Question

# If the base of a triangle becomes three times its height. What is the new area of a triangle?

Hint:

### Use formula Area of the triangle =

## The correct answer is: 3/2 h2

### Let Base = b and Height = h

It is given that base = 3 height

⇒ b = 3h

Area of the triangle =

=

=h^{2}

### Related Questions to study

### The side of an equilateral triangle is 16 units. What will be the double of its area ?

Area of an equilateral triangle = a

^{2}= 16

^{2}

= 64 cm

^{2}

Double of the area = 2 64

= 128 = 221.7 cm

^{2}

### The side of an equilateral triangle is 16 units. What will be the double of its area ?

Area of an equilateral triangle = a

^{2}= 16

^{2}

= 64 cm

^{2}

Double of the area = 2 64

= 128 = 221.7 cm

^{2}

### In triangle ABC, AB = 8cm. If the altitudes corresponding to AB and BC are 4 cm and 5 cm respectively. Find the measure of BC.

Area of the triangle with altitude corresponding to AB

=

= = 16 cm

^{2}

With Base = BC , Height, h = AE = 5 cm

Area of the triangle with altitude corresponding to BC is

= 16 cm

^{2}

= 16

BC = 6.4 cm

### In triangle ABC, AB = 8cm. If the altitudes corresponding to AB and BC are 4 cm and 5 cm respectively. Find the measure of BC.

Area of the triangle with altitude corresponding to AB

=

= = 16 cm

^{2}

With Base = BC , Height, h = AE = 5 cm

Area of the triangle with altitude corresponding to BC is

= 16 cm

^{2}

= 16

BC = 6.4 cm

### The base and corresponding altitude of a parallelogram are 18 cm and 6 cm respectively. Find its area

Area of the parallelogram = b h

= 18 6 = 108 cm

^{2}

### The base and corresponding altitude of a parallelogram are 18 cm and 6 cm respectively. Find its area

Area of the parallelogram = b h

= 18 6 = 108 cm

^{2}

### The ratio of the bases of two triangles is a : b. If the ratio of their corresponding altitudes is c : d, find the ratio of their areas (in the same order).

⇒ Bases of the triangles = ax , bx

Similarly, it is given that ratio of altitudes is c : d

⇒ Altitudes of the triangle = cy , dy

Area of first triangle =

=

Area of second triangle =

Ratio of the areas of two triangle

=

=

Hence, ratio of area of two triangles is ac : bd

### The ratio of the bases of two triangles is a : b. If the ratio of their corresponding altitudes is c : d, find the ratio of their areas (in the same order).

