# Area Problems

#### Chapter 25

5 Steps - 3 Clicks

# 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 aÂ 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 aÂ 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

â‡’ $$x$$ = 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 = $$\frac{1}{2}(diagonal)^2$$

â‡’ Area of square = $$\frac{1}{2}(6)^2$$

â‡’ Area of square = $$\frac{1}{2}36$$

â‡’ 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) = $$xy$$

â‡’ 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 $$s$$ = $$\frac{1}{2}(a + b + c)$$

â‡’ $$s$$ = $$\frac{1}{2}(10 + 13 + 15)$$

â‡’ $$s$$ = 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 $$(cm)^2$$