 ## Area Perimeter Formula Sheet

Students who are preparing for any competitive exam must know about the Area Perimeter Formula Sheet. In every competitive exam, questions related to Area Perimeter, time and distance formulas are being asked in Quantitative Aptitude section. The candidates who are known to Area Perimeter Formula and shortcut tricks can crack any exam easily. And the aspirants who don’t prepare the formulas will face difficulty in the examination. So for your convenience, we are providing the short tricks, examples and important formulas below.

If you want a good rank in the merit list then you have to do more hard work. For preparation, you can 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. Keep on practicing the formulas and important topics daily so that you will achieve success. Make a time table so that you will complete your preparation before the exams dates and stay in link with www.priatejobshub.in and achieve your goals to the apex.

Shortcut Tricks for Competitive Exam

The main problem in the competitive exams is time limit. Time duration is very less as compared to no. of questions asked in the paper. Try to solve the questions speedily because in recent times many students were not able to complete the question paper on a given time period. Don’t waste your time on any particular question, leave it and move to the other question and in last 10-15 minutes try to attempt that questions.

### Area Perimeter Formula Sheet

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?
• 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: 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 (p + p) × h (p, p 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.

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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.

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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

2). A rectangular lawn of length 50 m and breadth 35 m is to be surrounded externally by a path which is 2 m wide. Find the cost of turfing the path at the rate of \$3 per m². Solution:

Length of the lawn = 50 m
Breadth of the lawn = 35 m
Area of the lawn = (50 × 35) m²
= 1750 m²

Length of lawn including the path = [50 + (2 + 2)] m = 54 cm
Breadth of the lawn including the path = [35 + (2 + 2)] m = 39 m
Area of the lawn including the path = 54 × 39 m² = 2106 m²
Therefore, area of the path = (2106 - 1750) m² = 356 m²
For 1 m², the cost of turfing the path = \$ 3
For 356 m², the cost of turfing the path = \$3 × 356 = \$1068

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/ × base × height
= ¹/ × 12 × 9
= 54 cm²

Want help in Preparation? Get Tips: Practice Which Helps In Preparation

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.

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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 r and r be the outer and inner radii of ring.
Then 2πr = 308
2 × 22/7 r = 308
r = (308 × 7)/(2 × 22)
r = 49 m
2πr = 220
2 × 22/7 × r = 220
r = (220 × 7)/(2 × 22)
r = 35 m
Therefore, width of the track = (49 - 35) m = 14 m

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Final words:

We wish to say all the very best to the students who are going to be appear in the competitive exam. Aspirants don’t get confused among the formulas, read it and understand it. Write the formulas in your notebook for practice and remember the short tricks at the time of examination. Stay confident and do your best in exams.

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