⇒ Bases of the triangles = ax , bx

Similarly, it is given that ratio of altitudes is c : d

⇒ Altitudes of the triangle = cy , dy

Area of first triangle =

=

Area of second triangle =

Ratio of the areas of two triangle

=

=

Hence, ratio of area of two triangles is ac : bd

### The sides of triangle are 11 cm, 15 cm and 16 cm. What is the measure of altitude to the largest side?

⇒ Length of the side a = 3x , b = 4x and c = 5 x

Perimeter of the triangle = 144 m

3x + 4x + 5x = 144

12x = 144 ⇒ x = 12

Now Using Pythagoras theorem,

(5x)^{2} = (3x)^{2} + (4x)^{2}

25x^{2} = 9x^{2} + 16x^{2}

25x^{2} = 25x^{2} i.e. Pythagoras holds true

⇒ Given triangle is a right angled triangle

Base, b = 3x =3(12) = 36 m and Height, h = 4x =4(12) = 48 m

⇒ Area of the triangle =

### The sides of triangle are 11 cm, 15 cm and 16 cm. What is the measure of altitude to the largest side?

⇒ Length of the side a = 3x , b = 4x and c = 5 x

Perimeter of the triangle = 144 m

3x + 4x + 5x = 144

12x = 144 ⇒ x = 12

Now Using Pythagoras theorem,

(5x)^{2} = (3x)^{2} + (4x)^{2}

25x^{2} = 9x^{2} + 16x^{2}

25x^{2} = 25x^{2} i.e. Pythagoras holds true

⇒ Given triangle is a right angled triangle

Base, b = 3x =3(12) = 36 m and Height, h = 4x =4(12) = 48 m

⇒ Area of the triangle =

### The sides of triangle are 11 cm, 15 cm and 16 cm. What is the measure of altitude to the largest side?

Using Heron’s formula

Area of triangle = where s =

s = = 21

Area of triangle=

Since we have to find altitude to the largest side, base of the triangle = 16 cm

Also, area of triangle =

⇒

⇒ ( = 2.64)

Using Heron’s formula

Area of triangle = where s =

s = = 21

Area of triangle=

Since we have to find altitude to the largest side, base of the triangle = 16 cm

Also, area of triangle =

⇒

⇒ ( = 2.64)

### When a solid metal cube is completely submerged in a cylindrical vessel Containing milk with 30 cm diameter the level of milk rises by cm. Find the length of the edge of the metal cube.

We use principle of Archimedes to find the length of the cube.

Explanations:

Step 1 of 1:

Let the length of the edge of the metal cube be a.

Volume of cube =

*a*

^{3}

Given, r = 30/2 cm and h =

By the principle of Archimedes,

Volume of risen milk = volume of cube

, discarding the negative value since distance cannot be negative.

Final Answer:

The length of the edge of the cube is 10cm.

### When a solid metal cube is completely submerged in a cylindrical vessel Containing milk with 30 cm diameter the level of milk rises by cm. Find the length of the edge of the metal cube.

We use principle of Archimedes to find the length of the cube.

Explanations:

Step 1 of 1:

Let the length of the edge of the metal cube be a.

Volume of cube =

*a*

^{3}

Given, r = 30/2 cm and h =

By the principle of Archimedes,

Volume of risen milk = volume of cube

, discarding the negative value since distance cannot be negative.

Final Answer:

The length of the edge of the cube is 10cm.

### The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. Find the radius and volume of the cylinder.

We plug in the values in formulae and solve the problem.

Explanations:

Step 1 of 2:

Let the radius of vessel base be

We have, cm

Height h = 25cm

Step 2 of 2:

Volume of the vessel =

Final Answer:

The radius is 21cm and volume of the cylinder is 34650cm

^{3}.

### The circumference of the base of a cylindrical vessel is 132 cm and its height is 25 cm. Find the radius and volume of the cylinder.

We plug in the values in formulae and solve the problem.

Explanations:

Step 1 of 2:

Let the radius of vessel base be

We have, cm

Height h = 25cm

Step 2 of 2:

Volume of the vessel =

Final Answer:

The radius is 21cm and volume of the cylinder is 34650cm

^{3}.

### The volume of a metallic cylindrical pipe is 748 cubic.cm Its length is 14 cm., and its external radius is 9 cm. Find its thickness?

We find the internal radius and subtract it from external radius to get the thickness.

Explanations:

Step 1 of 1:

Let the internal radius of the pipe be r .

Given, external radius R = 9cm

Length = height of the pipe h = 14cm

We have volume of pipe = 748

Final Answer:

The thickness of the pipe is 8 cm.

### The volume of a metallic cylindrical pipe is 748 cubic.cm Its length is 14 cm., and its external radius is 9 cm. Find its thickness?

We find the internal radius and subtract it from external radius to get the thickness.

Explanations:

Step 1 of 1:

Let the internal radius of the pipe be r .

Given, external radius R = 9cm

Length = height of the pipe h = 14cm

We have volume of pipe = 748

Final Answer:

The thickness of the pipe is 8 cm.

### The difference between the outer and inner curved surface areas of a 14 cm long cylinder is 88 sq.cm Find the outer and inner radii of the cylinder given that the volume of the metal is 176 cubic cm

Forming the equations based on the given information, we will get two equations. Then we will find the radii by solving those two equations.

