e. y = 100 is the only option that satisfies this condition.

Hence, y = 100 is the line having zero slope.

Maths-

### The cost of an ice cream scoop is $11 and the cost of additional toppings is $1.50. Find the cost of an ice cream with 4 toppings.

Hint:-

1. Cost of additional toppings = number of toppings × cost of each topping

2. Total cost = Cost of 1 ice cream scoop + cost of additional toppings.

Step-by-step solution:-

We are given that-

Cost of 1 ice cream scoop = $11

Cost of each additional topping = $1.50

Let x be the number of toppings and y be the cost of entire ice-cream.

Cost of additional toppings = Cost of each topping × total number of toppings

∴ Cost of additional toppings = 1.50 x .................................................................................................. (Equation i)

Now, we know that-

cost of an icecream = y = Cost of 1 icecream scoop + cost of additional toppings

∴ y = 11 + 1.50 x ........ (From given information & Equation i) ....................... (Equation ii)

Now, we need to find the cost of an icecream with 4 additional toppings.

∴ x = 4

We substitute x = 4 in Equation ii-

y = 11 + 1.50x

∴ y = 11 + 1.50 (4)

∴ y = 11 + 6

∴ y = 17

Final Answer:-

∴ Total cost of an ice cream with 4 additional toppings is $17.

1. Cost of additional toppings = number of toppings × cost of each topping

2. Total cost = Cost of 1 ice cream scoop + cost of additional toppings.

Step-by-step solution:-

We are given that-

Cost of 1 ice cream scoop = $11

Cost of each additional topping = $1.50

Let x be the number of toppings and y be the cost of entire ice-cream.

Cost of additional toppings = Cost of each topping × total number of toppings

∴ Cost of additional toppings = 1.50 x .................................................................................................. (Equation i)

Now, we know that-

cost of an icecream = y = Cost of 1 icecream scoop + cost of additional toppings

∴ y = 11 + 1.50 x ........ (From given information & Equation i) ....................... (Equation ii)

Now, we need to find the cost of an icecream with 4 additional toppings.

∴ x = 4

We substitute x = 4 in Equation ii-

y = 11 + 1.50x

∴ y = 11 + 1.50 (4)

∴ y = 11 + 6

∴ y = 17

Final Answer:-

∴ Total cost of an ice cream with 4 additional toppings is $17.

### The cost of an ice cream scoop is $11 and the cost of additional toppings is $1.50. Find the cost of an ice cream with 4 toppings.

Maths-General

Hint:-

1. Cost of additional toppings = number of toppings × cost of each topping

2. Total cost = Cost of 1 ice cream scoop + cost of additional toppings.

Step-by-step solution:-

We are given that-

Cost of 1 ice cream scoop = $11

Cost of each additional topping = $1.50

Let x be the number of toppings and y be the cost of entire ice-cream.

Cost of additional toppings = Cost of each topping × total number of toppings

∴ Cost of additional toppings = 1.50 x .................................................................................................. (Equation i)

Now, we know that-

cost of an icecream = y = Cost of 1 icecream scoop + cost of additional toppings

∴ y = 11 + 1.50 x ........ (From given information & Equation i) ....................... (Equation ii)

Now, we need to find the cost of an icecream with 4 additional toppings.

∴ x = 4

We substitute x = 4 in Equation ii-

y = 11 + 1.50x

∴ y = 11 + 1.50 (4)

∴ y = 11 + 6

∴ y = 17

Final Answer:-

∴ Total cost of an ice cream with 4 additional toppings is $17.

1. Cost of additional toppings = number of toppings × cost of each topping

2. Total cost = Cost of 1 ice cream scoop + cost of additional toppings.

Step-by-step solution:-

We are given that-

Cost of 1 ice cream scoop = $11

Cost of each additional topping = $1.50

Let x be the number of toppings and y be the cost of entire ice-cream.

Cost of additional toppings = Cost of each topping × total number of toppings

∴ Cost of additional toppings = 1.50 x .................................................................................................. (Equation i)

Now, we know that-

cost of an icecream = y = Cost of 1 icecream scoop + cost of additional toppings

∴ y = 11 + 1.50 x ........ (From given information & Equation i) ....................... (Equation ii)

Now, we need to find the cost of an icecream with 4 additional toppings.

∴ x = 4

We substitute x = 4 in Equation ii-

y = 11 + 1.50x

∴ y = 11 + 1.50 (4)

∴ y = 11 + 6

∴ y = 17

Final Answer:-

∴ Total cost of an ice cream with 4 additional toppings is $17.

