Maths-
General
Easy
Question
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.
Hint:
We use principle of Archimedes to find the length of the cube.
The correct answer is: The length of the edge of the cube is 10cm.
Explanations:
Step 1 of 1:
Let the length of the edge of the metal cube be a.
Volume of cube = a3
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.
Related Questions to study
Maths-
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.
Hint:
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 34650cm3.
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,
Height h = 25cm
Step 2 of 2:
Volume of the vessel =
Final Answer:
The radius is 21cm and volume of the cylinder is 34650cm3.
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.
Maths-General
Hint:
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 34650cm3.
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,
Height h = 25cm
Step 2 of 2:
Volume of the vessel =
Final Answer:
The radius is 21cm and volume of the cylinder is 34650cm3.
Maths-
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?
Hint:
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.
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?
Maths-General
Hint:
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.
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.
Maths-
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
Hint:
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
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
Step 2 of 3:
Also, given volume with R – volume with r = 176
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
Maths-General
Hint:
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
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
Step 2 of 3:
Also, given volume with R – volume with r = 176
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
Maths-
The sides of a triangle are 16 cm, 30 cm and 34 cm. What is its area ?
It is given that a = 16 cm , b = 30 cm and c = 34 cm
Using Heron’s formula
Area of triangle =
where s = 
s =
= 40
Area of triangle
=
Using Heron’s formula
Area of triangle =
s =
Area of triangle
=
=
= 240 cm2
The sides of a triangle are 16 cm, 30 cm and 34 cm. What is its area ?
Maths-General
It is given that a = 16 cm , b = 30 cm and c = 34 cm
Using Heron’s formula
Area of triangle =
where s = 
s =
= 40
Area of triangle
=
Using Heron’s formula
Area of triangle =
s =
Area of triangle
=
=
= 240 cm2
Maths-
The base of an isosceles right angled triangle is 30 cm. Find its area.
It is given that Base of the triangle = 30 cm
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 cm2
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 cm2
The base of an isosceles right angled triangle is 30 cm. Find its area.
Maths-General
It is given that Base of the triangle = 30 cm
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 cm2
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 cm2
Maths-
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.
Hint:-
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.
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.
Maths-General
Hint:-
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.
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.
Maths-
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.
It is given that radius of the triangle = 4 cm
Base of the triangle = 2r = 8 cm
Height of the triangle = r = 4 cm
Area of the triangle =
= 16 cm2
Area of semi-circle =
= (3.14) 42
= 25.12 cm2
Area not occupied by the triangle
= Area of semi-circle – Area of triangle
= 25.12 – 16 = 9.12 cm2
Base of the triangle = 2r = 8 cm
Height of the triangle = r = 4 cm
Area of the triangle =
= 16 cm2
Area of semi-circle =
= 25.12 cm2
Area not occupied by the triangle
= Area of semi-circle – Area of triangle
= 25.12 – 16 = 9.12 cm2
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.
Maths-General
It is given that radius of the triangle = 4 cm
Base of the triangle = 2r = 8 cm
Height of the triangle = r = 4 cm
Area of the triangle =
= 16 cm2
Area of semi-circle =
= (3.14) 42
= 25.12 cm2
Area not occupied by the triangle
= Area of semi-circle – Area of triangle
= 25.12 – 16 = 9.12 cm2
Base of the triangle = 2r = 8 cm
Height of the triangle = r = 4 cm
Area of the triangle =
= 16 cm2
Area of semi-circle =
= 25.12 cm2
Area not occupied by the triangle
= Area of semi-circle – Area of triangle
= 25.12 – 16 = 9.12 cm2
Maths-
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?
Hint:
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


cm3
Final Answer:
The volume of the cylinder is 616cm3.
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
Given, h = 16cm
Volume of the cylinder
Final Answer:
The volume of the cylinder is 616cm3.
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?
Maths-General
Hint:
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


cm3
Final Answer:
The volume of the cylinder is 616cm3.
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
Given, h = 16cm
Volume of the cylinder
Final Answer:
The volume of the cylinder is 616cm3.
Maths-
Sides of triangle are 100 m, 120 m and 140 m. Find its area
It is given that a = 100 , b = 120 and c = 140
Using Heron’s formula
Area of triangle =
where s = 
s =
= 180
Area of triangle
=
=
= 100
= 2400 cm2
= 5878.76 cm2 (
= 2.44)
Using Heron’s formula
Area of triangle =
s =
Area of triangle
=
=
= 100
= 2400 cm2
= 5878.76 cm2 (
Sides of triangle are 100 m, 120 m and 140 m. Find its area
Maths-General
It is given that a = 100 , b = 120 and c = 140
Using Heron’s formula
Area of triangle =
where s = 
s =
= 180
Area of triangle
=
=
= 100
= 2400 cm2
= 5878.76 cm2 (
= 2.44)
Using Heron’s formula
Area of triangle =
s =
Area of triangle
=
=
= 100
= 2400 cm2
= 5878.76 cm2 (
Maths-
Find the equation of the given line . Application
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
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
Find the equation of the given line . Application
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
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
Maths-
Find the area of largest triangle that can be inscribed in a semi-circle of radius 9 cm.
