Maths-

General

Easy

Question

# Find the area of triangle whose base is 10 cm and height is 5 cm

Hint:

### Use formula directly

## The correct answer is: 25 cm2

### It is given that base, b = 10 cm and height, h = 5 cm

Area of triangle =

=

### Related Questions to study

Maths-

### Find the area of a piece of cardboard which is in shape of an equilateral triangle of side 12 cm.

It is given that side of the triangle =12 cm

Area of equilateral triangle = (side)

= (12)

= 36 cm

(Substituting the value of = 1.73, in the above equation)

= 62.35 cm

Area of equilateral triangle = (side)

^{2}= (12)

^{2 }= (144)= 36 cm

^{2 }(Substituting the value of = 1.73, in the above equation)

= 62.35 cm

^{2}### Find the area of a piece of cardboard which is in shape of an equilateral triangle of side 12 cm.

Maths-General

It is given that side of the triangle =12 cm

Area of equilateral triangle = (side)

= (12)

= 36 cm

(Substituting the value of = 1.73, in the above equation)

= 62.35 cm

Area of equilateral triangle = (side)

^{2}= (12)

^{2 }= (144)= 36 cm

^{2 }(Substituting the value of = 1.73, in the above equation)

= 62.35 cm

^{2}Maths-

### The difference between the sides of right angle triangle containing the right angle is 7 cm and its area is 60 cm^{2} . Calculate the perimeter of 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 right angle triangle containing the right angle is 7 cm and its area is 60 cm^{2} . Calculate the perimeter of 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-

### The Sum of parallel sides of a trapezoidal window is 800 inch and its area is 40000 sq inches. What is the height of the window?

Ans :- 105 m.

Explanation :-

Given Sum of lengths of parallel sides = 800 inch

Area of trapezium = 40,000 sq. inches

Let height or depth of cross section be h m

Therefore , the height of the window is 100 inch .

Explanation :-

Given Sum of lengths of parallel sides = 800 inch

Area of trapezium = 40,000 sq. inches

Let height or depth of cross section be h m

Therefore , the height of the window is 100 inch .

### The Sum of parallel sides of a trapezoidal window is 800 inch and its area is 40000 sq inches. What is the height of the window?

Maths-General

Ans :- 105 m.

Explanation :-

Given Sum of lengths of parallel sides = 800 inch

Area of trapezium = 40,000 sq. inches

Let height or depth of cross section be h m

Therefore , the height of the window is 100 inch .

Explanation :-

Given Sum of lengths of parallel sides = 800 inch

Area of trapezium = 40,000 sq. inches

Let height or depth of cross section be h m

Therefore , the height of the window is 100 inch .

Maths-

### George took a nonstop flight from Dallas to Los Angeles, a total flight distance of 1,233 miles. The plane flew at a speed of 460 miles per hour for the first 75 minutes of the flight and at a speed of 439 miles per hour for the remainder of the flight. To the nearest minute, for how many minutes did the plane fly at a speed of 439 miles per hour?

The total distance between Los Angeles and Dallas is 1233 miles.

So, total distance covered = 1233 miles.

Initially,

Speed of the plane = 460 miles per hour

Time taken = 75 minutes

We convert the time in hours as the speed is given in miles per hour.

As, 60 minutes = 1 hour

So, we get, 75 minutes = = = 1.25 hours

Thus, time taken = 1.5 hours

We know,

Rewriting the equation , we have

Thus, the distance travelled initially is given by

Thus, the distance covered by the flight in the first 75 minutes while the speed was 460 miles per hour = 575 miles

Next,

Remaining distance to travel = 1233 – 575 = 658 miles

For the second part of the journey,

Speed of the plane = 439 miles per hour

Distance travelled = 658 miles

Rewriting the above formula, we get

Using the above quantities, we get

Converting the time into minutes, we have,

Thus, the number of minutes for which the plane flies at speed 439 miles per hour = 90 minutes

.

So, total distance covered = 1233 miles.

Initially,

Speed of the plane = 460 miles per hour

Time taken = 75 minutes

We convert the time in hours as the speed is given in miles per hour.

