Area: It is defined as the surface enclosed by its sides.
Square: It is defined as a four sided shape that is made up of four straight sides that are the same lengths and that has four right angles.
Triangle: The plane figure formed by connecting three points not in a straight line by straight line segments like a three sided polygon.

(i) Sum of angles of a triangle is 180 degrees.
(ii) The sum of any two sides of a triangle is greater than the third side.
(iii) The line joining the mid - point of a side of a triangle to the opposite vertex is called the meridian.
(iv) The point where three medians of a triangle meet, is called centroid. The centroid divides each of the medians in the ratio 2 : 1.
(v) In an isosceles triangle, the altitude from the vertex bisects the base.
(vi) The median of a triangle divides it into two triangles of the same area.
(vii) The area of a triangle formed by joining the mid - points of the sides of a given triangle is one - fourth of the area of the given triangle.

Rectangle: The plane figure formed with four straight sides and four right angles, especially one with unequal adjacent sides, in contrast to a square.
Circle: A line that is curved so that its ends meet and every point on the line is the same distance from the centre.
Quadrilateral: Quadrilateral means four sided flat shape.

(i) The diagonals of a parallelogram bisect each other.
(ii) Each diagonal of a parallelogram divides it into two triangles of the same area.
(iii) The diagonals of a rectangle are equal and bisect each other.
(iv) The diagonals of a square are equal and bisect each other at right angles.
(v) The diagonals of a rhombus are unequal and bisect each other at right angles.
(vi) A parallelogram and a rectangle on the same base and between the same parallels are equal in area.
(vii) Of all the parallelogram of given sides, the parallelogram which is a rectangle has the greatest area.

Example: Given a square where the length of each side (edge) is 5cm. Find the area of this square.
Solution:

[latex]A = L^{2}[/latex]
= [latex]5^{2}[/latex]
= 5 x 5
= [latex]25 \ cm^{2}[/latex]

Example: Given a rectangle with the length of 4ft and the width of 3ft. Find its area.
Solution:

[latex]A = lw[/latex]
= (4)(3)
= [latex]12ft^{2}[/latex]

Example: Given a circle with the radius 3cm. Find its area. Take [latex]\pi[/latex] = 3.14.
Solution:

[latex]A = \pi r^{2}[/latex]
= (3.14)[latex](3)^{2}[/latex]
= (3.14)(9)
= 28.26 [latex]cm^{2}[/latex]

Example: Given a parallelogram with the base 5 in and the height 3 in. Find the area of this parallelogram.
Solution:

[latex]A = bh[/latex]
= (5)(3)
= [latex]15 in^{2}[/latex]

Example: Given a triangle with the base 4cm and the height 2cm. Find its area.
Solution:

[latex]A = \frac{1}{2}bh[/latex]
= [latex]\frac{1}{2}(4)(2)[/latex]
= [latex]\frac{1}{2}(8)[/latex]
= [latex]4cm^{2}[/latex]

Example: Given a trapezoid with the height of 4 in, and two parallel sides of 2 in and 3 in respectively. Calculate its area.
Solution:

[latex]A = (\frac{a + b}{2})h[/latex]
= [latex](\frac{2 + 3}{2})4[/latex]
= [latex](\frac{5}{2})4[/latex]
= [latex]10in^{2}[/latex]

1 Pythagoras theorem: In a right-angled triangle, [latex](hypotenuse)^2[/latex] = [latex](base)^2[/latex] + [latex](height)^2[/latex].
2: Area of a triangle = (length x breadth) Therefore, length = [latex]\frac{area}{breadth}[/latex] and breadth = [latex]\frac{area}{length}[/latex].
3: Perimeter of a rectangle = 2(length x breadth).
4: Area of a square = [latex](side)^2 [/latex] = [latex]\frac{1}{2}(diagonal)^2[/latex].
5: Area of four walls of a room = 2(length x breadth) x height.
6: Area of a triangle = [latex]\frac{1}{2}[/latex] x base x height.
7: Area of a triangle = [latex]\sqrt{s(s - a)(s - b)(s - c)}[/latex], where [latex]a, b, c[/latex] are sides of the triangle and [latex]s[/latex] = [latex]\frac{1}{2}(a + b + c)[/latex].
8: Area of an equilateral triangle = [latex]\frac{\sqrt{3}}{4} * (side)^2[/latex].
9: Radius of incircle of an equilateral triangle of side [latex]a[/latex] is [latex]\frac{a}{2\sqrt{3}}[/latex].
10: Radius of circumference of an equilateral triangle of side [latex]a[/latex] is [latex]\frac{a}{\sqrt{3}}[/latex].
11: Radius of incircle of a triangle of area [latex]\bigtriangleup[/latex] and semi-perimeter [latex]s[/latex] = [latex]\frac{\bigtriangleup}{s}[/latex].
12: Area of a parallelogram = base x height.
13: Area of a trapezium = [latex]\frac{1}{2}[/latex] x (sum of parallel sides) x distance between them.
14: Area of a circle = [latex]\pi R^2[/latex], where R is the radius.
15: Circumference of a circle = 2[latex]\pi[/latex]R.
16: Length of an arc = [latex]\frac{2\pi R\theta}{360}[/latex], where [latex]\theta[/latex] is the central angle.
17: Area of a sector = [latex]\frac{1}{2}[/latex](arc x R) = [latex]\frac{\pi R^2\theta}{360}[/latex].
18: Area of a semi circle = [latex]\frac{\pi R^2}{2}[/latex].
19: Area of isosceles triangle = [latex]\frac{a}{4}\sqrt{4b^2 - a^2}[/latex]square units, where [latex]a, b[/latex] are linear units.
20: Circumference of a semi-circle = [latex]\pi R[/latex].
21: Side of a rhombus = [latex]\frac{1}{2}\sqrt{(d_{1})^2 + (d_{2})^2}[/latex] linear units, where [latex]d_{1}[/latex] and [latex]d_{2}[/latex] are the lengths of diagonals.