It is given that radius of the circle = 9 cm
⇒ Diameter = 2
r = 2
9 = 18 cm
As we can see in the picture
Base of the triangle = diameter = 18 cm
Height of the triangle = radius = 9 cm
Area of the triangle =
=
⇒ Diameter = 2
As we can see in the picture
Base of the triangle = diameter = 18 cm
Height of the triangle = radius = 9 cm
Area of the triangle =
=
Find the area of largest triangle that can be inscribed in a semi-circle of radius 9 cm.
Maths-General
It is given that radius of the circle = 9 cm
⇒ Diameter = 2
r = 2
9 = 18 cm
As we can see in the picture
Base of the triangle = diameter = 18 cm
Height of the triangle = radius = 9 cm
Area of the triangle =
=
⇒ Diameter = 2
As we can see in the picture
Base of the triangle = diameter = 18 cm
Height of the triangle = radius = 9 cm
Area of the triangle =
=
Maths-
An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of triangle.
Let length of the third side = b
It is given that perimeter of triangle = 30 cm
12 + 12 + x = 30
x = 30 – 24 = 6 cm
Area of an isosceles triangle =
where a is length of equal side and b is third side
So, a = 12 and b = 6 cm
Area of an isosceles triangle =
=
It is given that perimeter of triangle = 30 cm
12 + 12 + x = 30
x = 30 – 24 = 6 cm
Area of an isosceles triangle =
So, a = 12 and b = 6 cm
Area of an isosceles triangle =
=
= 34.86 cm2
An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of triangle.
Maths-General
Let length of the third side = b
It is given that perimeter of triangle = 30 cm
12 + 12 + x = 30
x = 30 – 24 = 6 cm
Area of an isosceles triangle =
where a is length of equal side and b is third side
So, a = 12 and b = 6 cm
Area of an isosceles triangle =
=
It is given that perimeter of triangle = 30 cm
12 + 12 + x = 30
x = 30 – 24 = 6 cm
Area of an isosceles triangle =
So, a = 12 and b = 6 cm
Area of an isosceles triangle =
=
= 34.86 cm2
Maths-
A line passes through point (3, 1) and has slope = -2/3. Find its equation .Application
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
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
A line passes through point (3, 1) and has slope = -2/3. Find its equation .Application
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
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
Maths-
Find the area of a triangle, two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.
Let third side be x
It is given that Perimeter of triangle = 42 cm
18 + 10 + x = 42
x = 42 – 28 = 14
So, length of third side is 14 cm
Using Heron’s formula,
Area of triangle =
where s = 
s =
= 21
Area of triangle =
=
= 21
cm2
(
= 3.31)
It is given that Perimeter of triangle = 42 cm
18 + 10 + x = 42
x = 42 – 28 = 14
So, length of third side is 14 cm
Using Heron’s formula,
Area of triangle =
s =
Area of triangle =
=
= 21
(
= 69.65 cm2
Find the area of a triangle, two sides of which are 18 cm and 10 cm and the perimeter is 42 cm.
Maths-General
Let third side be x
It is given that Perimeter of triangle = 42 cm
18 + 10 + x = 42
x = 42 – 28 = 14
So, length of third side is 14 cm
Using Heron’s formula,
Area of triangle =
where s = 
s =
= 21
Area of triangle =
=
= 21
cm2
(
= 3.31)
It is given that Perimeter of triangle = 42 cm
18 + 10 + x = 42
x = 42 – 28 = 14
So, length of third side is 14 cm
Using Heron’s formula,
Area of triangle =
s =
Area of triangle =
=
= 21
(
= 69.65 cm2
Maths-
An isosceles right angled triangle has area 8 cm2 . Find the length of its hypotenuse.
We know that base and height have equal length in right angled isosceles triangle.
Let base = height = x
It is given that Area of the triangle = 8 cm2
= 8
x2 = 8
x2 = 16 ⇒ x = 4 cm
So, base = 4 cm = height
Using Pythagoras theorem,
Hypotenuse2 = Base2 + Height2 = 42 + 42 = 2
82
Hypotenuse = 4
cm
(
= 1.41)
= 5.64 cm
Let base = height = x
It is given that Area of the triangle = 8 cm2
x2 = 16 ⇒ x = 4 cm
So, base = 4 cm = height
Using Pythagoras theorem,
Hypotenuse2 = Base2 + Height2 = 42 + 42 = 2
Hypotenuse = 4
(
= 5.64 cm
An isosceles right angled triangle has area 8 cm2 . Find the length of its hypotenuse.
Maths-General
We know that base and height have equal length in right angled isosceles triangle.
Let base = height = x
It is given that Area of the triangle = 8 cm2
= 8
x2 = 8
x2 = 16 ⇒ x = 4 cm
So, base = 4 cm = height
Using Pythagoras theorem,
Hypotenuse2 = Base2 + Height2 = 42 + 42 = 2
82
Hypotenuse = 4
cm
(
= 1.41)
= 5.64 cm
Let base = height = x
It is given that Area of the triangle = 8 cm2
x2 = 16 ⇒ x = 4 cm
So, base = 4 cm = height
Using Pythagoras theorem,
Hypotenuse2 = Base2 + Height2 = 42 + 42 = 2
Hypotenuse = 4
(
= 5.64 cm