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 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.

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.

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

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
= 

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

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.

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

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.

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)

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

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 = 
= 

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 = 
= 

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

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)

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