On the webpage "perimeter and area" we are going to see formula for each and every shapes in math.
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. |
Square unit.(sq) Measures two dimensions.
Example : 36 in² or 36 inches squared
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 = √s(s-a)(s-b)(s-c)
Perimeter of scalene triangle = a + b + c
here "s = (a + b + c)/2"
Quadrilateral
Area of quadrilateral =(1/2) x d x (h₁+h₂)
Perimeter of quadrilateral = Sum of all four sides
Parallelogram
Area of parallelogram = b x h
Perimeter of quadrilateral = Sum of all four side
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 Semi circle= (1/2) Π r²
Perimeter of semi circle = (Π+2)r
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
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
Length x breadth = 147
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:
radius = 35 cm
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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