As, 60 minutes = 1 hour

So, we get, 75 minutes = = = 1.25 hours

Thus, time taken = 1.5 hours

We know,

Rewriting the equation , we have

Thus, the distance travelled initially is given by

Thus, the distance covered by the flight in the first 75 minutes while the speed was 460 miles per hour = 575 miles

Next,

Remaining distance to travel = 1233 – 575 = 658 miles

For the second part of the journey,

Speed of the plane = 439 miles per hour

Distance travelled = 658 miles

Rewriting the above formula, we get

Using the above quantities, we get

Converting the time into minutes, we have,

Thus, the number of minutes for which the plane flies at speed 439 miles per hour = 90 minutes

.

### George took a nonstop flight from Dallas to Los Angeles, a total flight distance of 1,233 miles. The plane flew at a speed of 460 miles per hour for the first 75 minutes of the flight and at a speed of 439 miles per hour for the remainder of the flight. To the nearest minute, for how many minutes did the plane fly at a speed of 439 miles per hour?

Maths-General

The total distance between Los Angeles and Dallas is 1233 miles.

So, total distance covered = 1233 miles.

Initially,

Speed of the plane = 460 miles per hour

Time taken = 75 minutes

We convert the time in hours as the speed is given in miles per hour.

As, 60 minutes = 1 hour

So, we get, 75 minutes = = = 1.25 hours

Thus, time taken = 1.5 hours

We know,

Rewriting the equation , we have

Thus, the distance travelled initially is given by

Thus, the distance covered by the flight in the first 75 minutes while the speed was 460 miles per hour = 575 miles

Next,

Remaining distance to travel = 1233 – 575 = 658 miles

For the second part of the journey,

Speed of the plane = 439 miles per hour

Distance travelled = 658 miles

Rewriting the above formula, we get

Using the above quantities, we get

Converting the time into minutes, we have,

Thus, the number of minutes for which the plane flies at speed 439 miles per hour = 90 minutes

.

So, total distance covered = 1233 miles.

Initially,

Speed of the plane = 460 miles per hour

Time taken = 75 minutes

We convert the time in hours as the speed is given in miles per hour.

As, 60 minutes = 1 hour

So, we get, 75 minutes = = = 1.25 hours

Thus, time taken = 1.5 hours

We know,

Rewriting the equation , we have

Thus, the distance travelled initially is given by

Thus, the distance covered by the flight in the first 75 minutes while the speed was 460 miles per hour = 575 miles

Next,

Remaining distance to travel = 1233 – 575 = 658 miles

For the second part of the journey,

Speed of the plane = 439 miles per hour

Distance travelled = 658 miles

Rewriting the above formula, we get

Using the above quantities, we get

Converting the time into minutes, we have,

Thus, the number of minutes for which the plane flies at speed 439 miles per hour = 90 minutes

.

Maths-

### A Cylindrical container 21 m in radius and 72 m high is full of water. The water is emptied into a rectangular tank 63 m long and 24 m wide. Find the height of the water level in the tank?

Hint:

The volume of water remains same in both containers. So, we equate the volume of water in both the containers to get the height of water in rectangular tank.

Explanations:

Step 1 of 2:

Let h be the height of water in the tank.

Volume of water in cylindrical container = m

Volume of water in rectangular tank = m

Step 2 of 2:

Given, volume water in container = volume of water in tank

66 m

Final Answer:

The height of the water level in the tank is 66 m.

The volume of water remains same in both containers. So, we equate the volume of water in both the containers to get the height of water in rectangular tank.

Explanations:

Step 1 of 2:

Let h be the height of water in the tank.

Volume of water in cylindrical container = m

^{3}Volume of water in rectangular tank = m

^{3}Step 2 of 2:

Given, volume water in container = volume of water in tank

66 m

Final Answer:

The height of the water level in the tank is 66 m.

### A Cylindrical container 21 m in radius and 72 m high is full of water. The water is emptied into a rectangular tank 63 m long and 24 m wide. Find the height of the water level in the tank?

Maths-General

Hint:

The volume of water remains same in both containers. So, we equate the volume of water in both the containers to get the height of water in rectangular tank.

Explanations:

Step 1 of 2:

Let h be the height of water in the tank.

Volume of water in cylindrical container = m

Volume of water in rectangular tank = m

Step 2 of 2:

Given, volume water in container = volume of water in tank

66 m

Final Answer:

The height of the water level in the tank is 66 m.

The volume of water remains same in both containers. So, we equate the volume of water in both the containers to get the height of water in rectangular tank.

Explanations:

Step 1 of 2:

Let h be the height of water in the tank.

Volume of water in cylindrical container = m

^{3}Volume of water in rectangular tank = m

^{3}Step 2 of 2:

Given, volume water in container = volume of water in tank

66 m

Final Answer:

The height of the water level in the tank is 66 m.

