# Area Problems

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# Area Problems

### Introduction

Area Problems deals with parameters like sides, length, area, breadth, centroid, median, perimeter, height etc. of all the geometrical shapes like square, rectangle, triangle, circle and so on.

### Methods

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.
(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:
$A = L^{2}$
= $5^{2}$
= 5 x 5
= $25 \ cm^{2}$

Example: Given a rectangle with the length of 4ft and the width of 3ft. Find its area.
Solution:
$A = lw$
= (4)(3)
= $12ft^{2}$

Example: Given a circle with the radius 3cm. Find its area. Take $\pi$ = 3.14.
Solution:
$A = \pi r^{2}$
= (3.14)$(3)^{2}$
= (3.14)(9)
= 28.26 $cm^{2}$

Example: Given a parallelogram with the base 5 in and the height 3 in. Find the area of this parallelogram.
Solution:
$A = bh$
= (5)(3)
= $15 in^{2}$

Example: Given a triangle with the base 4cm and the height 2cm. Find its area.
Solution:
$A = \frac{1}{2}bh$
= $\frac{1}{2}(4)(2)$
= $\frac{1}{2}(8)$
= $4cm^{2}$

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:
$A = (\frac{a + b}{2})h$
= $(\frac{2 + 3}{2})4$
= $(\frac{5}{2})4$
= $10in^{2}$

### Formulae

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

### Samples

1. The area of four walls of a room is 600$m^2$ and its length is twice its breadth. If the height of the room is 11 m, then find the area of the floor in $m^2$?
Solution:
Let, breadth of a room = $x$ m
Given,
Length of the room = 2$x$ m
Height of the room = 11 m
Total area of 4 waits of room = 2(2$x$ + $x$) x 11 $m^2$ = 66$m^2$
By hypothesis, 66$x$ = 660
⇒ $x$ = 10 m
Hence, area of floor = 2$x$ + $x$ = 2$x^2$ = 2$(10)^2$ = 2 x 100 = 200$m^2$

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 = $\frac{60}{4}$ = 15 cm
Side of second square = $\frac{44}{4}$ = 11 cm
Area of third square = $(15)^2 - (11)^2$ $(cm)^2$
= (225 - 121)$(cm)^2$
= 104 $(cm)^2$
Side of third square = $\sqrt{104}$ 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 = $\frac{1}{2}(diagonal)^2$
⇒ Area of square = $\frac{1}{2}(6)^2$
⇒ Area of square = $\frac{1}{2}36$
⇒ Area of square = 18 $m^2$
Therefore, Area of square = 18 $m^2$

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,
$x$ and $y$ be the length and breadth of the rectangle respectively.
Then, $x$ - 5 = $y$ + 4
$x$ - $y$ = 9 -------- (i)
Now, Area of rectangle = length x breadth and
Area of square = ($x$ - 5)($y$ + 4)
Therefore, ($x$ - 5)($y$ + 4) = $xy$
⇒ 4$x$ - 5$y$ = 20 -------- (ii)
By soling (i) and (ii),
$x$ = 25 and $y$ = 16.
Therefore, Perimeter of the rectangle = 2($x$ + $y$) = 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 $a$ = 10 cm, $b$ = 13 cm and $c$ = 15 cm
Now, consider $s$ = $\frac{1}{2}(a + b + c)$
⇒ $s$ = $\frac{1}{2}(10 + 13 + 15)$
⇒ $s$ = 19
Therefore,
(s - a) = 19 - 10 = 9
(s - b) = 19 - 13 = 6
(s - c) = 19 - 15 = 4
Therefore, area = $\sqrt{s(s - a)(s - b)(s - c)}$ = $\sqrt{19 * 9 * 6 * 3}$ = $\sqrt{3078}$ = 55.47 $(cm)^2$