Formula Math Volume of a Cuboid

Smart Math formula Volume of a Cuboid

In geometry, a cuboid is a solid shaped figure formed by six faces. There are two definitions for a cuboid. In the more general definition of a cuboid, the only supplementary condition is that each of these six faces is a quadrilateral. Otherwise, the word “cuboid” is sometimes used for referring a shape of this type in which each of the faces is a rectangle, and in which each pair of adjacent faces meets in a right angle.

 A cuboid with length l units, width w units and height h units has a volume of V cubic units given by: V = l × w × h

•   A jewellery box that has the shape of a rectangular prism, has a height of 13 cm, a length of 35 cm and a width of 22cm. Find the volume of the jewellery box?

Math Solution:
V = l × w × h
V= 13 × 35 × 22 
V= 10010 cm³

•   Find the volume of a brick whose size is 30 cm by 25 cm by 10 cm.

Math Solution:
The volume of the brick is given by:
V = l × w × h
V = 30 × 25 × 10
V = 7500 cm³

So, the volume of the brick is 7500cm³.

•   Calculate the volume of a cuboid whose size is 8cm × 12cm × 6cm

Math Solution:
Volume of the cuboid is given by:
V = l × w × h
V = 8 × 12 × 6
V = 576 cm³

• Given that the dimension of a cuboidal shape beam is 10m in length, 60 cm in width and 25 cm in thickness. How much does Nick have to pay for it, if it costs $250.00/ m³?

Math Solution:
Notice that all the measurements should be expressed in the same units (100 cm = 1 m). Volume of the beam = length x breadth x height (thickness here)
V = 10 x (60/100) x (25/100) = 1.5 m³
The final price of the beam will be: 1.5 x 250 = $375.00.

•   A goods wagon shaped in the form of cuboid of measure 60m × 40m × 30m. How many cuboidal boxes can be stored in it if the volume of one box is 0.8 cubic meters.

Math Solution:
The volume of one box = 0.8 m³
The volume of the goods wagon is: 60 × 40 × 30 = 72000 m³
Number of boxes that can be stored in the goods wagon is therefore:
72000 / 0.8 = 90000
Hence the number of cuboidal boxes that can be stored in the goods wagon is 90,000.

•   Find the height of a cuboid whose volume is 300 cubic cm and the base area is 30cm.

Math Solution:   
V = l × w × h
Since the base area is defined as: l × w, the height is therefore:
h = V / base area
h = 300 / 30
h = 10 cm

•   A box of the size 60 cm × 40 cm × 30 cm. If the size of one chocolate bar is 24 cm × 12 cm × 4 cm each, how many bars can the box hold?

Math Solution:
The volume of the box = 60 × 40 × 30 = 72000 cm³
The volume of each bar = 24 × 12 × 4 = 1152 cm³

Therefore the number of chocolate bars that the box can hold is:
72000/1152 = 62

•  The surface areas of the three coterminous faces of a cuboid are 6, 15 and 10 cm² respectively. Find the volume of the cuboid.

Math Solution:
To begin with we need to determine the length, width and height of the cuboid.
l = √(6)
w = √(15)
h = √(10)

V = l × w × h
V = √(6) × √(15) × √(10)
V = 30 cm²

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Surface Area of a Cuboid Formula Math

The total surface area (TSA) of a cuboid is the sum of the areas of its 6 faces,
which is given by:



TSA = 2 (lw + wh + hl)
Remember the surface area is the total area of all the faces of a 3D shape.
The lateral surface area of a cuboid is given by:
LSA = 2 (lh + wh) = 2 h (l + w)

Smart Math Formula in aplication :
Find the total surface area of a cuboid with dimensions 8 cm by 6 cm by 5 cm.


TSA = 2 (lw + wh + hl)
TSA = 2 (8*6 + 6*5 + 5*8)
TSA = 2 (48 + 30 + 40)
TSA = 236
So, the total surface area of this cuboid is 236 cm2.
Find the surface area of a cuboid of dimensions 4.8 cm, 3.4 cm and 7.2 cm.


