Area:Â The amount of space occupied by an object two dimensionally is termed as its area.
For the given figure, red coloured portion shows area for different objects.
Perimeter:Â The distance around the boundary of the figure is termed as its perimeter.
For the given figure, the boundary of the square will be its perimeter.
Area and Perimeter of common shapes:
Shape 
Figure  Area  Perimeter 
Rectangle 
Length x Breadth 
2 (Length + Breadth) 

Square  (Side)^{2} 
4 x Length 

Triangle 
Â½ Base x Height 
a + b + c, where a, b and c are three sides of triangle 

Parallelogram 
Â 
Base x Height 
2 (a + b), where a and b are two different sidesÂ of parallelogram 
Circle 
Â 
Ï€ x (Radius)2 
2 x Ï€ x radius 
Â
Examples based on Area and Perimeter:
Example 1: Find the area of square whose side is 50m.
Solution:Â The area of square = (Side)^{ 2
}Given, side = 50 m.
then, area of square = (50)^{2} = 2500 m^{2}.
Example 2: Find the perimeter of a triangle with two sides of 15 cm and one side of 20 cm.
Solution:Â We know that, perimeter of triangle = sum of sides
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = (15 + 15 + 20) = 50 cm.
Example 3: A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m^{2}? (If required you can split the tiles in whatever way you want to fill up the corners).
Solution:Â Given, base = 24 cm, height = 10 cm.
Â Â Â Â Â Â Â Â Â Â Â Â area of parallelogram = base x height
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 24 x 10 = 240 cm^{2}.
Â Now, we know that 1 m = 100 cm. Therefore, 1080 m^{2 }= 1080 x 100 x 100 cm^{2}.
Â Number of tiles required = (Total area) / (area of one tile)
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = (1080 x 100 x 100)/(240) = 45000.
Â Thus, 45000 tiles will be required to cover a floor of area 1080 m^{2} using a tile of 240 cm^{2}.
Example 4: An ant is moving around a few food pieces of different shapes scattered on the floor. For which foodpiece would the ant have to take a longer round? Remember, circumference of a circle can be obtained by using the expression c = 2Ï€r, where r is the radius of the circle.
Solution:Â For the given problem, the figure for which the perimeter will be maximum will be the one for which ant will take longer round. So, let us find the perimeter for each figure.
(a) From the figure, diameter = 2.8 cm. Therefore, radius = 2.8 / 2 = 1.4 cm.
Now, perimeter of semicircle is Ï€ r = Ï€ x 1.4 cm = 4.4 cm.
On combining, total perimeter = (perimeter of one side) + (perimeter of semicircle)
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 2.8 cm + 4.4 cm = 7.3 cm
(b) Again, perimeter of semicircle is Ï€ r = Ï€ x 1.4 cm = 4.4 cm
Total perimeter = (perimeter of three sides) + (perimeter of semicircle)
Â Â Â Â Â Â Â Â Â Â Â Â Â = (1.5 cm + 2.8 cm + 1.5 cm) + 4.4 cm = 10.2 cm
(c) Again, perimeter of semicircle is Ï€ r = Ï€ x 1.4 cm = 4.4 cm
Total perimeter = (perimeter of two sides) + (perimeter of semicircle)
Â Â Â Â Â Â Â Â Â Â Â Â Â = (2 cm + 2 cm) + 4.4 cm = 8.4 cm
We can see that the perimeter for figure (b) is largest amongst all, hence, ant will take longer round to travel for (b) in comparison with the other two.
Now we will learn about surface area and volume of solids such as cube, cuboid andÂ cylinder.
Area:
1. Area of trapezium:
For any given trapezium, if a and b are two parallel sides and h is perpendicular distance between them (as shown in figure). Then, its area is given by
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Area of trapezium = h x (a +b)/2
In general, area of trapezium = height x (sum of parallel sides)/2.
Example 1: Calculate the area for the following given trapezium.
Solution:Â As per the formula, Area of trapezium = height x (sum of parallel sides)/2
Here, height = 4 cm and two sides are 5 cm & 10 cm. Substituting these values, we get,
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Area of trapezium = 4 x (5 + 10)/2 = 30 cm^{2}.
Example 2: The area of a trapezium is 20 cm^{2} and the length of one of the parallel sides is 8 cm and its height is 2 cm. Find the length of the other parallel side.
Solution:Â Given, area = 20 cm, length of one side = a = 8 cm, height = 2 cm.
Now, area of trapezium = height x (a + b)/2
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 20 = 2 x (8 + b)/2
Â On solving, Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â b = 12 cm.
Thus, the length of other parallel side for the given trapezium is 12 cm.
2. Area of Quadrilateral:
Any quadrilateral can be divided into two triangles by drawing one of the diagonals.
Then applying and calculating area of triangle formula on individual triangles, we obtain the area of quadrilateral.Consider the quadrilateral ABCD as shown above. On drawing diagonal AC, we have two triangles ABC and ACD.