Maths-

### The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used up when writing 310 words on an average. How many words would use up a bottle of ink containing one fifth of litre?

Hint:

We find the capacity of the pen barrel, and then find the number of words accordingly.

Explanations:

Step 1 of 2:

Given, length of the barrel = height of barrel, h = 7cm

Base radius of the barrel, r = 5/2 mm = 2.5 mm = 0.25cm (since 1cm = 10mm)

Capacity of the barrel = volume of the barrel

cm

So, 310 words use up 1.375 cm

Step 2 of 2:

1/5

Now, 1.375cm

200cm

Final Answer:

The required number of words is 45091.

We find the capacity of the pen barrel, and then find the number of words accordingly.

Explanations:

Step 1 of 2:

Given, length of the barrel = height of barrel, h = 7cm

Base radius of the barrel, r = 5/2 mm = 2.5 mm = 0.25cm (since 1cm = 10mm)

Capacity of the barrel = volume of the barrel

cm

^{3}So, 310 words use up 1.375 cm

^{3}of ink.Step 2 of 2:

1/5

^{th}of litre = 1/51000cm^{3}= 200cm^{3}Now, 1.375cm

^{3}of ink is used up by 310 words200cm

^{3}of ink will be used up by = 45090.9 45091 wordsFinal Answer:

The required number of words is 45091.

### The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used up when writing 310 words on an average. How many words would use up a bottle of ink containing one fifth of litre?

Maths-General

Hint:

We find the capacity of the pen barrel, and then find the number of words accordingly.

Explanations:

Step 1 of 2:

Given, length of the barrel = height of barrel, h = 7cm

Base radius of the barrel, r = 5/2 mm = 2.5 mm = 0.25cm (since 1cm = 10mm)

Capacity of the barrel = volume of the barrel

cm

So, 310 words use up 1.375 cm

Step 2 of 2:

1/5

Now, 1.375cm

200cm

Final Answer:

The required number of words is 45091.

We find the capacity of the pen barrel, and then find the number of words accordingly.

Explanations:

Step 1 of 2:

Given, length of the barrel = height of barrel, h = 7cm

Base radius of the barrel, r = 5/2 mm = 2.5 mm = 0.25cm (since 1cm = 10mm)

Capacity of the barrel = volume of the barrel

cm

^{3}So, 310 words use up 1.375 cm

^{3}of ink.Step 2 of 2:

1/5

^{th}of litre = 1/51000cm^{3}= 200cm^{3}Now, 1.375cm

^{3}of ink is used up by 310 words200cm

^{3}of ink will be used up by = 45090.9 45091 wordsFinal Answer:

The required number of words is 45091.

Maths-

### Find the equation of line that passes through the point (2, -9) and which is perpendicular to the line x = 5.Application

Hint:-

1. Slope of verticle lines are undefined & slop of horizontal lines are 0.

2. Slopes of Perpendicular lines are negative reciprocals of each other.

3. Equation of a line in slope point form is-

(y-y1) = m (x-x1)

Step-by-step solution:-

Let l be the line for which slope is to be found.

The given equation of line x = 5 represents a verticle line (because its y-coordinate = 0)

Now, line l is perpendicular to the given line (x = 5).

∴ line l is a horizontal line i.e. parallel to x-axis.

Slope of a horizontal line i.e. parallel to x-axis is always 0

∴ Slope of line l = m = 0 …...................................................................................................... (Equation i)

We are given that line l passes through the point (2,-9) and we know its slope.

i.e. x1 = 2 & y1 = -9 & m = 0

We can use slope point form of an equation to find the equation of line l-

(y - y1) = m (x-x1)

∴ [y - (-9)] = 0 (x-2)

∴ y + 9 = 0

∴ y = -9

1. Slope of verticle lines are undefined & slop of horizontal lines are 0.

2. Slopes of Perpendicular lines are negative reciprocals of each other.

3. Equation of a line in slope point form is-

(y-y1) = m (x-x1)

Step-by-step solution:-

Let l be the line for which slope is to be found.

The given equation of line x = 5 represents a verticle line (because its y-coordinate = 0)

Now, line l is perpendicular to the given line (x = 5).

∴ line l is a horizontal line i.e. parallel to x-axis.

