9.2-Area of a Polygon

9.2-Area of a Polygon Important Formulae

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9.2 - Area of a Polygon
  • Area of a polygon can be calculated by dividing it into simpler shapes like triangles and rectangles.
  • For a regular polygon, the area is given by the formula: $A = \dfrac{1}{4} n a^2 \cot \left(\dfrac{\pi}{n}\right)$ where $n$ is the number of sides and $a$ is the length of a side.
  • The area of an irregular polygon can be calculated using the method of triangulation or the shoelace theorem.
  • For a triangle (a type of polygon), the area is $A = \dfrac{1}{2} \times base \times height$.
  • The area of a quadrilateral can be calculated using the formula: $A = \dfrac{1}{2} \times (d_1 \times d_2)$ where $d_1$ and $d_2$ are the diagonals.

9.2 - Area of a Polygon

In this section, we will learn how to calculate the area of a polygon. A polygon is a closed figure with straight sides. Examples include triangles, quadrilaterals, pentagons, hexagons, etc. The area of a polygon refers to the space enclosed within its sides. The formula for the area depends on the type of polygon and its specific properties.

For a regular polygon, where all sides and angles are equal, we can calculate the area using the following formula:

Formula for Area of a Regular Polygon:

The area $A$ of a regular polygon with $n$ sides, each of length $a$, and apothem (the perpendicular distance from the center to the midpoint of a side) $p$ is given by:

$A = \frac{1}{2} \times n \times a \times p$

Where:

  • $n$ = number of sides of the polygon
  • $a$ = length of one side of the polygon
  • $p$ = apothem of the polygon

For example, if we have a regular hexagon with side length 6 cm and an apothem of 5 cm, the area can be calculated as:

$A = \frac{1}{2} \times 6 \times 6 \times 5 = 90$ cm²

For polygons that are not regular, the area calculation method varies depending on the specific shape and the available information.

Area of a Triangle

The area of a triangle is calculated using the formula:

$A = \frac{1}{2} \times \text{base} \times \text{height}$

Where:

  • Base = the length of the base of the triangle
  • Height = the perpendicular distance from the base to the opposite vertex

For instance, if the base of a triangle is 8 cm and its height is 5 cm, then the area of the triangle is:

$A = \frac{1}{2} \times 8 \times 5 = 20$ cm²

Area of a Quadrilateral

For a quadrilateral, the area can be calculated in different ways depending on the type of quadrilateral:

  • Rectangle or Square: The area of a rectangle or square is given by:

$A = \text{length} \times \text{width}$

For a rectangle with length 10 cm and width 4 cm, the area is:

$A = 10 \times 4 = 40$ cm²

  • Parallelogram: The area of a parallelogram is:

$A = \text{base} \times \text{height}$

For a parallelogram with a base of 12 cm and height of 6 cm, the area is:

$A = 12 \times 6 = 72$ cm²

  • Trapezium: The area of a trapezium (trapezoid) is given by:

$A = \frac{1}{2} \times (\text{sum of parallel sides}) \times \text{height}$

For a trapezium with parallel sides of lengths 10 cm and 6 cm, and height 4 cm, the area is:

$A = \frac{1}{2} \times (10 + 6) \times 4 = 32$ cm²

Using the Shoelace Theorem

For an irregular polygon with known coordinates of its vertices, the area can be calculated using the Shoelace Theorem. Suppose the coordinates of the vertices of the polygon are $(x_1, y_1), (x_2, y_2), \dots, (x_n, y_n)$, where the vertices are listed in order. The area is calculated using the following formula:

$A = \frac{1}{2} \left| \sum_{i=1}^{n-1} (x_i \times y_{i+1} - y_i \times x_{i+1}) + (x_n \times y_1 - y_n \times x_1) \right|$

Where $n$ is the number of vertices. The terms $(x_i, y_i)$ and $(x_{i+1}, y_{i+1})$ represent the coordinates of consecutive vertices.