So, area of quadrilateral ABCD = area of Î” ABC + area of Î” ACD
As per the area of triangle formula, we can write,
Area of quadrilateral ABCD = (1/2 x AC x h_{1}) + (1/2 x AC x h_{2})
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 1/2 x AC x (h_{1 }+ h_{2})
Suppose the length of diagonal AC is d, then,
Area of quadrilateral ABCD = 1/2 x d x (h_{1 }+ h_{2})
Example 1: Find the area of some quadrilateral whose diagonal length is 8 cm. And the lengths of two perpendiculars on given diagonal from the vertices are 4 cm and 3 cm.
Solution:Â Here, d = 8 cm. Let the length of two perpendiculars be h_{1} = 4 cm and h_{2} = 3 cm.
As per the formula of quadrilateral, we can write,
Area of quadrilateral ABCD = 1/2 x d x (h_{1 }+ h_{2})
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 1/2 x 8 x (4 + 3) = 28 cm^{2}.
Example 2: Find the area of quadrilateral ABCD as shown in figure below.Solution:Â Given, d = 5 cm, h_{1} = 2cm, h_{2 }= 1 cm.
Now, Area of quadrilateral ABCD = 1/2 x d x (h_{1 }+ h_{2})
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 1/2 x 5 x (2 + 1) = 7.5 cm^{2}
3. Area of Rhombus:
In rhombus, the diagonals are perpendicular bisectors of each other. Thus, total rhombus will be divided into two equal triangles.Area of rhombus ABCD = area of Î” ABC + area of Î” ACD
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = (1/2 x AC x OD) + (1/2 x AC x OB)
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = (1/2 x AC x BD)
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = (1/2 x d_{1} x d_{2})
In general, area of rhombus is half the product of its diagonals.
Example: Find the area of a rhombus in which the length of its two diagonals is 4cm and 6cm.
Solution:Â Area of rhombus ABCD = (1/2 x d_{1} x d_{2})
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 1/2 x 4 x 6 = 12 cm^{2}.
4. Area of a Polygon:
For any polygon, try to divide the polygon into possible number of triangles, quadrilaterals, trapeziums, rhombus, etc. and then find area of each of them. At last, add all the calculated individual area which will give the area of polygon.
For example: Consider the polygon ABCDE given in the figure below.Now, we draw a diagonal AD and two perpendiculars BF and CG on it. Thus, we divide the polygon ABCDE into four parts.
Finally, area of ABCDE = area of Î”AFB + area of trapezium BCGF + area of Î”CDG + area of Î”ADE.
Example: Find the area of octagon shown in figure below.Solution:Â The given octagon can be divided into three parts as shown in figure below:
From figure, area of part A & C will be same which resembles a trapezium. And part B is a rectangle.
Area of part A (Trapezium) = h x (sum of two parallel sides)/2
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 2 x (10 + 3)/2 = 13 cm^{2}.
Area of part B (Rectangle) = length x breadth
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 10 x 3 = 30 cm^{2}.
Thus, area of Octagon = 2 x area of part A + area of part B
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 2 x 13 + 30 = 56 cm^{2}.Â Â Â Â
Solid Shapes:
The three dimensional objects having width, depth and height are known as solid shapes.
Example: Cube, cuboid, cylinder, etc.
1. Surface area of Cuboid:
A cuboid is made of six parts out of which four parts are same and remaining two parts are same.
From figure, we can write surface area of cuboid as sum of area of six individual parts.
Thus, area of cuboid = l x h + b x h + l x b + b x h + l x h + l x b
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 2 (lb + bh + hl)
Example 1: Find total surface area of cuboid having length, breadth and height as 10cm, 5cm and 2cm, respectively.
Solution:Â Given, l = 10cm, b = 5cm, h = 2cm
We know that, area of cuboid = 2 (lb + bh + hl)
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 2 (10x5 + 5x2 + 2x10) = 160 cm^{2}.
Example 2: The internal measures of a cuboidal room are 20 m Ã— 10 m Ã— 6 m. Find the total cost of whitewashing all walls of a room including ceiling, if the cost of white washing is Rs 10 per m^{2} .
Solution:Â Given, length = 20m, breadth = 10m, h = 6 m.
Surface area of room = 2 (lb + bh + hl)
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 2 (20 x 10 + 10 x 6 + 6 x 20) = 760 m^{2}.
Cost of white washing per m^{2} = Rs 10
Hence, total cost of white washing entire room = 760 x 10 = 7600 Rs.
2. Surface area of Cube:
A cube is made of six equal parts.
We know that area of square = (side)^{ 2}. In cube, there are six equal square sides.
Thus, area of cube = 6 x a^{2}; where a is the length of any side.
Example 1: Find total surface area of cube having its one length as 2cm.
Solution:Â Here, a = 2cm (Given)
We know that, area of cube = 6 x a^{2
}Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 6 x 2^{2} = 24 cm^{2}.