Maths-

### The base of an isosceles triangle is 24 cm and its area is 192 cm^{2} . Find its perimeter

It is given that base, b = 24 cm and

Area of an isosceles triangle = 192 cm

So, length of two equal sides = 20 cm

Now, perimeter of the triangle = sum of all sides

= 20 + 20 + 24 = 64 cm

Area of an isosceles triangle = 192 cm

^{2}So, length of two equal sides = 20 cm

Now, perimeter of the triangle = sum of all sides

= 20 + 20 + 24 = 64 cm

### The base of an isosceles triangle is 24 cm and its area is 192 cm^{2} . Find its perimeter

Maths-General

It is given that base, b = 24 cm and

Area of an isosceles triangle = 192 cm

So, length of two equal sides = 20 cm

Now, perimeter of the triangle = sum of all sides

= 20 + 20 + 24 = 64 cm

Area of an isosceles triangle = 192 cm

^{2}So, length of two equal sides = 20 cm

Now, perimeter of the triangle = sum of all sides

= 20 + 20 + 24 = 64 cm

Maths-

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

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)

Final Answer:-

∴ x = 5 is the equation of the given line.

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)

Final Answer:-

∴ x = 5 is the equation of the given line.

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

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)

Final Answer:-

∴ x = 5 is the equation of the given line.

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)

Final Answer:-

∴ x = 5 is the equation of the given line.

Maths-

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

Hint:-

Equation of a line in slope point form-

(y-y1) = m × (x - x1)

Step-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)

Step-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).

Maths-General

Hint:-

Equation of a line in slope point form-

(y-y1) = m × (x - x1)

Step-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)

Step-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-

### Write an equation of the line with the slope -1 and y-intercept 10

Hint:-

1. Equation of a line in slope intercept form-

y = mx + b

2. Y-intercept is that point on a line where the line intersects with the y-axis i.e. the point at which x-coordinate is 0.

Step-by-step solution:-

As per given information-

The given line has a slope of -1 and value of y-intercept 10.

∴ m = -1 & b = 10

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

y = mx + b

∴ y = -1 * x + 10

∴ y = -x + 10

Final Answer:-

∴ y = -x + 10 is the equation of the given line

1. Equation of a line in slope intercept form-

y = mx + b

2. Y-intercept is that point on a line where the line intersects with the y-axis i.e. the point at which x-coordinate is 0.

Step-by-step solution:-

As per given information-

The given line has a slope of -1 and value of y-intercept 10.

∴ m = -1 & b = 10

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

y = mx + b

∴ y = -1 * x + 10

∴ y = -x + 10

Final Answer:-

∴ y = -x + 10 is the equation of the given line

### Write an equation of the line with the slope -1 and y-intercept 10

Maths-General

Hint:-

1. Equation of a line in slope intercept form-

y = mx + b

2. Y-intercept is that point on a line where the line intersects with the y-axis i.e. the point at which x-coordinate is 0.

Step-by-step solution:-

As per given information-

The given line has a slope of -1 and value of y-intercept 10.

∴ m = -1 & b = 10

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

y = mx + b

∴ y = -1 * x + 10

∴ y = -x + 10

Final Answer:-

∴ y = -x + 10 is the equation of the given line

1. Equation of a line in slope intercept form-

y = mx + b

2. Y-intercept is that point on a line where the line intersects with the y-axis i.e. the point at which x-coordinate is 0.

Step-by-step solution:-

As per given information-

The given line has a slope of -1 and value of y-intercept 10.

∴ m = -1 & b = 10

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

y = mx + b

∴ y = -1 * x + 10

∴ y = -x + 10

Final Answer:-

∴ y = -x + 10 is the equation of the given line

Maths-

### A line passes through point (3, 1) and has slope = -2/3. Find its equation.

Hint:-

Equation of a line in slope point form-

(y-y1) = m × (x - x1)

Step-by-step solution:-

As per given information-

The given line passes through the point (3,1) and has a slope of -2/3.

∴ x1 = 3, y1 = 1 & m = - 2/3

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

(y-y1) = m × (x -x1)

∴ y - 1 = - 2/3 × (x - 3)

∴ y - 1 = -2/3 x - (- 2/3) × 3

∴ y - 1 = - 2/3 x - (- 2)

∴ y - 1 = -2/3 x + 2

∴ y = - 2/3 x + 2 + 1

∴ y = - 2/3 x + 3

Final Answer:-

∴ y = -2/3 x + 3 is the equation of the given line.

