PERIMETER AND AREA

On the webpage "perimeter and area" we are going to see formula for each and every shapes in math.

DIFFERENCE BETWEEN PERIMETER AND AREA

 The continuous line forming the boundary of a closed geometrical figure is known as perimeter. The region occupied by the closed figure is known as area.  Measuring unit of area

Square unit.(sq) Measures two dimensions.

Example : 36 in² or 36 inches squared

Measuring unit of perimeter

Linear unit measures one dimension.

Example : 36 in. or 36 inches

Square Area of square = a²

Perimeter of square = 4a

here "a" stands for side length of square.

Rectangle Area of rectangle = L x W

Perimeter of rectangle = 2(L + w)

Here L and w represents length and width of rectangle.

Triangle Area of Equilateral triangle = (√3/4) a²

Perimeter of Equilateral triangle = 3a

Here "a" represents side length of triangle.

Scalane triangle Area of scalene triangle ﻿

Area of scalene triangle = √s(s-a)(s-b)(s-c)

Perimeter of scalene triangle = a + b +  c

here "s = (a + b + c)/2" Area of Quadrilateral﻿

Area of quadrilateral =(1/2) x d x (h₁+h₂)

Perimeter of quadrilateral = Sum of all four sides

Parallelogram Area of Parallelogram﻿

Area of parallelogram = b x h

Perimeter of quadrilateral = Sum of all four side

Rhombus Area of rhombus ﻿

Area of rhombus =(1/2) x (d₁ x d₂)

Trapezium

Area of trapezoid =(1/2) (a + b) x h

Circle Area of circle = Π r ²

Circumference of circle = 2 Π r

Semi circle Area of semicircle﻿

Area of Semi circle= (1/2) Π r²

Perimeter of semi circle = (Π+2)r Area of quadrant

Area of quadrant = (1/4) Π r²

Perimeter of quadrant = [(Π/2) + 2] r

Sector Area of the sector = (θ/360) x Π r ² square units

(or)  Area of the sector = (1/2) x l r square units

Length of arc = (θ/360) x 2Πr

Perimeter of sector = L + 2r

Example problems to find area and perimeter

Problem 1:

Find the area of the square whose diagonal is measuring 4 cm.

Solution: The diagonal AC divides the square into two right triangles.Δ ACB and Δ ADC. In triangle ACB right angles at B.

So the side which is opposite to this side is called the hypotenuse side. Since it is square the length of 4 sides will be equal.

By using Pythagorean theorem

AC² = AB² + BC²

4² = x² + x²

16 = 2x²

2x² = 16

x² = 16/2

x² = 8

x = √8

x = √2 x 2 x 2

x = 2√2 cm

Therefore the length of all sides = 2√2 cm

Area of square = a²

= (2√2)²

= 2²(√2)²

= 4 (2)

= 8 cm²

Now we are going to the next example of the topic "Perimeter and area".

Problem 2:

A square is of area 64 cm². What is its perimeter?

Solution:

Area of a square = 64 cm²

a² = 64 cm²

a = √ 64

a = √8 x 8

a = 8 cm

Now we have to find the perimeter

Perimeter of the square = 4a

= 4 (8)

= 32 cm

Therefore perimeter of the square is 32 cm

Now we are going to the next example of the topic "Perimeter and area".

Problem 3:

The length and breadth of the rectangle are in the ratio 5:2 .If the area of the rectangle is measuring 147 cm². Then find the length and breadth of the rectangle.

Solution:

Let 5x and 2x be the length and breadth of the rectangle respectively.

Area of the rectangle = 147 cm²

5x x 2x = 147

7x² = 147

x² = 147/7

x² = 21

x = √21

Length = 5x = 5√21 cm

Breadth = 2x = 2√21 cm

Now we are going to the next example of the topic "Perimeter and area".

Problem 4:

The perimeter of a rectangle is 50 cm. The length is 15 cm. What is the area?

Solution:

Perimeter of the rectangle = 50 cm

2 (L + B) = 50

L + B = 50/2

L + B = 25

Here the length is 15 cm

15 + B = 25

B = 25 - 15

B = 10 cm

Therefore the required breadth is 10 cm

Now we are going to the next example of the topic "Perimeter and area".

Problem 5:

Find the area of rhombus if the diagonal is measuring 15 cm and 10 cm.

Solution:

Here d₁ = 15 cm d₂ = 10 cm

Area of the rhombus = (1/2) x (d₁ x d₂)

= (1/2) x 15 x 10

= 15 x  5

= 75 cm²

Now we are going to the next example of the topic "Perimeter and area".

Problem 6:

In a trapezium the measurement of one parallel side two more than the other parallel side and the height is 4 cm. The area of the trapezium is 64 cm². Find the two parallel sides.

Solution:

Area of the trapezium = 64 cm²

Let the two parallel sides be "a" and "b"

One parallel side is two more than the other parallel side. So we can assume b as a + 2

Area of a trapezium = (1/2) (a + b) x h

(1/2) (a + b) x h  =  64

(1/2) (a + a + 2) x 4 = 64

(1/2) (2a + 2) x 4 = 64

2 (2a + 2) = 64

2a + 2 = 64/2

2a + 2 = 32

2a = 32 - 2

2a = 30

a = 30/2

a = 15 cm

b = a + 2

b = 15 + 2

b = 17 cm

Therefore the two parallel sides are 15 cm and 17 cm.

Now we are going to the next example of the topic "Perimeter and area".

Problem 7:

Find the area of the sector whose radius is 35 cm and perimeter is 147 cm.

Solution: Perimeter of sector = 147 cm

L + 2r  =  147 cm

Here r = 35 cm

L + 2 (35) = 147

L + 70 = 147

L = 147 - 70

L = 77 cm

Now we have the length of an arc and radius. So we have to use the second formula to find the area of the given sector.

Area of the sector = (1/2) x l r square units

= (1/2) x 77 x 35

=  38.5 x 35

= 1347.5 square units

Now we are going to the next example of the topic "Perimeter and area".

Problem 9:

Find the length of arc if the perimeter of sector is 45 cm and radius is 10 cm.

Solution:

Perimeter of sector = 45 cm

Perimeter of sector = L + 2r

L + 2r = 45

here r = 10 cm

L + 2 (10) = 45

L + 20 = 45

L = 45 - 10

L = 35 cm

Now we are going to the next example of the topic "Perimeter and area".

Problem 10 :

Find the circumference of the semi-circle whose diameter is 7 cm.

Solution :

r = diameter/2

r = 7/2

r = 3.5

Now we can apply the formula

Circumference of semi circle = (Π+2)r

here r = 3.5 and Π = 22/7

= [(22/7)  + 2]x 3.5

= (22+14) x 0.5

= 36 x 0.5

= 18 cm

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