Explanations:

Step 1 of 3:

Given, outer CSA – inner CSA = 88

, where = length of cylinder, R = outer radius, r = inner radius

…(i)

Step 2 of 3:

Also, given volume with R – volume with r = 176

…(ii)

Step 3 of 3:

Adding (i) and (ii), we get

Putting R = 5/2 in equation (ii), we get

Final Answer:

The outer and inner radii of the cylinder are 2.5 cm and 1.5 cm

### The difference between the outer and inner curved surface areas of a 14 cm long cylinder is 88 sq.cm Find the outer and inner radii of the cylinder given that the volume of the metal is 176 cubic cm

Forming the equations based on the given information, we will get two equations. Then we will find the radii by solving those two equations.

Explanations:

Step 1 of 3:

Given, outer CSA – inner CSA = 88

, where = length of cylinder, R = outer radius, r = inner radius

…(i)

Step 2 of 3:

Also, given volume with R – volume with r = 176

…(ii)

Step 3 of 3:

Adding (i) and (ii), we get

Putting R = 5/2 in equation (ii), we get

Final Answer:

The outer and inner radii of the cylinder are 2.5 cm and 1.5 cm

### The sides of a triangle are 16 cm, 30 cm and 34 cm. What is its area ?

Using Heron’s formula

Area of triangle = where s =

s = = 40

Area of triangle

=

^{2}

### The sides of a triangle are 16 cm, 30 cm and 34 cm. What is its area ?

Using Heron’s formula

Area of triangle = where s =

s = = 40

Area of triangle

=

^{2}

### The base of an isosceles right angled triangle is 30 cm. Find its area.

We know that in a right isosceles triangle, base and height are equal and hypotenuse is the third side.

So, Area of the triangle =

= 450 cm

^{2}

### The base of an isosceles right angled triangle is 30 cm. Find its area.

We know that in a right isosceles triangle, base and height are equal and hypotenuse is the third side.

So, Area of the triangle =

= 450 cm

^{2}

### Driver A charges $25 fees and $0.10 per extra mile driven. Driver B charges no fees but charges $0.60 per mile driven. Find the Application distance in miles for which the two drivers will cost the same money. Also, find the cost.

1. Extra charges = number of miles driven × rate per mile

2. Total cost = Constant fees + Extra charges.

Step-by-step solution:-

Let x be the number of miles driven and y be the total fees charged.

We know that-

Extra charges = Number of miles driven × Rate per mile

∴ Extra charges = x * rate per mile ..................................................... (Equation i)

Now, for Driver A-

Total fees charged = y = Constant fees + Extra charges

∴ Total fees charged = y = Constant fees + x × rate per mile ................... (From Equation i)

∴ Total fees charged = y = 25 + x * 0.10 .................................................... (From given information)

∴ Total fees charged = y = 25 + 0.10 x ....................................................... (Equation ii)

For Driver B-

Total fees charged = y = Constant fees + Extra charges

∴ Total fees charged = y = Constant fees + x × rate per mile ................... (From Equation i)

∴ Total fees charged = y = 0 + x × 0.60 .................................................... (From given information)

∴ Total fees charged = y = 0.60 x ....................................................... (Equation iii)

Since we need to find the distance for which cost of both drivers is the same-

Equating Equations ii & iii, we get-

25 + 0.10 x = 0.60 x

∴ 25 = 0.60 x - 0.10 x

∴ 25 = 0.50 x

∴ 25 / 0.50 = x ...................................................................................... (Dividing both sides by 0.50)

∴ 50 = x

For a distance of 50 miles, the cost charged by both the drivers will be the sam.

Now, at 50 miles i.e. x = 50,

y = 0.60 x ….............................................. (Using Equation iii)

∴ y = 0.60 × 50

∴ y = 30

Note:-

Alternatively, we can find the total cost by using Equation ii,

y = 25 + 0.10 x

∴ y = 25 + 0.10 × 50

∴ y = 25 + 5

∴ y = 30.

Hence, the answer remains the same.