AREAS, CC BY-SA 4.0, via Wikimedia Commons

9.2-बहुपद का क्षेत्रफल

बहुपद (Polygon) एक समतल आकृति है जिसमें अधिकतम तीन या तीन से अधिक सीधी रेखाएं मिलकर एक बंद रूप बनाती हैं। बहुपद का क्षेत्रफल ज्ञात करने के लिए विभिन्न विधियाँ होती हैं, जो उसकी परिभाषा और प्रकार पर निर्भर करती हैं।

जब हम एक नियमित बहुपद (Regular Polygon) की बात करते हैं, तो इसका मतलब है कि इसके सभी भुजाएँ और आंतरिक कोण समान होते हैं। एक असामान्य बहुपद (Irregular Polygon) में भुजाएँ और कोण असमान हो सकते हैं। हम दोनों प्रकार के बहुपदों का क्षेत्रफल अलग-अलग तरीके से निकालते हैं।

1. नियमित बहुपद का क्षेत्रफल

एक नियमित बहुपद का क्षेत्रफल निकालने के लिए हम इसे छोटे-छोटे त्रिकोणों में विभाजित करते हैं। इस तरह के बहुपद में, प्रत्येक त्रिकोण का आधार बहुपद की भुजा होती है और ऊँचाई केंद्र से उस भुजा तक की दूरी होती है।

रोजगार सूत्र के रूप में, यदि एक नियमित बहुपद की भुजा $a$ और उसका मध्यक (apothem) $r$ हो, तो क्षेत्रफल $A$ निम्नलिखित सूत्र से निकाला जा सकता है:

$A = \dfrac{1}{2} \times P \times r$

जहाँ,
$P$ = परिधि (Perimeter) = संख्या of sides $\times$ length of one side = $n \times a$
$r$ = मध्यक (Apothem)

2. असामान्य बहुपद का क्षेत्रफल

असामान्य बहुपद का क्षेत्रफल निकालने के लिए हमें इसे त्रिकोणों या अन्य सरल रूपों में विभाजित करना पड़ता है। इस विधि में बहुपद को छोटे-छोटे भागों में बाँट कर हर भाग का क्षेत्रफल निकालते हैं और फिर उन्हें जोड़ते हैं।

यदि बहुपद के सभी शीर्षांक $A_1, A_2, A_3, ..., A_n$ दिए गए हैं और हम इनको एक समकोण त्रिकोण के द्वारा जोड़ते हैं, तो क्षेत्रफल का सूत्र निम्नलिखित होता है:

$A = \dfrac{1}{2} \times \sum_{i=1}^{n} (x_i \times y_{i+1} - x_{i+1} \times y_i)$

यहां, $(x_i, y_i)$ और $(x_{i+1}, y_{i+1})$ बहुपद के दो आसन्न शीर्षांक होते हैं। यह सूत्र "Shoelace Theorem" (जूते की डोरी प्रमेय) के रूप में प्रसिद्ध है।

3. त्रिकोण और बहुपद का क्षेत्रफल

बहुपद के क्षेत्रफल को कई बार हम त्रिकोणों के जोड़ के रूप में निकाल सकते हैं। एक बहुपद के प्रत्येक त्रिकोण का क्षेत्रफल निकालकर, उन्हें जोड़ कर हम बहुपद का समग्र क्षेत्रफल प्राप्त कर सकते हैं। उदाहरण के लिए, किसी नियमित बहुपद में, बहुपद को उसके केंद्र से जुड़े त्रिकोणों में विभाजित किया जा सकता है, और फिर प्रत्येक त्रिकोण का क्षेत्रफल निकाला जा सकता है।

इसके लिए, यदि बहुपद के केंद्र से किसी भुजा तक की दूरी $r$ और भुजा का आयतन $a$ हो, तो प्रत्येक त्रिकोण का क्षेत्रफल $A_t$ होगा:

$A_t = \dfrac{1}{2} \times a \times r$

फिर, सभी त्रिकोणों के क्षेत्रफल को जोड़ने पर, हम बहुपद का कुल क्षेत्रफल प्राप्त कर सकते हैं।

The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are 1 m and 1.2 m and perpendicular distance between them is 0.8 m.

Solution:

Area of Trapezium

The area of a trapezium is given by the formula:

$$ \text{Area} = \frac{1}{2} \times (a + b) \times h $$

Where:

  • $$ a $$ = length of the first parallel side = 1 m
  • $$ b $$ = length of the second parallel side = 1.2 m
  • $$ h $$ = perpendicular distance between the parallel sides = 0.8 m

Substituting the values:

$$ \text{Area} = \frac{1}{2} \times (1 + 1.2) \times 0.8 $$

$$ \text{Area} = \frac{1}{2} \times 2.2 \times 0.8 $$

$$ \text{Area} = \frac{1}{2} \times 1.76 $$

$$ \text{Area} = 0.88 \, \text{m}^2 $$

The area of a trapezium is 34 cm2 and the length of one of the parallel sides is 10 cm and its height is 4 cm. Find the length of the other parallel side.