Final Answer:-∴ y = -2/3 x + 3 is the equation of the given line.

Equation of a line in slope point form-

(y-y1) = m × (x - x1)

Step-by-step solution:-

As per given information-

The given line passes through the point (3,1) and has a slope of -2/3.

∴ x1 = 3, y1 = 1 & m = - 2/3

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

(y-y1) = m × (x -x1)

∴ y - 1 = - 2/3 × (x - 3)

∴ y - 1 = -2/3 x - (- 2/3) × 3

∴ y - 1 = - 2/3 x - (- 2)

∴ y - 1 = -2/3 x + 2

∴ y = - 2/3 x + 2 + 1

∴ y = - 2/3 x + 3

Final Answer:-

∴ y = -2/3 x + 3 is the equation of the given line.

Final Answer:-∴ y = -2/3 x + 3 is the equation of the given line.

### A line passes through point (3, 1) and has slope = -2/3. Find its equation.

Maths-General

Hint:-

Equation of a line in slope point form-

(y-y1) = m × (x - x1)

Step-by-step solution:-

As per given information-

The given line passes through the point (3,1) and has a slope of -2/3.

∴ x1 = 3, y1 = 1 & m = - 2/3

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

(y-y1) = m × (x -x1)

∴ y - 1 = - 2/3 × (x - 3)

∴ y - 1 = -2/3 x - (- 2/3) × 3

∴ y - 1 = - 2/3 x - (- 2)

∴ y - 1 = -2/3 x + 2

∴ y = - 2/3 x + 2 + 1

∴ y = - 2/3 x + 3

Final Answer:-

∴ y = -2/3 x + 3 is the equation of the given line.

Final Answer:-∴ y = -2/3 x + 3 is the equation of the given line.

Equation of a line in slope point form-

(y-y1) = m × (x - x1)

Step-by-step solution:-

As per given information-

The given line passes through the point (3,1) and has a slope of -2/3.

∴ x1 = 3, y1 = 1 & m = - 2/3

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

(y-y1) = m × (x -x1)

∴ y - 1 = - 2/3 × (x - 3)

∴ y - 1 = -2/3 x - (- 2/3) × 3

∴ y - 1 = - 2/3 x - (- 2)

∴ y - 1 = -2/3 x + 2

∴ y = - 2/3 x + 2 + 1

∴ y = - 2/3 x + 3

Final Answer:-

∴ y = -2/3 x + 3 is the equation of the given line.

Final Answer:-∴ y = -2/3 x + 3 is the equation of the given line.

Maths-

### A metal pipe has internal radius of 2.5 cm, the pipe is 2mm thick all around. Find the weight in kilogram of 2 meters of the pipe if 1 cubic cm of the metal weighs 7.7 gm.

Hint:

We find the volume of the pipe. Then we multiply it by the given parameter to find the weight and express it in the required unit.

Explanations:

Step 1 of :

Given, inter radius be r = 2.5cm and outer radius be R = (2.5+0.2)= 2.7cm (since 1cm=10mm)

Length be h = 2m = 200 cm (since 1m = 100cm)

Therefore, volume of metal in pipe =

cm

Step 2 of 2:

Given, 1 cm

653.71 cm

Final Answer:

The required weight of pipe is 5.03kg

We find the volume of the pipe. Then we multiply it by the given parameter to find the weight and express it in the required unit.

Explanations:

Step 1 of :

Given, inter radius be r = 2.5cm and outer radius be R = (2.5+0.2)= 2.7cm (since 1cm=10mm)

Length be h = 2m = 200 cm (since 1m = 100cm)

Therefore, volume of metal in pipe =

cm

^{3}Step 2 of 2:

Given, 1 cm

^{3}of the metal weighs 7.7gm653.71 cm

^{3}of the metal in pipe will weigh 653.717.7 = 5033.57gm = 5.03kgFinal Answer:

The required weight of pipe is 5.03kg

### A metal pipe has internal radius of 2.5 cm, the pipe is 2mm thick all around. Find the weight in kilogram of 2 meters of the pipe if 1 cubic cm of the metal weighs 7.7 gm.

Maths-General

Hint:

We find the volume of the pipe. Then we multiply it by the given parameter to find the weight and express it in the required unit.

