Perimeter = Sum of all side lengths

Perimeter of a rectangle = 2 × length + 2 × breadth

P = 2(1 + b) ⇒ (i) l = `P/2` − b, i.e., length = `"Perimeter"/2` − breadth

                         (ii) l = `P/2` − l, i.e., breadth = `"Perimeter"/2` − length

Perimeter of Square = Total boundary of the square

                                   = Side + Side + Side + Side

P = 4 × Side

Or: P = 4s (where 's' represents the side length)

side = ` "perimeter"/"4"`

Always include the correct linear unit (cm, m, mm, km, etc.)

Perimeter of a Triangle = 3 × length of a side.

For an equilateral triangle with side length = s

P = s + s + s

   = 3 × s

Perimeter = 3 × Length of one side

Area =  Amount of space inside a flat shape

Area of triangle = `(1/2) × "base" × "height" = 1/2 × b × h`.

Consider the points A (a,0), B( 0,b) and C (1,1) .

` Here (x_1=a,y_1=0).(x_2 = 0,y_2=b) and (x_3=1,y_3=1).`

It is given that the points are collinear. So,

`x_1 (y_2-y_3)+x_2(y_3-y_1) +x_3(y_1-y_2) =0`

`⇒  a(b-1)+0(1-0)+1(0-b)=0`

`⇒ ab-a-b=0`

Dividing the equation by ab:

`⇒ 1-1/b-1/a=0`

`⇒ 1-(1/a+1/b)=0`

`⇒(1/a+1/b)=1`

Therefore, the given points are collinear if  `(1/a+1/b)=1`

The given points are A (7, 10), B(-2, 5) and C(3, -4).

`AB= sqrt((-2-7)^2 +(5-10)^2) = sqrt((-9)^2 +(-5)^2) = sqrt((81+25)) = sqrt(106)`

`BC = sqrt((3-(-2))^2 +(-4-5)^2) = sqrt((5)^2 +(-9)^2 )= sqrt((25+81) )= sqrt(106)`

`AC = sqrt((3-7)^2 +(-4-10)^2) = sqrt(( -4)^2 +(-14)^2) = sqrt(16+196) = sqrt(212)`

Since, AB and BC are equal, they form the vertices of an isosceles triangle

Also,`(AB)^2 + (BC)^2 = ( sqrt(106))^2 +( sqrt(106)^2) = 212`

and `(AC)^2 = (sqrt(212))^2 = 212.

`Thus , (AB)^2 + (BC)^2 = (AC)^2`

This show that  ΔABC is right- angled at B. Therefore, the pointsA (7, 10), B(-2, 5) and C(3, -4). are the vertices of an isosceles rightangled triangle.

The given points are A(2, 4), B(2, 6) and C(2 +`sqrt(3)`,5) Now 

`AB =sqrt(((2-2)^2 +(4-6)^2 )) = sqrt((0)^2 +(-2)^2)`

    `= sqrt((0+4) =2`

`BC = sqrt((2-2- sqrt(3))^2 + (6-5)^2 ) = sqrt((- sqrt(3))^2 +(1)^2)`

`= sqrt(3+1) = 2`

`AC = sqrt((2-2-sqrt(3))^2 + (4-5)^2 ) = sqrt((- sqrt(3))^2 +(-1)^2)`

`= sqrt(3+1) =2`

Hence, the points A(2, 4), B(2, 6) and C(2 +`sqrt(3)`,5) are the vertices of an equilateral triangle

The vertices of the triangle are A(7, -3), B(5,3) and C(3,-1)

`"Coordinates of" D = ((5+3)/2,(3-1)/2) = (4,1)`

For the area of the triangle ADC, let

`A (x_1,y_1)=A(7,-3), D(x_2,y_2) =D(4,1) and C (x_3,y_3) = C(3,-1)`. Then

`"Area of"  Δ ADC = 1/2 [ x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]`

`=1/2 [7(1+1)+4(-1+3)+3(-3-1)]`

`=1/2[14+8-12}=5` sq. unit

Now, for the area of triangle ABD, let

`A(x_1,y_1) = A(7,-3), B(x_2,y_2) = B(5,3) and D (x_3,y_3) = D (4,1). `Then

`"Area of"  Δ ADC = 1/2 [ x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]`

`=1/2 [7(3-1)+5(1+3)+4(-3-3)]`

`=1/2[14+20-24] = 5` sq. unit 

Thus, Area (ΔADC)  = Area (ΔABD) = 5. sq units

Hence, AD divides  ΔABC into two triangles of equal areas.