- Perimeter: A perimeter is a path that surrounds an area. The term may be used either for the path or its length – it can be thought of as the length of the outline of a shape. The perimeter of a circular area is called its circumference. It is measured in cm, m, etc.
- Area: Area is a quantity that expresses the extent of a two-dimensional surface or shape in the plane. Area can be understood as the amount of material with a given thickness that would be necessary to fashion a model of the shape, or the amount of paint necessary to cover the surface with a single coat. This is measured in square unit like cm², m², etc.
- Volume: Volume is the amount of space enclosed by a shape or object, how much 3-dimensional space (length, width, and height) it occupies. This is measured in cubic unit like cm³, m³, etc.
Units related to area
- 1 Hactare = 10000 metre square
- 1 kilometre square = 1000000 metre square
- 1 Decametre = 100 metre square
- 1 Decimeter square = 1/100 metre square
- 1 Centimeter square = 1/10000 metre square
- 1 Milimeter square = 1/000000 metre square
Units related to volume
- 1 litre = 1000 cm³
- 1 Hectometer³ = 10000 meter³
- 1 Decameter³ = 100 meter³
- 1 Meter³ = 1000000 cm³
- 1 Decimeter³ = 100 cm³
- 1 Milimeter³ = 1/100 cm³
Perimeter (P) = a + b + c
and a, b and c are three sides of the triangle.
Also, ; where b is base and h is altitude
Perimeter = 3a
Area ; where a is a side
Area and h² = p² + b² (Pythagoras triplet)
where p is perpendicular, b is base and h is hypotenuse
Example 1: The base of a triangular field is 880 m and its height 550 m. Find the area of the field. Also calculate the charges for supplying water to the field at the rate of Rs. 24.25 per sq. hectometer.
Solution: Area of the field
sq.m. = 24.20 sq.hm
Cost of supplying water to 1 sq. hm = Rs. 24.25
∴ Cost of supplying water to the whole field
= 24.20 × 24.25 = Rs. 586.85
Example 2: Find the area of a triangle whose sides are 50 m, 78 m, 112 m respectively and also find the perpendicular from the opposite angle on the side 112 m.
Solution: Here a = 50 m, b = 78 m, c = 112m
s – a = 120 – 50 = 70 m
s – b = 120 – 78 = 42 m
s – c = 120 – 112 = 8 m
∴ Area = 1680 sq.m.
Area = Length × breadth
Perimeter = 2(Length + breadth)
Example 3: The length and breadth of a rectangle are in the ratio 9 : 5. If its area is 720 m², find its perimeter.
Solution: Let the length and breadth of a rectangle are 9x m and 5x m respectively.
In a rectangle, area = length × breadth
∴ 720 = 9x × 5x
or x² = 16
⇒ x = 4
Thus, length = 9 × 4 = 36 m
and breadth = 5 × 4 = 20 m
Therefore, perimeter of rectangle = 2(36 + 20) = 112 m
Area = (side) ²
Perimeter = 4 × side
Diagonal = × side
Example 4: How many squares are there in a 5 inch by 5 inch square grid, if the grid is made up one inch by one inch squares?
Solution: Required no. of squares
Area = base × height
Perimeter = 2 × (Side A + Side B)
Perimeter = 4 a
Area where a is side and d¹ and d²are diagonals.
Example 5: The perimeter of a rhombus is 146 cm and one of its diagonals is 55 cm. Find the other diagonal and the area of the rhombus.
Solution: Let ABCD be the rhombus in which AC = 55 cm.
∴ BO = 24 cm
Hence, the other diagonal BD = 48 cm
Now, Area of the rhombus
= 1320 sq. cm.
Perimeter = AD + DC + BC + AB
Area = ½ × (DP + BQ) × AC
Example 6: Find the area of a quadrilateral piece of ground, one of whose diagonals is 60 m long and the perpendicular from the other two vertices are 38 and 22 m respectively.
Perimeter = a + b + m + n
Area ;where a and b are two parallel sides;
m and n are two non-parallel sides; h is perpendicular to b
Example 6: A 5100 sq. cm trapezium has the perpendicular distance between the two parallel sides 60 m. If one of the parallel sides be 40m then find the length of the other parallel side.
⇒ 5100 ⇒ 170 = 40 + x
∴ other parallel side = 170 – 40 = 130 m
Area = π × (Radus) ²
Circumference = 2π × Radius
π or 3.14
Example 7: If area of a circular jogging track is 3850 sq. metres. What is the circumference of the jogging track?