Slope of a horizontal line i.e. parallel to x-axis is always 0

∴ Slope of line l = m = 0 …...................................................................................................... (Equation i)

We are given that line l passes through the point (2,-9) and we know its slope.

i.e. x1 = 2 & y1 = -9 & m = 0

We can use slope point form of an equation to find the equation of line l-

(y - y1) = m (x-x1)

∴ [y - (-9)] = 0 (x-2)

∴ y + 9 = 0

∴ y = -9

### Find the equation of line that passes through the point (2, -9) and which is perpendicular to the line x = 5.Application

Maths-General

Hint:-

1. Slope of verticle lines are undefined & slop of horizontal lines are 0.

2. Slopes of Perpendicular lines are negative reciprocals of each other.

3. Equation of a line in slope point form is-

(y-y1) = m (x-x1)

Step-by-step solution:-

Let l be the line for which slope is to be found.

The given equation of line x = 5 represents a verticle line (because its y-coordinate = 0)

Now, line l is perpendicular to the given line (x = 5).

∴ line l is a horizontal line i.e. parallel to x-axis.

Slope of a horizontal line i.e. parallel to x-axis is always 0

∴ Slope of line l = m = 0 …...................................................................................................... (Equation i)

We are given that line l passes through the point (2,-9) and we know its slope.

i.e. x1 = 2 & y1 = -9 & m = 0

We can use slope point form of an equation to find the equation of line l-

(y - y1) = m (x-x1)

∴ [y - (-9)] = 0 (x-2)

∴ y + 9 = 0

∴ y = -9

1. Slope of verticle lines are undefined & slop of horizontal lines are 0.

2. Slopes of Perpendicular lines are negative reciprocals of each other.

3. Equation of a line in slope point form is-

(y-y1) = m (x-x1)

Step-by-step solution:-

Let l be the line for which slope is to be found.

The given equation of line x = 5 represents a verticle line (because its y-coordinate = 0)

Now, line l is perpendicular to the given line (x = 5).

∴ line l is a horizontal line i.e. parallel to x-axis.

Slope of a horizontal line i.e. parallel to x-axis is always 0

∴ Slope of line l = m = 0 …...................................................................................................... (Equation i)

We are given that line l passes through the point (2,-9) and we know its slope.

i.e. x1 = 2 & y1 = -9 & m = 0

We can use slope point form of an equation to find the equation of line l-

(y - y1) = m (x-x1)

∴ [y - (-9)] = 0 (x-2)

∴ y + 9 = 0

∴ y = -9

Maths-

### Identify the parallel lines and perpendicular lines from the given set. Application

2x + y = 1

9x + 3y = 6

y = 3x

y = -3x

2y = 4x +6

Y = - x/2

Hint:-

1. Standard form of equation of a straight line is y = mx + c.

2. Slopes of parallel lines are equal.

3. Slopes of perpendicular lines are negative reciprocals of each other.

Step-by-step solution:-

We will simplify the given equations and compare the same with standard form of a straight line to find the value of m.

a. 2x + y = 1

∴ y = -2x + 1

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = -2 ......................... (Equation i)

b. 9x + 3y = 6

∴ 3y = -9x + 6

∴ y = -3x + 2 ............................ (Dividing both sides by 3)

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = -3 ......................... (Equation ii)

c. y = 3x

∴ y = 3x + 0

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = 3 ......................... (Equation iii)

We know that slopes of perpendicular lines are negative reciprocals of each other.

and we observe that-

Slope of line e (2y = 4x + 6) = 2 ...................................................... (From Equation v)

∴ Slope of line e (2y = 4x + 6) = -1/ -1/2 …........................................ (Multiplying and dividing by -1/2)

∴ Slope of line e (2y = 4x + 6) = -1/ Slope of line f (y = -x/2) ........... (From Equation vi)

∴ Slope of line e and f are negative reciprocals of each other

∴ line e (2y = 4x +6) and line f (Y = - x/2) are perpendicular lines.

Also, We know that slopes of parallel lines are equal.

and we observe that-

Slope of line b (9x + 3y = 6) = Slope of line d (y = -3x) = -3 .............. (From Equation ii & iv)

∴ line b (9x + 3y = 6) and line d (y = -3x) are parallel lines.

d. y = -3x

∴ y = -3x + 0

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = -3 ......................... (Equation iv)

e. 2y = 4x +6

∴ y = 2x + 3 .................................... (Dividing both sides by 2)

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = 2 ......................... (Equation v)

1. Standard form of equation of a straight line is y = mx + c.

2. Slopes of parallel lines are equal.

3. Slopes of perpendicular lines are negative reciprocals of each other.

Step-by-step solution:-

We will simplify the given equations and compare the same with standard form of a straight line to find the value of m.

