UPSC, CDS preparation , Area and Perimeters of plane figures with diagram - Pk hindi tech :- Technologies se related jaankari hindi me

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 UPSC,  CDS preparation , Area and Perimeters of plane figures with diagram

UPSC, CDS preparation , Area and Perimeters of plane figures with diagram

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   UPSC ,CDS 

AREA AND PERIMETER OF  PLANE FIGURES.

 

 --**Square**-



 let each side of   square be( a) unit.

(1)-  Area = side* side = 1/2 (Diagonal)²

(2)- Perimeter of square = 4 × side

(3)- Diagonal of square = √2 × a



--**Rectangle**--


Let (l )and (b) be the length and breadth of a rectangle respectively, then

(1)- Area of rectangle = length × breadth

(2)- Perimeter of rectangle = 2 ( Length + Breadth ) units


(3)- Diagonal of rectangle = √ [(Length)²+ ( Breadth)²

(4)- Area of track ( coloured region) = ( L1 B1 - L2 B2)


--**Quadrilateral**-- 

(1) Let ABCD is a quadrilateral in which DM = H1 and BN = H2 are perpendicular on diagonal AC from other two vertices B and D , then
  

 Area of quadrilateral = 1/2 × diagonal × (H1 + H2)

(2) Let D1 and D2 are the two diagonals and (Θ) is the angle between them , then
         



           Area of quadrilateral = 1/2 ×D1× D2 × sinΘ  .



--** Parallelogram**--



Let adjacent sides of a parallelogram are  "b" and "a" respectively, and "h" is the corresponding altitude of side a , then

(1):- Area of parallelogram = b × h = a ×h    sq. unit

(2):- Perimeter of parallelogram = 2(a+b)   units

* Each diagonal of parallelogram divides it into two triangles of equal areas.



--**Rhombus**--



Let the length of each side of a rhombus is  "a" and length of both diagonal are "d1' and "d2", then

(1) Area of Rhombus = 1/2 d1 d2   sq. unit

(2) side of rhombus = 1/2 √(d1² +  d2² )

(3) Perimeter = 4 × side = 4a 

* Diagonals of  a rhombus bisects each other at right angles.


--** Trapezium**--

Let the length of  parallel sides of a trapezium are "a"  and  "b", and distance between them is  h1  , then 


(1) Area of trapezium = 1/2 ( CD + AB ) × h    sq. unit




--**Scalene Triangle**--

  Let d=the sides of triangle are a ,b ,and c and  h be the corresponding altitude to side a ,then


(1):-  Perimeter of  scalene triangle is , 2S ( 2 of semi - perimeter) = a+ b+ c units

(2):- Semi - perimeter of scalene triangle , S = ( a + b + c ) / 2
                                                 
(3):- Area of  triangle =√{ S (S-a) (S-b) ( S-c) }

(4):- The radius of Incircle  =  2 × ( area) / Perimeter

(5):- The radius of Circumcircle =  a b c / 4( area)



--** Right Angled Triangle**--

(1):- Perimeter = ( b+ h + P) units


(2):- Area of right angled triangle = 1/2 × b × h



--**Isosceles Triangle **--

(1):- Perimeter =(  a + b + b)  = a + 2b = (1/ 4) a   √(4b² - a²)

(2) :- Area = (s-b) √s(s-a)  sq unit 

(3):- Area of a right isosceles triangle in which equal sides form right angle is given by ,

            Area =  1/2   a ²  sq unit


(4) If  ( theta) is the angle between two sides of isosceles traingle, then
    
            Area = 1/2  a²  sinΘ



--**Equilateral Triangle**--


 Let "a" be the side of an equilateral triangle , then-


(1):- height ( Altitude) of an equilateral triangle = (√3 ÷2) × a

(2):- Area of an equilateral triangle = (√3 ÷ 4) × a

(3):- Perimeter of an equilateral traingle = 3 a 

(4):- The radius of incircle = a ÷2√3

(5):- The radius of Circum circle = a ÷ √3



 --** Circle**--

Let the radius of circle = r , then


(1):- Circumference of Circle = 2πr = πD

(2):- Area of Circle = πr² s    sq unit

(3):- Distance covered by a wheel in one revolution = Circumference of the wheel = 2πr



--**Circular Ring**--

 If  "R" and "r" be the outer and inner radii of a ring , then 

 area of ring = π(R² - r²)  sq unit.

 



--** Semi- Circle**--

(1):- Area of Semi - circle = 1/2 πr² sq. unit


(2):- Perimeter of Semi - Circle = (πr + D) units



--** Quadrant of a circle**--

(1):- The perimeter of the quadrant of circle = 1/4 × 2πr + 2r units
                                                                                    
                                                           = (πr ÷2 )+ 2r = (π /2 + 2 ) r  units

(2):- Area of quadrant = 1/4 ( area of circle) = 1/4 × πr²  sq. unit


* If two diameters are perpendicular to each other , the they divide the circle into four quadrants.



---** Area of sectors**---

 If   Θ is the angle at the centre of a circle of a radius r , then 

(1):- Length of an ARC  PQ = 2πrΘ ÷ 360 degree

(2):- Area of setor OPRQO = πr²Θ ÷ 360 degree

(3):- Area of segment PRQP = (Area of sector OPRQO) - Area of triangle OPQ

                                                     = (πr²Θ ÷ 360 degree ) - 1/2 r²sinΘ

(4):- Area of major segment QPSQ = Area of circle - Area of minor segment PRQP 


--**Regular polygon**--


*The polygon having all sides and angles equal is called regular polygon.

 let n = number of sides
       a = length of sides
      r= radius of inscribed circle.
      R = radius of circumcircle.

(1) :- Area of regular polygon = n/4 a² cot(π/n)
                                                   = nr² tan (π/n)
                                                   = n/2 R² Sin (2π/n)

(2) :- Perimeter of regular polygon = n× a = 2nr sin( π/n)

(3) :- Inradius = a/2 cot (π/n)

(4) :- Circumradius = a/2 cosec( π/n)




---***Some important results***---


(1) Angle inscribed by minute hand in 60 minutes = 360 degrees

(2) angle inscribed by hour hand im 12 hour = 360 degrees

(3) Angle inscribed by minute hand in 1 min = 6 degree

(4) If the length of a square/ rectangle is inscreased or decreased by X% and the breadth is increased or decreased by Y%, then the net effect 
      on the area is given by ,

              [ ± x ± y  + (±x) (±y)  /1oo] %

(5) If the side of a square / rectangle / triangle is doubled , then the area is increased by 300 % , that is the area becomes four times of itself.

(6) If the radius of a circle is decreased by x% then area is decreased by 
           [ -2x + x²/ 100] %

     And if the radius is inscreased by x% then area is inscreased by 
         [ 2x + x²/ 100] %

(7) If the room of dimension ( l × b )  m is to be paved with square tiles, then

     (a) The side of the largest square tile =  HCF of   l and b

    (b) The least number of tiles required =  (l × b ) ÷ (HCF of   l and b)²

(8) 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   √2r

(9) The area of the largest triangle inscribed in a semi - circle of radius  r  is  .








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