Solution: Let the radius of the circular jogging track be r metre.
or r² =
∴ r = = 35 metre
∴ Circumference = = 220 metre
Example 8: The radius of a wheel is 42 cm. How many revolutions will it make in going 26.4 km?
Solution: Distance travelled in one revolution = Circumference of the wheel
= 2πr = = 264 cm
∴ No. of revolutions required to travel 26.4 km
= = 10000
Perimeter = πr + 2r Area
Sector of a Circle
Area of sector
Length of an arc (ℓ) =
Area of segment = Area of sector – Area of triangle OAB
Perimeter of segment
= length of the arc + length of segment AB
Example 9: Find the area of sector of a circle whose radius is 6 cm when:
- the angle at the centre is 35°
- when the length of arc is 22 cm
1. Area of sector
2. Here length of arc ℓ = 22 cm
Area of sector
Area of ring
Example 10: The circumference of a circular garden is 1012 m. Find the area of outsider road of 3.5 m width runs around it. Calculate the area of this road and find the cost of gravelling the road at Rs 32 per 100 sqm.
Solution: A = πr², C = 2πr = 1012
∴ Area of garden
Area of the road = area of bigger circle – area of the garden
Now, radius of bigger circle
= 161 + 3.5 m
∴ Area of bigger circle
Thus, area of the road
= Rs. 1145.76
Three Dimensional Figures
A cuboid is a three dimensional box.
If L, B and H are length, breadth and height of the cuboid, then Volume = L × B × H
Surface area = 2 (L × B + B × H + H + L)
A cube is a cuboid which has all its edges equal.
If a is the each side of cube, then
Volume = a × a × a = a³
Whole surface area = 2(a × a + a × a + a × a) = 6a²
Diagonal of cube
A prism is a solid which can have any polygon at both its ends.
Lateral or curved surface area = Perimeter of base × height
Total surface area = Lateral surface area + 2 (area of the end)
Volume = Area of base × height
Right Circular Cylinder
It is a solid which has both its ends in the form of a circle.
If radius of cylinder is r and height or length is h, then
Volume = πr²h
Lateral surface Area = 2πrh
Whole surface area = (2πrh + 2πr²)
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Example 11: A road roller of diameter 1.75 m and length 1 m has to press a ground of area 1100 sqm. How many revolutions does it make?
Solution: Area covered in one revolution = curved surface area
∴ Number of revolutions
A pyramid is a solid which can have any polygon at its base and its edges converge to single apex.
Lateral or curved surface area= 1/2 (perimeter of base) × slant height
Total surface area = lateral surface area + area of the base
Volume = 1/3 (area of the base) × height
Example 12: A right pyramid, 12 cm high, has a square base each side of which is 10 cm. Find the volume of the pyramid.
Solution: Area of the base = 10 × 10 = 100 sq.cm.
Height = 12 cm
∴ Volume of the pyramid
= 400 cu.cm.
It is a solid in the form of a ball with radius r.
If r is the radius of sphere, then volume =
Surface Area = 4πr²
It is a solid half of the sphere.
Surface area = 2πr²
Total surface area = 2πr² + πr² = 3πr²
Right Circular Cone
It is a solid which has a circle as its base and a slanting lateral surface that converges at the apex.
If base-radius, height and vertical height of a cone is r, h and ℓ respectively, then
Lateral surface area = πrl
Total surface area = πrl + πr²
Vertical height = ℓ =
Example 13: A frustum of a right circular cone has a diameter of base 10 cm, top of 6 cm, and a height of 5 cm; find the area of its whole surface and volume.
Solution: Here r¹ = 5 cm, r² = 3 cm and h = 5 cm.
∴ cm = 5.385 cm
∴ Whole surface of the frustum
= 242.25 sq.cm.
Volume = = 256.67 cu. cm.
Frustum of a Cone
When a cone cut the left over part is called the frustum of the cone.
Curved surface area = πℓ (r₁ + r₂)
Total surface area
Example 14: The height of a bucket is 45 cm. The radii of the two circular ends are 28 cm and 7 cm, respectively. The volume of the bucket is :
- 38610 cm³
- 48600 cm³
- 48510 cm³
- None of these
Solution : (3): Here r¹ = 7 cm, r² = 28 cm and h = 45 cm
Volume of the bucket
Hence, the required volume
= 48510 cm³
Area of Pathways
Running across the middle of a rectangle
A = a (ℓ + b) – a²; where ℓ → length
b → breadth,a →width of the pathway.
A = (l + 2a) (b + 2a) – lb;where l → length
b → breadtha → width of the pathway
A = lb – (l – 2a) (b – 2a) ; where l → length
b → breadth
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