Math Solution:
Area of Face 1: 4.8 × 7.2 = 34.56 cm²
Area of Face 2: 3.4 × 7.2 = 24.48 cm²
Area of Face 3: 4.8 × 3.4 = 16.32 cm²

Adding the area of these 3 faces gives 75.36 cm², since each face is duplicated on the opposite side, the total surface area of the cuboid will be:
TSA = 2 (75.36) = 150.72 cm²
The length, width and height of a cuboid are 10cm, 8cm and 7cm respectively. Find the lateral surface area of a cuboid.


Math Solution:
Lateral surface area of cuboid is given by:
LSA = 2h(l+w)
Note :
l = length = 10 cm
w = width = 8 cm
h = height = 7 cm

Insert these values into the formula we will get:
LSA = 2 ×7(10 + 8)
LSA = 14 × 18
LSA = 252 cm2

The length, breadth and height of a cuboid are 16cm, 14cm and 10cm respectively. Find the total surface area of the cuboid.

Math Solution:
The total surface area of a cuboid is given by:
TSA = 2 (l*b + b*h + h*l)

 Given that:
l = 16cm
b = 14cm
h = 10cm

Substituting the values in the equation we will get
TSA = 2 (16*4 + 14*10 + 10*16)
TSA = 2(224 + 140 + 160)
TSA = 2 * 524
TSA = 1048 cm2

Given a cereal box whose length is 20 cm, height is 30 cm and width is 8 cm. Find the surface area of the box.

Math Solution:
To find the surface are of the box we need to find the area of each rectangular face and add them all up.
The area of the front face is: 20 x 30 = 600 cm2.
The area of the top face is: 20 x 8 = 160 cm2.
The area of the side face is: 8 x 30 = 240 cm2.

Now add these values together we will get: 600 + 160 + 240 = 1000 cm2.

And the total surface area is therefore 1000 x 2 = 2000 cm2.

Find the surface area of a cuboid whose sides are 3cm by 6cm by 10cm.

Math Solution:
Surface area of the cuboid is given by:
TSA = 2 (16*4 + 14*10 + 10*16)
TSA = 2(3 x 6 + 6 x 10 + 3 x 10)
TSA = 2(18 + 60 + 30)
TSA = 216 cm2

That is smart math formula in aplication of Surface Area of a Cuboid. Formula Surface Area of a Cuboid Math

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Formula Math Of Cylinder

Mathematics formula math of Cylinder :

Definition of A cylinder as a solid figure that is bound by a curved surface and two flat surfaces. The surface area of a cylinder can be found by breaking it down into 2 parts:

1.  The two circles that make up the caps of the cylinder.
2.  The side of the cylinder, which when "unrolled" is a rectangle.

 Formula Math :
The area of each end cap can be found from the radius r of the circle, which is given by:
A = πr²

Thus the total area of the caps is 2πr².

The area of a rectangle is given by:
A = height × width

The width is the height h of the cylinder, and the length is the distance around the end circles, or in other words the perimeter/circumference of the base/top circle and is given by:
P = 2πr

Thus the rectangle's area is rewritten as:
A = 2πr × h

Combining these parts together we will have the total surface area of a cylinder, and the final formula is given by:

A = 2πr² + 2πrh

Smart Math Formula Note :
π  =  Pi, approximately 3.142,  r  = the radius of the cylinder, h  = height of the cylinder

By factoring 2πr from each term we can simplify the formula to:

A = 2πr(r + h)

The lateral surface area of a cylinder is simply given by: LSA = 2πr × h.
Smart Math Formula  in the application :

• Find the surface area of a cylinder with a radius of 4 cm, and a height of 3 cm.

Smart Solution:
SA = 2 × π × r² + 2 × π × r × h
SA = 2 × 3.14 × 4² +  2 × 3.14 × 4 × 3
SA = 6.28 × 16 + 6.28 × 12
SA = 100.48 + 75.36
SA = 175.84
Surface Area = 175.84 cm²

•  Find the surface area of the cylinder with a radius of 5.5cm and height of 10cm.

