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__Area Perimeter
Formula Sheet__

__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.
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!!**Here we are providing the complete details for solving the questions based on Are Perimeter Formula-**

__Shortcut Tips & Tricks:__

**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 Perimeter?__

**Check this out-**

__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 (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

**: Dear participants to get confirm success in the exam in such questions keep preparing with above given content.**

__Reminder____Take a Look on Below Table__

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