Explanations:

Step 1 of :

Given, inter radius be r = 2.5cm and outer radius be R = (2.5+0.2)= 2.7cm (since 1cm=10mm)

Length be h = 2m = 200 cm (since 1m = 100cm)

Therefore, volume of metal in pipe =

cm

Step 2 of 2:

Given, 1 cm

653.71 cm

Final Answer:

The required weight of pipe is 5.03kg

We find the volume of the pipe. Then we multiply it by the given parameter to find the weight and express it in the required unit.

Explanations:

Step 1 of :

Given, inter radius be r = 2.5cm and outer radius be R = (2.5+0.2)= 2.7cm (since 1cm=10mm)

Length be h = 2m = 200 cm (since 1m = 100cm)

Therefore, volume of metal in pipe =

cm

^{3}Step 2 of 2:

Given, 1 cm

^{3}of the metal weighs 7.7gm653.71 cm

^{3}of the metal in pipe will weigh 653.717.7 = 5033.57gm = 5.03kgFinal Answer:

The required weight of pipe is 5.03kg

Maths-

### Top of a metallic plate is in the form of a parallelogram. If one of its sides is 8 inches and its altitude is 3 inches. Find the area of the metallic plate

Ans :- 150 sq. inches.

Explanation :-

Given , Base = 8 inches ; height (or ) altitude = 3 inches

Area of parallelogram = base × height = 8 × 3 = 24 sq. inches .

Therefore, the area of the metallic plate is 24 sq. inches .

Explanation :-

Given , Base = 8 inches ; height (or ) altitude = 3 inches

Area of parallelogram = base × height = 8 × 3 = 24 sq. inches .

Therefore, the area of the metallic plate is 24 sq. inches .

### Top of a metallic plate is in the form of a parallelogram. If one of its sides is 8 inches and its altitude is 3 inches. Find the area of the metallic plate

Maths-General

Ans :- 150 sq. inches.

Explanation :-

Given , Base = 8 inches ; height (or ) altitude = 3 inches

Area of parallelogram = base × height = 8 × 3 = 24 sq. inches .

Therefore, the area of the metallic plate is 24 sq. inches .

Explanation :-

Given , Base = 8 inches ; height (or ) altitude = 3 inches

Area of parallelogram = base × height = 8 × 3 = 24 sq. inches .

Therefore, the area of the metallic plate is 24 sq. inches .

Maths-

### Find the equation of the given line.

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. Equation of a line in slope point form-

(y-y1) = m × (x-x1)

Step-by-step solution:-

From the given diagram, we can see that-

rise = increase in the y-coordinates of the given line = 4 units and

run = increase in the x-coordinates of the given line = 2 units

∴ Slope of the given line = m = rise / run

∴ Slope of the given line = m = 4 / 2

∴ Slope of the given line = m = 2

We observe from the given diagram that-

The given line passes through origin (0,0)

i.e. x1 = 0; y1 = 0 and m = 2

Hence, we can use the slope point formula of a line to find the equation of the given line-

(y-y1) = m × (x - x1)

∴ y - 0 = 2 × (x - 0)

∴ y = 2 × x

∴ y = 2x

Final Answer:-

∴ y = 2x is the equation of the given line.

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. Equation of a line in slope point form-

(y-y1) = m × (x-x1)

Step-by-step solution:-

From the given diagram, we can see that-

rise = increase in the y-coordinates of the given line = 4 units and

run = increase in the x-coordinates of the given line = 2 units

∴ Slope of the given line = m = rise / run

∴ Slope of the given line = m = 4 / 2

∴ Slope of the given line = m = 2

We observe from the given diagram that-

The given line passes through origin (0,0)

i.e. x1 = 0; y1 = 0 and m = 2

Hence, we can use the slope point formula of a line to find the equation of the given line-

(y-y1) = m × (x - x1)

∴ y - 0 = 2 × (x - 0)

∴ y = 2 × x

∴ y = 2x

Final Answer:-

∴ y = 2x is the equation of the given line.

### Find the equation of the given line.

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. Equation of a line in slope point form-

(y-y1) = m × (x-x1)

Step-by-step solution:-

From the given diagram, we can see that-

rise = increase in the y-coordinates of the given line = 4 units and

run = increase in the x-coordinates of the given line = 2 units

∴ Slope of the given line = m = rise / run

∴ Slope of the given line = m = 4 / 2

∴ Slope of the given line = m = 2

We observe from the given diagram that-

The given line passes through origin (0,0)

i.e. x1 = 0; y1 = 0 and m = 2

Hence, we can use the slope point formula of a line to find the equation of the given line-

(y-y1) = m × (x - x1)

∴ y - 0 = 2 × (x - 0)

∴ y = 2 × x

∴ y = 2x

Final Answer:-

∴ y = 2x is the equation of the given line.