a. 2x + y = 1

∴ y = -2x + 1

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = -2 ......................... (Equation i)

b. 9x + 3y = 6

∴ 3y = -9x + 6

∴ y = -3x + 2 ............................ (Dividing both sides by 3)

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = -3 ......................... (Equation ii)

c. y = 3x

∴ y = 3x + 0

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = 3 ......................... (Equation iii)

We know that slopes of perpendicular lines are negative reciprocals of each other.

and we observe that-

Slope of line e (2y = 4x + 6) = 2 ...................................................... (From Equation v)

∴ Slope of line e (2y = 4x + 6) = -1/ -1/2 …........................................ (Multiplying and dividing by -1/2)

∴ Slope of line e (2y = 4x + 6) = -1/ Slope of line f (y = -x/2) ........... (From Equation vi)

∴ Slope of line e and f are negative reciprocals of each other

∴ line e (2y = 4x +6) and line f (Y = - x/2) are perpendicular lines.

Also, We know that slopes of parallel lines are equal.

and we observe that-

Slope of line b (9x + 3y = 6) = Slope of line d (y = -3x) = -3 .............. (From Equation ii & iv)

∴ line b (9x + 3y = 6) and line d (y = -3x) are parallel lines.

d. y = -3x

∴ y = -3x + 0

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = -3 ......................... (Equation iv)

e. 2y = 4x +6

∴ y = 2x + 3 .................................... (Dividing both sides by 2)

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = 2 ......................... (Equation v)

### Identify the parallel lines and perpendicular lines from the given set. Application

2x + y = 1

9x + 3y = 6

y = 3x

y = -3x

2y = 4x +6

Y = - x/2

Maths-General

Hint:-

1. Standard form of equation of a straight line is y = mx + c.

2. Slopes of parallel lines are equal.

3. Slopes of perpendicular lines are negative reciprocals of each other.

Step-by-step solution:-

We will simplify the given equations and compare the same with standard form of a straight line to find the value of m.

a. 2x + y = 1

∴ y = -2x + 1

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = -2 ......................... (Equation i)

b. 9x + 3y = 6

∴ 3y = -9x + 6

∴ y = -3x + 2 ............................ (Dividing both sides by 3)

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = -3 ......................... (Equation ii)

c. y = 3x

∴ y = 3x + 0

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = 3 ......................... (Equation iii)

We know that slopes of perpendicular lines are negative reciprocals of each other.

and we observe that-

Slope of line e (2y = 4x + 6) = 2 ...................................................... (From Equation v)

∴ Slope of line e (2y = 4x + 6) = -1/ -1/2 …........................................ (Multiplying and dividing by -1/2)

∴ Slope of line e (2y = 4x + 6) = -1/ Slope of line f (y = -x/2) ........... (From Equation vi)

∴ Slope of line e and f are negative reciprocals of each other

∴ line e (2y = 4x +6) and line f (Y = - x/2) are perpendicular lines.

Also, We know that slopes of parallel lines are equal.

and we observe that-

Slope of line b (9x + 3y = 6) = Slope of line d (y = -3x) = -3 .............. (From Equation ii & iv)

∴ line b (9x + 3y = 6) and line d (y = -3x) are parallel lines.

d. y = -3x

∴ y = -3x + 0

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = -3 ......................... (Equation iv)

e. 2y = 4x +6

∴ y = 2x + 3 .................................... (Dividing both sides by 2)

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = 2 ......................... (Equation v)

1. Standard form of equation of a straight line is y = mx + c.

2. Slopes of parallel lines are equal.

3. Slopes of perpendicular lines are negative reciprocals of each other.

Step-by-step solution:-

We will simplify the given equations and compare the same with standard form of a straight line to find the value of m.

a. 2x + y = 1

∴ y = -2x + 1

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = -2 ......................... (Equation i)

b. 9x + 3y = 6

∴ 3y = -9x + 6

∴ y = -3x + 2 ............................ (Dividing both sides by 3)

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = -3 ......................... (Equation ii)

c. y = 3x

∴ y = 3x + 0

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = 3 ......................... (Equation iii)

We know that slopes of perpendicular lines are negative reciprocals of each other.

and we observe that-

Slope of line e (2y = 4x + 6) = 2 ...................................................... (From Equation v)

∴ Slope of line e (2y = 4x + 6) = -1/ -1/2 …........................................ (Multiplying and dividing by -1/2)

∴ Slope of line e (2y = 4x + 6) = -1/ Slope of line f (y = -x/2) ........... (From Equation vi)

∴ Slope of line e and f are negative reciprocals of each other

∴ line e (2y = 4x +6) and line f (Y = - x/2) are perpendicular lines.

