Revision: Perimeter and Area Maths SSC (English Medium) 7th Standard Maharashtra State Board

Polygon: Polygon refers to a closed 2D shape which is made up of a finite number of line segments, but the perimeter is a one-dimensional measurement.

• Cube: A cube is a three-dimensional solid object bounded by six square faces, facets, or sides, with three meetings at each vertex. The cube is the only regular hexahedron (i.e., a solid figure with six plane faces) and is one of the five Platonic solids. It has 6 faces, 12 edges, and 8 vertices.
• Lateral surface area of the cube: Out of the six faces of a cube, we only find the area of the four faces, leaving the bottom and top faces. In such a case, the area of these four faces is called the lateral surface area of the cube.

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

The perimeter of a regular polygon = (length of one side) × number of sides.

The perimeter of an Irregular polygon = Sum of all sides of Irregular polygons.

Area =  Amount of space inside a flat shape

Area of square = side × side

                          = s × s

                          = s²

Area of square = (side)²

 ⇒ its side = \[\sqrt{Area}\]

Area of a rectangle = length × breadth

Written as: A = l × b

l = `A/b` i.e.,  length = `"Area"/"Breadth"`

and, b = `A/l`  i.e., breadth  = `"Area"/"Length"`

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

• Total surface area of cuboid = 2(lb + bh + lh)
• The lateral surface area of a cuboid = 2h(l + b)

• Cuboid: A cuboid is a convex polyhedron bounded by six quadrilateral faces, whose polyhedral graph is the same as that of a cube. A cuboid looks like a rectangular box. It has 6 faces. Each face has 4 edges. Each face has 4 corners (called vertices).
• Surface of a cuboid: the outer surface of a cuboid is made up of six rectangles (in fact, rectangular regions, called the faces of the cuboid), whose areas can be found by multiplying the length by breadth for each of them separately and then adding the six areas together.
• Lateral surface area of the cuboid: Out of the six faces of a cuboid, we only find the area of the four faces, leaving the bottom and top faces. In such a case, the area of these four faces is called the lateral surface area of the cuboid.

• Total surface area of the cube = 6a².
• Lateral surface area of a cube = 4a².

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.