Example 2: Find the side of a cube whose surface area is 864 cm^{2}.
Solution:Â Given, surface area of cube = 864 cm^{2}.Â Let â€˜aâ€™ be the length of the cube.
We know that, area of cube = 6 x a^{2
}Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 864 = 6 x a^{2
}Thus, Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â a = 12 cm.
3. Surface area of Cylinder:From the above figure, we can see that cylinder is made of two equal circles and a rectangle in circular form.
So, we can write, area of cylinder = area of circular base + area of curved rectangle + area of circular top
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = Ï€r^{2} + 2Ï€rh + Ï€r^{2
}Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 2Ï€r (r + h)
Example 1: Find the total surface area of cylinder given below.Solution:Â Here, r = 7cm and h = 4cm. (Given)
We know that, area of cylinder = 2Ï€r (r + h)
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 2 x Ï€ x 7 (7 + 4) = 484 cm^{2}.
Example 2: Find the height of a cylinder whose radius is 7 cm and the total surface area is 968 cm^{2} .
Solution:Â Given, r = 7cm, area = 968 cm^{2}.
We know that, area of cylinder = 2Ï€r (r + h)
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â 968 = 2 x Ï€ x 7 (7 + h)
On solving, Â Â Â Â Â Â Â Â Â Â Â Â Â Â h = 15 cm.
Example 3: A company packages its milk powder in cylindrical container whose base has a diameter of 20 cm and height 30 cm. Company places a label around the surface of the container (as shown in the figure). If the label is placed 2 cm from top and bottom, what is the area of the label?Solution:Â Here, the height of label will be (304) = 26 cm. And the radius of the label will be 20/10 = 10 cm.
Now, area of label = 2Ï€rh
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 2 x Ï€ x 10 x 26 = 1634.28 cm^{2}.
Volume:
The amount of space occupied by an object three dimensionally is termed as its volume.
1. Volume of Cuboid:The volume of cuboid = length x width/breadth x height = lbh.
Example 1: Find volume of cuboid having length, breadth and height as 10cm, 5cm and 2cm, respectively.
Solution:Â Here, l = 5cm, b = 7cm, h = 2cm (Given).
We know that, volume of cuboid = l x b x h
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 5 x 7 x 2 = 70 cm^{3}.
Example 2: A godown is in the form of a cuboid of measures 50 m Ã— 40 m Ã— 30 m. How many cuboidal boxes can be stored in it if the volume of one box is 10 m^{3}?
Solution:Â Given, volume of one box = 10 m^{3
}Volume of godown = 50 Ã— 40 Ã— 30
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 60000 m^{3
}Thus, number of boxes that can be stored in the godown = 60000 / 10
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 6000.
2. Volume of Cube:In cube, all the sides are equal.
So, the volume of cube = length x length x length = l^{3}.
Example 1: Find volume of cube having its one length as 5cm.
Solution:Â Here, a = 5cm (Given)
We know that, volume of cube = l^{3
}Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = 5^{3} = 125 cm^{3}.
Example 2: If each edge of a cube is doubled, then how many times will its volume increase?
Solution:Â Let the length of cube be â€˜lâ€™. Thus, double length will be 2l.
For length l, volume of cube = l^{3}.
For length 2l, volume of cube = (2l)^{3} = 8l^{3}.
Thus, on doubling the length of cube, its volume will increase by 8 times.
3. Volume of Cylinder:The volume of cylinder = Ï€r^{2}h.
Example 1: Find the volume of cylinder having height of 10cm and radius of 4cm.
Solution:Â Here, r = 4cm and h = 10cm. (Given)
We know that, volume of cylinder = Ï€r^{2}h
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = Ï€ x (4)^{2 }x 10 = 503 cm^{3}.
Example 2: A rectangular piece of paper 10 cm Ã— 5 cm is folded without overlapping to make a cylinder of height 5 cm. Find the volume of the cylinder.
Solution:Â Here, the length of the paper will become the perimeter of the base of the cylinder and width becomes height.
Now, perimeter of base of cylinder = 2Ï€r = 10
Thus, r = 1.6 cm.Â Now, volume of cylinder = Ï€r^{2}h
Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = Ï€ x 1.6 x 1.6 x 5 = 40.23 cm^{3}.
Volume and Capacity:
The difference between volume and capacity are as under:
(a) Volume â€“ it refers to the amount of space occupied by an object.
(b) Capacity â€“ it refers to the quantity that an object holds.
Note: If a water tin holds 100 cm^{3} of water then the capacity of the water tin is 100 cm^{3}. Capacity is also measured in terms of litres. The relation between litre and cm^{3Â }is,
1 mL = 1 cm^{3}
1 L = 1000 cm^{3}
Thus, Â Â Â 1 m^{3Â }= 1000000 cm^{3Â }= 1000 L.
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Helpful
Area of 4walls can be called as lsa(lateral surface area) or csa(curved surface area)
Amazing notes
Lateral Surface area
Thanx for notes but where is lateral surface area?
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