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. Equation of a line in slope point form-

(y-y1) = m × (x-x1)

Step-by-step solution:-

From the given diagram, we can see that-

rise = increase in the y-coordinates of the given line = 4 units and

run = increase in the x-coordinates of the given line = 2 units

∴ Slope of the given line = m = rise / run

∴ Slope of the given line = m = 4 / 2

∴ Slope of the given line = m = 2

We observe from the given diagram that-

The given line passes through origin (0,0)

i.e. x1 = 0; y1 = 0 and m = 2

Hence, we can use the slope point formula of a line to find the equation of the given line-

(y-y1) = m × (x - x1)

∴ y - 0 = 2 × (x - 0)

∴ y = 2 × x

∴ y = 2x

Final Answer:-

∴ y = 2x is the equation of the given line.

Maths-

### The top view of a swimming pool is in the form of a parallelogram with a base and height of 60 feet and 45 feet respectively. Find the area of a top view of the swimming pool.

Ans :- 150 sq. inches.

Explanation :-

Given , Base = 60 feet ; height = 45 feet

Area of parallelogram = base × height = 60 × 45 = 2,700 sq. feet .

Therefore, the area of the top view of the swimming pool is 2,700 sq. feet.

Explanation :-

Given , Base = 60 feet ; height = 45 feet

Area of parallelogram = base × height = 60 × 45 = 2,700 sq. feet .

Therefore, the area of the top view of the swimming pool is 2,700 sq. feet.

### The top view of a swimming pool is in the form of a parallelogram with a base and height of 60 feet and 45 feet respectively. Find the area of a top view of the swimming pool.

Maths-General

Ans :- 150 sq. inches.

Explanation :-

Given , Base = 60 feet ; height = 45 feet

Area of parallelogram = base × height = 60 × 45 = 2,700 sq. feet .

Therefore, the area of the top view of the swimming pool is 2,700 sq. feet.

Explanation :-

Given , Base = 60 feet ; height = 45 feet

Area of parallelogram = base × height = 60 × 45 = 2,700 sq. feet .

Therefore, the area of the top view of the swimming pool is 2,700 sq. feet.

Maths-

### Calculate the area of equilateral triangle whose height is 25 cm

It is given that height, h = 25 cm

Let AB = x

We know that all angles are equal to 60°.

Sin A =

Sin 60° =

Since all sides are equal BC =

We know that Area of triangle =

=

= 208.33

= 360.84 cm

NOTE – We can use Pythagoras theorem too. Since perpendicular of an equilateral triangle bisects the side, implies that BD = . So, in triangle ABD

AB

Find x and then use Area of triangle =

Let AB = x

We know that all angles are equal to 60°.

Sin A =

Sin 60° =

Since all sides are equal BC =

We know that Area of triangle =

=

= 208.33

= 360.84 cm

^{2}( ∴ = 1.73 )NOTE – We can use Pythagoras theorem too. Since perpendicular of an equilateral triangle bisects the side, implies that BD = . So, in triangle ABD

AB

^{2}= AD^{2}+ BD^{2}Find x and then use Area of triangle =

### Calculate the area of equilateral triangle whose height is 25 cm

Maths-General

It is given that height, h = 25 cm

Let AB = x

We know that all angles are equal to 60°.

Sin A =

Sin 60° =

Since all sides are equal BC =

We know that Area of triangle =

=

= 208.33

= 360.84 cm

NOTE – We can use Pythagoras theorem too. Since perpendicular of an equilateral triangle bisects the side, implies that BD = . So, in triangle ABD

AB

Find x and then use Area of triangle =

Let AB = x

We know that all angles are equal to 60°.

Sin A =

Sin 60° =

Since all sides are equal BC =

We know that Area of triangle =

=

= 208.33

= 360.84 cm

^{2}( ∴ = 1.73 )NOTE – We can use Pythagoras theorem too. Since perpendicular of an equilateral triangle bisects the side, implies that BD = . So, in triangle ABD

AB

^{2}= AD^{2}+ BD^{2}Find x and then use Area of triangle =