### Driver A charges $25 fees and $0.10 per extra mile driven. Driver B charges no fees but charges $0.60 per mile driven. Find the Application distance in miles for which the two drivers will cost the same money. Also, find the cost.

1. Extra charges = number of miles driven × rate per mile

2. Total cost = Constant fees + Extra charges.

Step-by-step solution:-

Let x be the number of miles driven and y be the total fees charged.

We know that-

Extra charges = Number of miles driven × Rate per mile

∴ Extra charges = x * rate per mile ..................................................... (Equation i)

Now, for Driver A-

Total fees charged = y = Constant fees + Extra charges

∴ Total fees charged = y = Constant fees + x × rate per mile ................... (From Equation i)

∴ Total fees charged = y = 25 + x * 0.10 .................................................... (From given information)

∴ Total fees charged = y = 25 + 0.10 x ....................................................... (Equation ii)

For Driver B-

Total fees charged = y = Constant fees + Extra charges

∴ Total fees charged = y = Constant fees + x × rate per mile ................... (From Equation i)

∴ Total fees charged = y = 0 + x × 0.60 .................................................... (From given information)

∴ Total fees charged = y = 0.60 x ....................................................... (Equation iii)

Since we need to find the distance for which cost of both drivers is the same-

Equating Equations ii & iii, we get-

25 + 0.10 x = 0.60 x

∴ 25 = 0.60 x - 0.10 x

∴ 25 = 0.50 x

∴ 25 / 0.50 = x ...................................................................................... (Dividing both sides by 0.50)

∴ 50 = x

For a distance of 50 miles, the cost charged by both the drivers will be the sam.

Now, at 50 miles i.e. x = 50,

y = 0.60 x ….............................................. (Using Equation iii)

∴ y = 0.60 × 50

∴ y = 30

Note:-

Alternatively, we can find the total cost by using Equation ii,

y = 25 + 0.10 x

∴ y = 25 + 0.10 × 50

∴ y = 25 + 5

∴ y = 30.

Hence, the answer remains the same.

### The largest triangle is inscribed in a semi-circle of radius 4 cm. Find the area inside the semi-circle which is not occupied by the triangle.

Base of the triangle = 2r = 8 cm

Height of the triangle = r = 4 cm

Area of the triangle =

= 16 cm

^{2}

Area of semi-circle = = (3.14) 4

^{2}

= 25.12 cm

^{2}

Area not occupied by the triangle

= Area of semi-circle – Area of triangle

= 25.12 – 16 = 9.12 cm

^{2}

### The largest triangle is inscribed in a semi-circle of radius 4 cm. Find the area inside the semi-circle which is not occupied by the triangle.

Base of the triangle = 2r = 8 cm

Height of the triangle = r = 4 cm

Area of the triangle =

= 16 cm

^{2}

Area of semi-circle = = (3.14) 4

^{2}

= 25.12 cm

^{2}

Area not occupied by the triangle

= Area of semi-circle – Area of triangle

= 25.12 – 16 = 9.12 cm

^{2}

### A rectangular paper of 22 cm by 16 cm size can be exactly wrapped to cover the curved surface of a cylinder of 16 cm height. Find the volume of the cylinder?

Observe that the paper is wrapped by its length. So, the circumference of the cylinder is equal to the length of the paper. We use this information to get the volume.

Explanations:

Step 1 of 1:

Let the base radius of the cylinder be r .

We have

cm

Given, h = 16cm

Volume of the cylinder

cm

^{3}

Final Answer:

The volume of the cylinder is 616cm

^{3}.

### A rectangular paper of 22 cm by 16 cm size can be exactly wrapped to cover the curved surface of a cylinder of 16 cm height. Find the volume of the cylinder?

Observe that the paper is wrapped by its length. So, the circumference of the cylinder is equal to the length of the paper. We use this information to get the volume.

Explanations:

Step 1 of 1:

Let the base radius of the cylinder be r .

We have

cm

Given, h = 16cm

Volume of the cylinder

cm

^{3}

Final Answer:

The volume of the cylinder is 616cm

^{3}.