Area Perimeter Formula Sheet - Shortcut Tips & Tricks


Area Perimeter Formula Sheet

To get superb marks in the entire competitive exams you have superb grip in the section known as Quantitative Aptitude. The questions
based on Area Perimeter Formula Sheet are also comes from this section. Almost every exam possesses questions related to Area Perimeter, time and distance formulas to check the mental / mathematical skills of the appliers in the Quantitative Aptitude section specially. Such job grabbers who are known to Area Perimeter Formula and shortcut tricks they have more probability to crack any exam easily. The main thing is preparation of the questions related to the formulas to solve difficulty in the examination.

 Area Perimeter Formula Sheet

To list your name in the declared cut off after competitive exam it is necessary for aspirants   to do more hard work as per requirement of the time. Applicants are advised to buy the previous year question papers and sample papers from the books shop or you can download it from the official website without any charges to do better then the best preparation to beat a exam. Also remind all the formulas and Shortcut Tips & Tricks daily so that you will achieve success. Make a time table so that you will complete your preparation before the exams dates. So for your convenience, we are providing the short tricks, examples and important formulas below. More details in favour of Area Perimeter Formula Sheet are declared below for all the visitors of this article. Best of luck to candidates for success in the exam!!

Shortcut Tips & Tricks: Here we are providing the complete details for solving the questions based on Are Perimeter Formula-

What is Perimeter? The length of the boundary of a closed figure is called the perimeter of the plane figure. The units of perimeter are same as that of length, i.e., m, cm, mm, etc.


What is Area? Check this out-
A part of the plane enclosed by a simple closed figure is called a plane region and the measurement of plane region enclosed is called its area.
Area is measured in square units.
The units of area and the relation between them are given below:
Area Perimeter Formula Sheet


Different geometrical shapes formula of area and perimeter

Perimeter and Area of Rectangle:
The perimeter of rectangle = 2(l + b).
Area of rectangle = l × b; (l and b are the length and breadth of rectangle)
Diagonal of rectangle = √(l² + b²)
Perimeter and Area of the Square
Perimeter of square = 4 × S.
Area of square = S × S.
Diagonal of square = S√2; (S is the side of square)


Perimeter and Area of the Triangle:
Perimeter of triangle = (a + b + c); (a, b, c are 3 sides of a triangle)
Area of triangle = √(s(s - a) (s - b) (s - c)); (s is the semi-perimeter of triangle)
S = 1/2 (a + b + c)
Area of triangle = 1/2 × b × h; (b base, h height)
Area of an equilateral triangle = (a²√3)/4; (a is the side of triangle)

Perimeter and Area of the Parallelogram:
Perimeter of parallelogram = 2 (sum of adjacent sides)
Area of parallelogram = base × height
Perimeter and Area of the Rhombus
Area of rhombus = base × height
Area of rhombus = 1/2 × length of one diagonal × length of other diagonal
Perimeter of rhombus = 4 × side


Perimeter and Area of the Trapezium:
Area of trapezium = 1/2 (sum of parallel sides) × (perpendicular distance between them)
                                = 1/2 (p₁ + p₂) × h (p₁, p₂ are 2 parallel sides)

Circumference and Area of Circle

Circumference of circle = 2πr
                                       = πd
                       Where, π = 3.14 or π = 22/7
                                     r is the radius of circle
                                     d is the diameter of circle
Area of circle = πr²
Area of ring = Area of outer circle - Area of inner circle.


Illustrations

Perimeter and Area of rectangle

1). Find the perimeter and area of the rectangle of length 17 cm and breadth 13 cm.
Solution:
Given: length = 17 cm, breadth = 13 cm
Perimeter of rectangle = 2 (length + breadth)
                                    = 2 (17 + 13) cm
                                    = 2 × 30 cm 
                                    = 60 cm
We know that the area of rectangle = length × breadth
                                                         = (17 × 13) cm2
                                                        = 221 cm2

2). Find the breadth of the rectangular plot of land whose area is 660 m2 and whose length is 33 m. Find its perimeter.
Solution:
We know that the breadth of the rectangular plot = Area/length
                                                                         = 660m233m660m233m
                                                                          = 20 m
Therefore, the perimeter of the rectangular plot = 2 (length + breadth)
                                                                             = 2(33 + 20) m
                                                                             = 2 × 53 m
                                                                             = 106 m