Also, We know that slopes of parallel lines are equal.

and we observe that-

Slope of line b (9x + 3y = 6) = Slope of line d (y = -3x) = -3 .............. (From Equation ii & iv)

∴ line b (9x + 3y = 6) and line d (y = -3x) are parallel lines.

d. y = -3x

∴ y = -3x + 0

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = -3 ......................... (Equation iv)

e. 2y = 4x +6

∴ y = 2x + 3 .................................... (Dividing both sides by 2)

Comparing the above equation with standard form of a line i.e. y = mx + c, we get- m = 2 ......................... (Equation v)

Maths-

### A line l touches the Y-axis at point 3 and has slope –3. Find the slope of the line perpendicular to the line l.

Hint:-

1. The slope of a line can be defined as the change in y coordinates of any 2 points on that line corresponding to the change in the x coordinates of those 2 points. This is generally referred to as the rise to run ratio of the given line i.e. how much did the y-coordinates rise vis-a-vis how long a distance was covered by the x-coordinates. Slope = m = rise / run = y2-y1 / x2-x1

2. Slopes of perpendicular lines are negative reciprocals of each other.

Step-by-step solution:-

Slope of line l = -3 …...................... (Given).

The given line is perpendicular to line l ….................. (Given)

We know that slopes of perpendicular lines are negative reciprocals of each other.

∴ Slope of the given line = -1/ slope of line l

∴ Slope of the given line = -1/ -3

∴ Slope of the given line = 1/3

Final Answer:-

∴ Slope of the line which is perpendicular to line l is 1/3.

1. The slope of a line can be defined as the change in y coordinates of any 2 points on that line corresponding to the change in the x coordinates of those 2 points. This is generally referred to as the rise to run ratio of the given line i.e. how much did the y-coordinates rise vis-a-vis how long a distance was covered by the x-coordinates. Slope = m = rise / run = y2-y1 / x2-x1

2. Slopes of perpendicular lines are negative reciprocals of each other.

Step-by-step solution:-

Slope of line l = -3 …...................... (Given).

The given line is perpendicular to line l ….................. (Given)

We know that slopes of perpendicular lines are negative reciprocals of each other.

∴ Slope of the given line = -1/ slope of line l

∴ Slope of the given line = -1/ -3

∴ Slope of the given line = 1/3

Final Answer:-

∴ Slope of the line which is perpendicular to line l is 1/3.

### A line l touches the Y-axis at point 3 and has slope –3. Find the slope of the line perpendicular to the line l.

Maths-General

Hint:-

1. The slope of a line can be defined as the change in y coordinates of any 2 points on that line corresponding to the change in the x coordinates of those 2 points. This is generally referred to as the rise to run ratio of the given line i.e. how much did the y-coordinates rise vis-a-vis how long a distance was covered by the x-coordinates. Slope = m = rise / run = y2-y1 / x2-x1

2. Slopes of perpendicular lines are negative reciprocals of each other.

Step-by-step solution:-

Slope of line l = -3 …...................... (Given).

The given line is perpendicular to line l ….................. (Given)

We know that slopes of perpendicular lines are negative reciprocals of each other.

∴ Slope of the given line = -1/ slope of line l

∴ Slope of the given line = -1/ -3

∴ Slope of the given line = 1/3

Final Answer:-

∴ Slope of the line which is perpendicular to line l is 1/3.

1. The slope of a line can be defined as the change in y coordinates of any 2 points on that line corresponding to the change in the x coordinates of those 2 points. This is generally referred to as the rise to run ratio of the given line i.e. how much did the y-coordinates rise vis-a-vis how long a distance was covered by the x-coordinates. Slope = m = rise / run = y2-y1 / x2-x1

2. Slopes of perpendicular lines are negative reciprocals of each other.

Step-by-step solution:-

Slope of line l = -3 …...................... (Given).

The given line is perpendicular to line l ….................. (Given)

We know that slopes of perpendicular lines are negative reciprocals of each other.

∴ Slope of the given line = -1/ slope of line l

∴ Slope of the given line = -1/ -3

∴ Slope of the given line = 1/3

Final Answer:-

∴ Slope of the line which is perpendicular to line l is 1/3.