Solution:

Solution:

The formula for the area of a trapezium is:

Area = $ \frac{1}{2} \times (a + b) \times h $

Where:

  • $a$ = length of one parallel side = 10 cm
  • $b$ = length of the other parallel side (to be found)
  • $h$ = height = 4 cm
  • Area = 34 cm$^2$

Substitute the known values into the formula:

34 = $ \frac{1}{2} \times (10 + b) \times 4 $

Now, simplify the equation:

34 = $ 2 \times (10 + b) $

34 = $ 20 + 2b $

Subtract 20 from both sides:

34 - 20 = $ 2b $

14 = $ 2b $

Divide both sides by 2:

$ b = \frac{14}{2} $

$ b = 7 $ cm

The length of the other parallel side is 7 cm.

Length of the fence of a trapezium shaped field ABCD is 120 m. If BC = 48m,CD = 17 m and AD= 40 m, find the area of this field.Side AB is perpendicular to the parallel sides AD and BC.

Solution:

Question:

Length of the fence of a trapezium shaped field ABCD is 120 m. If BC = 48 m, CD = 17 m and AD = 40 m, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.

Solution:

Given:

  • Length of BC = 48 m
  • Length of CD = 17 m
  • Length of AD = 40 m
  • Length of AB is perpendicular to the parallel sides AD and BC.

The formula for the area of a trapezium is:

Area = $ \frac{1}{2} \times (Sum \ of \ parallel \ sides) \times Height $

Here, the parallel sides are AD and BC, and the height is AB.

Let the height AB be $ h $. Then, we can find $ h $ using the Pythagorean theorem in the right-angled triangle ABD.

Using the Pythagoras theorem:

AD$^2$ = AB$^2$ + BD$^2$

Since BD = BC - CD = 48 m - 17 m = 31 m, we substitute the values:

40$^2$ = AB$^2$ + 31$^2$

1600 = AB$^2$ + 961

AB$^2$ = 1600 - 961 = 639

AB = $ \sqrt{639} $ ≈ 25.3 m

Now, the area of the trapezium is:

Area = $ \frac{1}{2} \times (48 + 17) \times 25.3 $

Area = $ \frac{1}{2} \times 65 \times 25.3 $

Area = 32.5 × 25.3 = 821.25 m$^2$

Therefore, the area of the field is 821.25 m$^2$.

The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.

Solution:

Area of the Quadrilateral Shaped Field

Given:

  • Diagonal of the quadrilateral = 24 m
  • Perpendicular from one opposite vertex = 8 m
  • Perpendicular from the other opposite vertex = 13 m

The area of a quadrilateral can be found using the formula:

Area = $\frac{1}{2} \times d \times (h_1 + h_2)$

Where:

  • $d$ = length of the diagonal = 24 m
  • $h_1$ = perpendicular from the first vertex = 8 m
  • $h_2$ = perpendicular from the second vertex = 13 m

Substituting the values:

Area = $\frac{1}{2} \times 24 \times (8 + 13)$

Area = $\frac{1}{2} \times 24 \times 21$

Area = $12 \times 21$

Area = 252 m2

The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.

Solution:

Question: The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.

The formula for the area of a rhombus is:

Area = $ \frac{1}{2} \times d_1 \times d_2 $

Where:

  • $d_1$ = length of the first diagonal = 7.5 cm
  • $d_2$ = length of the second diagonal = 12 cm

Substitute the given values into the formula:

Area = $ \frac{1}{2} \times 7.5 \times 12 $

Area = $ \frac{1}{2} \times 90 $

Area = 45 cm²

The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m$^2$ is Rs. 4.