Perimeter and Area of Square

1). Find the perimeter and area of a square of side 11 cm.
Solution:
We know that the perimeter of square = 4 × side
Side= 11 cm
Therefore, perimeter = 4 × 11 cm = 44 cm
Now, area of the square = (side × side) sq. units
                                        = 11 × 11 cm²
                                        = 121 cm²

2). The perimeter of a square is 52 m. Find the area of the square.
Solution:
Perimeter of square = 52 m
But perimeter of square = 4 × side
Therefore, 4 × side = 52 m
Therefore, side= 52/4 m = 13m
Now, the area of the square = (side × side)
Therefore, area of the square = 13 × 13 m² = 169 m².


Area of Path
1). A painting is painted on a cardboard 19 cm and 14 cm wide, such that there is a margin of 1.5 cm along each of its sides. Find the total area of the margin.
Solution:
Length of the cardboard = 19 cm
Breadth of the cardboard = 14 cm
Area of the cardboard = 19 × 14 cm² = 266 cm²
Length of the painting excluding the margin = [19 - (1.5 + 1.5)] cm = 16 cm
Breadth of the painting excluding the margin = 14 - (1.5 + 1.5) = 11 cm
Area of the painting excluding the margin = (16 × 11) cm² = 176 cm²
Therefore, area of the margin = (266 - 176) cm² = 90 cm²

Area and Perimeter of the Triangle

1). Find the area and height of an equilateral triangle of side 12 cm. (√3 = 1.73).
Solution:
Area of the triangle = √3 / 4 a² square units
= √3 / 4 × 12 × 12
= 36√3 cm²
= 36 × 1.732 cm²
= 62.28 cm²

Height of the triangle = √3 / 2 a units
= √3 / 2 × 12 cm
= 1.73 × 6 cm
= 10.38 cm

2). Find the area of right angled triangle whose hypotenuse is 15 cm and one of the sides is 12 cm.
Solution:
AB² = AC² - BC²
       = 15² - 12²
       = 225 - 144    
        = 81
Therefore, AB = 9
Therefore, area of the triangle = ¹/₂ × base × height
                                                 = ¹/₂ × 12 × 9
                                                 = 54 cm²

Area and Perimeter of the Parallelogram


1). The base of the parallelogram is thrice its height. If the area is 192 cm², find the base and height.
Solution:
Let the height of the parallelogram = x cm
then the base of the parallelogram = 3x cm
Area of the parallelogram = 192 cm²
Area of parallelogram = base × height
192 = 3x × x
3x² = 192
x² = 64
x = 8
Therefore, 3x = 3 × 8 = 24
Therefore, Base of the parallelogram is 24 cm and height is 8 cm.

2). A parallelogram has sides 12 cm and 9 cm. If the distance between its shorter sides is 8 cm, find the distance between its longer side.
Solution:
Adjacent sides of parallelogram = 2 cm and 9 cm
Distance between shorter sides = 8 cm
Area of parallelogram = b × h
                                   = 9 × 8 cm²
                                   = 72 cm²
Again, area of parallelogram = b × h
72 = 12 × h
h = 72/12
h = 6 cm
Therefore, the distance between its longer side = 6 cm.


Circumference and Area of Circle

1). Find the circumference and area of radius 7 cm.
Solution:
Circumference of circle = 2πr
                                     = 2 × 22/7 × 7
                                     = 44 cm
Area of circle = πr²
                      = 22/7 × 7 × 7 cm²
                      = 154 cm²

2). A race track is in the form of a ring whose inner circumference is 220 m and outer circumference is 308 m. Find the width of the track.
Solution:
Let r₁ and r₂ be the outer and inner radii of ring.
Then 2πr₁ = 308
2 × 22/7 r₁ = 308
r₁ = (308 × 7)/(2 × 22)
r₁ = 49 m
2πr₂ = 220
2 × 22/7 × r₂ = 220
r₂ = (220 × 7)/(2 × 22)
r₂ = 35 m
Therefore, width of the track = (49 - 35) m = 14 m
                                               
Reminder: Dear participants to get confirm success in the exam in such questions keep preparing with above given content.

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