Today we are going to learn about the Area and Perimeter of Plane Figures .
These include various figures like, Square , rectangle , quadrilateral , parallelogram , rhombus , trapezium , right angled Triangle , isosceles triangle , scalene triangle, equilateral triangle , circle , circular ring , semi circle , quadrant of circle , area of sector , regular polygon , etc.
All these are important plane figures and today we are going to study about their area and perimeter and how we can compute them easily .
Tip : get a pen and paper and write each formula for five to six times
so that you may easily remember the formulae of area and perimeter of plane figures .
So, lets begin ,
Area of the plane figure is the amount of surface enclosed by its boundary . It is measured in square units.
SQUARE
Let each side of a square be a unit .
Then,
Perimeter of square = 4 (SIDE) = 4a units
Diagonal of square = √2 (SIDE)= a√2 units
Area of square = SIDE * SIDE = a² sq units = (diagonal)² / 2 = d² / 2
Side of square = √area = √a² sq units
RECTANGLE
Let l and b be the length and breadth of a rectangle respectively, then
Area of rectangle = Length * Breadth = l * b
Perimeter of rectangle = 2 (length + breadth)
Diagonal of rectangle = √{(length)² + (breadth)²}
Area of track = (L1 *B1 - L2*B2) sq units
QUADRILATERAL
Let ABCD is a quadrilateral in which DM = h1 and BN = h2 are perpendiculars on diagonal AC from other two vertices B and D, then
Area of quadrilateral = Diagonal * (h1 + h2) / 2
= AC * (DM + BN) / 2 sq units
PARALLELOGRAM
Let adjacent sides of a parallelogram are b and a and b is corresponding altitude (height) of side a.
Area of the parallelogram = (Base * Heigtht) = a * b sq units
Perimeter of a parallelogram = 2 (Sum of adjacent side) = 2(a + b) units
Each diagonal of a parallelogram divides it into two triangles of equal area.
RHOMBUS
Let the length of each sides of a rhombus is a and length of both diagonals are d1 and d2 , then
Area of rhombus = d1 * d2 / 2 sq units
Side of rhombus =[ √{d1² + d2²} ] / 2
=> 4a² = d1² + d2²
Perimeter of rhombus = 4 * side units
Diagonals of a rhombus bisect each other.
TRAPZEIUM
Let the length of parallel sides of a trapezium are a and b and distance between them is h , then
Area of trapezium = (Sum of parallel sides) * (Distance between them) / 2
= (AB + CD) * h / 2 = (a + b) * h sq units .
RIGHT ANGLED TRIANGLE
A figure bounded by three straight lines is called a triangle.
Let perpendicular , base and hypotenuse of a right angled triangle (ABC) are p , b and h respectively then,
Perimeter of right angled triangle = AB + BC + CA = b + p + h units
Area of right angle triangle = Base * Altitude / 2
ISOSCELES TRIANGLE
Let sides of an isosceles triangle are a, b and b , then
Perimeter of isosceles triangle = a + b + b = a + 2b units
Area of isosceles triangle = (s - b) (√s(s - a))
where, a = Base and b = Equal sides
Area of a right isosceles triangle , in which equal sides from right angle then
Area = a² / 2 sq units
SCALENE TRIANGLE
Let the sides of a triangle are a, b, c and h be the corresponding height to side a , then
Perimeter of scalene triangle , 2s = a + b + c
Semi perimeter of scaler triangle = s = ( a + b + c ) / 2
Area of triangle = √s (s - a) (s - b) (s - c) [HERO'S FORMULA]
or area of triangle = a * h / 2
EQUILATERAL TRIANGLE
Let a be the side of an equilateral triangle , then
Height (altitude) of equilateral triangle = a√3 / 2
Area of equilateral triangle = a² √3 / 4
Perimeter of equilateral triangle = 3 * Side = 3a
CIRCLE
Let the radius of a circle be r, then
Circumference of circle = 2𝝿r , also 2r =D
Area of circle = 𝝿r²
Distance covered be a wheel in one revolution = Circumference of the wheel
CIRCULAR RING
If 'R' and 'r' be outer and inner radii of a ring , then the area of ring = 𝝿(R² - r²) sq units
SEMI CIRCLE
A diameter divides a circle into two equal parts . Each of these two arcs is called semi circle.
If r is the radius of a circle , then
Area of semi circle = 𝞹r² / 2 sq units
Perimeter of semi circle = (𝞹r + 2r) units
QUADRANT OF A CIRCLE
If r is the radius of a circle, then
Perimeter of the quadrant = (circumference of a circle) / 4 + 2r
= 2𝞹r / 4 + 2r
Area of the quadrant = (Area of circle) / 4
= 𝞹r² / 4 sq units
If two diameters are perpendicular to each other , then they divides the circle into four quadrants.
AREA OF SECTOR
If Θ be the angle at the centre of a circle of radius r , then
Length of the arc PQ = 2𝞹rΘ / 360⁰
Area of sector OPRQO = 𝞹r²Θ / 360⁰
Area of minor segment PRQP = Area of sector OPRQO - Area of △OPQ
= 𝞹r²Θ - r² sin Θ 360⁰ 2
Area of major segment QSPQ = Area of circle = Area of minor segment PRQP
REGULAR POLYGON
Let a be the side of a regular polygon.
Then ,
Area of regular polygon = 5√3 a² / 4 sq units
Area of regular hexagon = 3√3 a² / 2 sq units
Area of regular octagon = 2 (√2 + 1) a² sq units
SOME USEFUL RESULTS
* area of room = length * breadth
* area of 4 walls of a room = 2 (length + breadth) * height
* radius of circumcircle of an equilateral triangle of side 'a' = a / √3
* radius of incircle of triangle = ◭ / s , s = (a + b + c) / 2
* angle inscribed by minute hand in 60 min = 360⁰
* angle inscribed by hour hand in 12 h = 360⁰
* angle inscribed by minute hand in 1 min = 6⁰
* distance moved by a wheel in one revolution = circumference of the wheel
* If the length of a square / rectangle is increased by a% and the breadth is increased by y %, the net effect on the area is given by
net effect = [x + y + xy /100] %
* If the length and breadth of a square / rectangle are increased by x% and the breadth is decreased by y% the net effect on the area is given by
net effect = [x - y - xy /100] %
* If the length and breath of a square / rectangle are decreased by x% and y% respectively, the net effect on the area is given by
net effect = [-x -y + xy /100] %
* If the side of a square / rectangle / triangle is doubled the area is increased by 300%, i.e. the area becomes four times of itself.
*If the radius of a circle is decreased by x%, the net effect on the area is (-x² / 100)% , i.e. the area is decreased by (x² / 100)%.
*If a floor of dimensions (l*b) m is to be covered by a carpet of width wm at the rate X rs per metre , then the total amount required is rs (Xlb/w)
*If a room of dimensions (l*b) m is to be proved with square tiles , then
the side of the square tiel = HCF of l and b
the number of tiles required = l*b / (HCF of l and b)²
* area of a square inscribed in a circle of radius r is 2r² and the side of a square inscribed in a circle of radius r is √2 r .
* area of the largest triangle inscribed in a semi circle of radius r is r².
Hope you liked our article on how to find the area and perimeter of plane figures.
EXERCISES
THANKS FOR READING THIS AT Math Capsule.