Smart Solution:
The radius of cylinder = 5.5 cm.
The height of cylinder = 10 cm.
The total surface area of the cylinder is therefore:
TSA = 2πr(r+h)
TSA = 11π (5.5+10)
TSA = 170.5 π
TSA = 535.6 cm²

•   Find the total surface area of a cylindrical tin of radius 17 cm and height 3 cm.

Smart Solution:
Again as in the previous example:
TSA = 2πr(r+h)
TSA = 2π× 17(17+3)
TSA = 2π×17×20
TSA = 2136.56 cm²

•  Find the surface area of the cylinder with radius of 6 cm and height of 9 cm.

Smart Solution:
The radius of cylinder: r = 6 cm
The height of cylinder: h = 9 cm
Total surface area of cylinder is therefore:
TSA = 2πr(r + h)
TSA = 12π (6+9)
TSA = 180 π
TSA = 565.56 cm²

• Find the radius of cylinder whose lateral surface area is 150 cm² and its height is 9 cm.

Smart Solution:
Lateral surface area of cylinder is given by:
LSA = 2πrh
Given that:
LSA = 150cm²
h = 9cm
π is the constant and its value = 3.14

Substitute the values in the formula and find the value of r by isolating it from the equation:
LSA = 2πrh
150 = 2× π × r × 9
r = 150 / (2×9× π)
r = 2.65cm
So the radius of the cylinder is 2.65 cm.

That is  Smart Math aplication from Cylinder Formula Math.

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Easy Math of Linear (Equations and Inequalities) One Variable

Math of Linear (Equations and Inequalities) One Variable
Completion of the set of graphs of linear equations of the variables shown in a number line, in the form of dot (point).
example
Determine the completion of the set of equations
4 (2x + 3) = 10x + 8, if x
variables on the set of integers. Then, draw the number line
completion:
4 (2x + 3) = 10x + 8
8x + 12 = 10x + 8
8x + 12 - 12 = 10x + 8-12 (both sides minus 12)
8x = 10x - 4
8x - 10x = 10x - 4 - 10x (10x minus both sides)
-2x = -4
-2x: (-2) = -4 (-2) (both sides are divided by -2)
 x = 2
Thus, the solution set is {2}.

Easy Math of Linear Inequalities One Variable
In everyday life, surely you've come across or find sentences like the following.
a. Rebbecca weigh more than 52 kg.
b. Josh height 7 cm less than my height.
c. One of the requirements to be members of the Army is his height not less than 165 cm.
d. A bus can carry no more than 55 people.
How sentences are expressed in the form of mathematical sentence? To be able to learn to answer the following description.

1. understanding inequality
In order for you to understand the sense of inequality, try to remember back in elementary school about the matter in writing notation <,>,<,> , and not same.
a. 3 less than 5 written 3 <5.
b. 8 more than 4 written 8> 4.
c. x no more than 9 written x < 9.
d. Two times y is not less than 16 written 2y > 16.
Sentences 3 <5, 8> 4, x < 9, and  are called the inequality 2y >16 .
In general it can be written as follows.
An inequality is always marked by one of the following hyphen.
"<" For less than stated.
">" To declare over.
"<" to represent no more than or less than or equal to.
">" to represent not less than or more than or equal to.

2. Linear Inequalities One Variable
On the front you have learned that an equation is always marked with a hyphen "=". In this section you will learn the characteristics of an inequality.
example
Of the following forms, which specify a linear inequality with one variable.
a. x - 3 <5
b. A < 1 - 2b
c. x2 - 3x > 4
completion:
a. x - 3 <5
Inequality x - 3 <5 has one variable, namely x and rank 1, so x - 3 <5 is a variable linear inequality.
b. A < 1 - 2b
Inequality a < 1 - 2b has two variables, namely a and b, each of which rank 1.
Thus a < 1 - 2b is not a single variable linear inequalities.
c. x2 - 3x > 4
Due to the inequality x2 - 3x > 4 has variable x and x2, then x2 - 3x > 4 is not a single variable linear inequalities in Easy Math of Linear.

 

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