Solution:

Given:
  • Number of tiles = 3000
  • Shape of each tile = Rhombus
  • Diagonal 1 = 45 cm
  • Diagonal 2 = 30 cm
  • Cost per m$^2$ = Rs. 4
To Find:

Total cost of polishing the floor

Step 1: Area of one rhombus-shaped tile

The area of a rhombus is given by the formula:

$$ \text{Area} = \frac{1}{2} \times d_1 \times d_2 $$

Substituting the values of the diagonals:

$$ \text{Area} = \frac{1}{2} \times 45 \, \text{cm} \times 30 \, \text{cm} $$ $$ \text{Area} = \frac{1}{2} \times 1350 \, \text{cm}^2 $$ $$ \text{Area} = 675 \, \text{cm}^2 $$
Step 2: Convert area to m$^2$

1 m$^2$ = 10000 cm$^2$, so

$$ \text{Area in m}^2 = \frac{675}{10000} = 0.0675 \, \text{m}^2 $$
Step 3: Total area of 3000 tiles
$$ \text{Total area} = 3000 \times 0.0675 \, \text{m}^2 $$ $$ \text{Total area} = 202.5 \, \text{m}^2 $$
Step 4: Total cost of polishing
$$ \text{Total cost} = 202.5 \, \text{m}^2 \times 4 \, \text{Rs.} $$ $$ \text{Total cost} = 810 \, \text{Rs.} $$

Mohan wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the side along the road. If the area of this field is 10500 m$^2$ and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.

Solution:

Problem:

Mohan wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the side along the road. If the area of this field is 10500 m$^2$ and the perpendicular distance between the two parallel sides is 100 m, find the length of the side along the river.

Solution:

Let the length of the side along the road be $x$ meters. Then, the length of the side along the river is $2x$ meters.

The area $A$ of a trapezium is given by the formula:

$A = \frac{1}{2} \times (a + b) \times h$

Where $a$ and $b$ are the lengths of the parallel sides, and $h$ is the perpendicular distance between them. Here, $a = x$ (side along the road), $b = 2x$ (side along the river), and $h = 100$ m (the perpendicular distance).

Substituting the known values into the area formula:

10500 = $\frac{1}{2} \times (x + 2x) \times 100$

10500 = $\frac{1}{2} \times 3x \times 100$

10500 = $150x$

Now, solving for $x$:

$x = \frac{10500}{150} = 70$ meters

Therefore, the length of the side along the river is $2x = 2 \times 70 = 140$ meters.

Top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.

Solution:

Find the area of the octagonal surface

We are given a regular octagon. The area of a regular octagon can be calculated using the formula:

Area = $2(1 + \sqrt{2}) \times a^2$, where $a$ is the length of the side of the octagon.

If the length of the side $a$ is given, substitute it in the formula to find the area.

There is a pentagonal shaped park as shown in the figure. For finding its area Jyoti and Kavita divided it in two different ways.
Find the area of this park using both ways. Can you suggest some other way of finding its area?

Solution:

Area of the Pentagonal Shaped Park

The pentagonal shaped park can be divided in two different ways to find its area. We will explore both methods.

Method 1: Dividing into Triangles

In this method, we divide the pentagon into 5 triangles. The area of each triangle can be calculated using the formula:

Area of a triangle = $ \frac{1}{2} \times \text{Base} \times \text{Height} $

Once the area of each triangle is found, we add them up to get the total area of the pentagonal park.

Method 2: Dividing into a Rectangle and Triangles

In this method, we divide the pentagon into a rectangle and two triangles. The area of the rectangle is found using:

Area of rectangle = $ \text{Length} \times \text{Width} $

Then, calculate the area of the two triangles using the same formula as above and add them to the area of the rectangle to get the total area of the park.

Other Possible Methods

Another way to find the area of a pentagon is to use the formula:

Area of a regular pentagon = $ \frac{1}{4} \times \sqrt{5(5 + 2\sqrt{5})} \times \text{Side}^2 $

This formula is applicable when all sides and angles of the pentagon are equal.

Diagram of the adjacent picture frame has outer dimensions = 24 cm × 28 cm and inner dimensions 16 cm × 20 cm. Find the area of each section of the frame, if the width of each section is same.

Solution:

Diagram of the Picture Frame

Outer dimensions: $24 \, \text{cm} \times 28 \, \text{cm}$

Inner dimensions: $16 \, \text{cm} \times 20 \, \text{cm}$

Let the width of each section of the frame be $x$ cm.

Area of the outer rectangle = $24 \times 28 = 672 \, \text{cm}^2$

Area of the inner rectangle = $16 \times 20 = 320 \, \text{cm}^2$

Area of the frame = Area of the outer rectangle - Area of the inner rectangle

Area of the frame = $672 - 320 = 352 \, \text{cm}^2$

Each section of the frame has equal width $x$ cm. The width is the difference between the outer and inner dimensions divided by 2.

So, $x = \frac{(24 - 16)}{2} = 4 \, \text{cm}$.

Therefore, the width of each section is $4 \, \text{cm}$.