1. The area of four walls of a room is 600[latex]m^2[/latex] and its length is twice its breadth. If the height of the room is 11 m, then find the area of the floor in [latex]m^2[/latex]?
Solution:

Let, breadth of a room = [latex]x[/latex] m
Given,
Length of the room = 2[latex]x[/latex] m
Height of the room = 11 m
Total area of 4 waits of room = 2(2[latex]x[/latex] + [latex]x[/latex]) x 11 [latex]m^2[/latex] = 66[latex]m^2[/latex]
By hypothesis, 66[latex]x[/latex] = 660
⇒ [latex]x[/latex] = 10 m
Hence, area of floor = 2[latex]x[/latex] + [latex]x[/latex] = 2[latex]x^2[/latex] = 2[latex](10)^2[/latex] = 2 x 100 = 200[latex]m^2[/latex]

2. The perimeter of two squares are 60 cm and 44 cm. Find the perimeter of a third square whose area is equal to the difference of the areas of two squares?
Solution:

Given,
Side of first square = [latex]\frac{60}{4}[/latex] = 15 cm
Side of second square = [latex]\frac{44}{4}[/latex] = 11 cm
Area of third square = [latex](15)^2 - (11)^2[/latex] [latex](cm)^2[/latex]
= (225 - 121)[latex](cm)^2[/latex]
= 104 [latex](cm)^2[/latex]
Side of third square = [latex]\sqrt{104}[/latex] cm = 10.19 cm
Therefore, required perimeter = (10.19 x 4) cm = 40.76 cm.

3. Find the area of the square, one of whose diagonals is 6 m long?
Solution:

Given,
Diagonal = 6 m
Now, consider
Area of square = [latex]\frac{1}{2}(diagonal)^2[/latex]
⇒ Area of square = [latex]\frac{1}{2}(6)^2[/latex]
⇒ Area of square = [latex]\frac{1}{2}36[/latex]
⇒ Area of square = 18 [latex]m^2[/latex]
Therefore, Area of square = 18 [latex]m^2[/latex]

4. If the length of a certain rectangle is decreased by 5 cm and the width is increased by 4 cm, a square with the same area as the original rectangle would result . Find the perimeter of the original rectangle?
Solution:

Let,
[latex]x[/latex] and [latex]y[/latex] be the length and breadth of the rectangle respectively.
Then, [latex]x[/latex] - 5 = [latex]y[/latex] + 4
[latex]x[/latex] - [latex]y[/latex] = 9 -------- (i)
Now, Area of rectangle = length x breadth and
Area of square = ([latex]x[/latex] - 5)([latex]y[/latex] + 4)
Therefore, ([latex]x[/latex] - 5)([latex]y[/latex] + 4) = [latex]xy[/latex]
⇒ 4[latex]x[/latex] - 5[latex]y[/latex] = 20 -------- (ii)
By soling (i) and (ii),
[latex]x[/latex] = 25 and [latex]y[/latex] = 16.
Therefore, Perimeter of the rectangle = 2([latex]x[/latex] + [latex]y[/latex]) = 2(16 + 25) = 82 cm.

5. Find the area of the triangle whose base sides measure 10 cm, 13 cm and 15 cm?
Solution:

Given that,
Let base sides [latex]a[/latex] = 10 cm, [latex]b[/latex] = 13 cm and [latex]c[/latex] = 15 cm
Now, consider [latex]s[/latex] = [latex]\frac{1}{2}(a + b + c)[/latex]
⇒ [latex]s[/latex] = [latex]\frac{1}{2}(10 + 13 + 15)[/latex]
⇒ [latex]s[/latex] = 19
Therefore,
(s - a) = 19 - 10 = 9
(s - b) = 19 - 13 = 6
(s - c) = 19 - 15 = 4
Therefore, area = [latex]\sqrt{s(s - a)(s - b)(s - c)}[/latex] = [latex]\sqrt{19 * 9 * 6 * 3}[/latex] = [latex]\sqrt{3078}[/latex] = 55.47 [latex](cm)^2[/latex]