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## What are plane figures?

Plane figures are two-dimensional surfaces that have no height or thickness.

## Area of plane figures in maths

In Mathematics, problems on the area of plane figures can come in two ways. Firstly, you may just be given a plane shape that may be regular or irregular to determine its areas, such as finding the area of any polygon, circle or cross.

Secondly, you may be asked to find the area of a plane surface derived from a 3 dimensional solid. For instance, if you are required to find the area of the base of a triangular prism, it certainly means you have been asked to find the area of a triangular plane since the base of that prism is triangular. Be vigilant because your approach depends on the format of the question.

## Types of area of plane figures

Plane figures or plane shapes are classified into** regular** and **irregular** figures.

### Regular plane figures

We have regular plane shapes when the interior angles and length of sides are equal. Examples of regular plane shapes are, for instance, a square, an equilateral triangle and a regular polygon (regular pentagon, regular hexagon, regular heptagon, regular octagon, etc.).

Image of regular plane figures, StudySmarter Originals

### Irregular plane figures

**Irregular plane figures **occur** **when either the interior angles or the length of their sides are unequal. Examples of irregular plane figures are a rectangle, a parallelogram, a non-equilateral triangle (right, isosceles, scalene), irregular quadrilaterals, irregular pentagons, irregular hexagons, etc.

Images of irregular plane figures, StudySmarter Originals

## How to solve the area of plane figures?

We will look into finding the area of several plane shapes. It is important to note that the unit of area is square units e.g. $c{m}^{2},f{t}^{2},{m}^{2}$^{ }etc.

### Triangles

A triangle is a 3-sided plane shape. The general formula used in calculating the area of a triangle is the product of half of its base and the height:

${A}_{triangle}=\frac{1}{2}\times base\times height=\frac{1}{2}\times b\times h.$

A triangular lawn of base length 12 m and height 15 m is to be cleared. Calculate the time it takes to clear the lawn if it takes 4 minutes to clear a square meter of the lawn.

**Solution:**

Since the lawn is a triangle, we need to find its area. We recall that the area of a triangle is given by,

${A}_{triangularlawn}=\frac{1}{2}\times b\times h$

Here b=12 m and h=15 m. Substituting into the formula, we have

${A}_{traingularlawn}=\frac{1}{2}\times 12\times 15=6\times 15=90{m}^{2}$

We recall that $1{m}^{2}$ of lawn takes 4 minutes to be cleared, then $90{m}^{2}$ will take,

$4mins\times 90=360mins\phantom{\rule{0ex}{0ex}}$

But, 1 hr is 60 mins, thus, 360 mins are 6 hours.

Finally, it will take an individual 6 hours to clear the lawn.

### Quadrilaterals

A quadrilateral is a 4-sided plane shape where the sum of their interior angles is equal to 360 degrees. There are several quadrilaterals, each has a formula for calculating its area.

#### Rectangle

A rectangle has all interior angles equal to 90 degrees and has its opposite sides equal.

The area of a rectangle is given by,

$Are{a}_{rec\mathrm{tan}gle}=length\times breadth$#### Square

A square has all its interior angles equal to 90 degrees. Also, it has all its sides equal.

The area of the square is given by,

$Are{a}_{square}=length\times breadth={l}^{2}$A sandy rectangular garage with a length of 16 m and a width of 18 m is to be floored with 2 m of square-shaped tiles. How many tiles would be needed to complete the flooring?

**Solution:**

We first need to calculate the size of the rectangular garage. To do this we calculate its area,

${A}_{rec\mathrm{tan}gulargarage}=l\times b=16\times 18=288{m}^{2}$

Next, we should find the size of the tile by calculating the area of the square,

${A}_{squaretile}={l}^{2}={2}^{2}=4{m}^{2}$

To derive how many tiles would floor the garage, we have

$numberoftiles=\frac{Are{a}_{rec\mathrm{tan}gulargarage}}{Are{a}_{squaretile}}=\frac{288}{4}=72.$

Thus, 72 tiles would be needed to floor the garage.

#### Rhombus

A rhombus is also called a diamond. It also has all its sides equal. Its opposite sides are parallel. Its area is calculated by multiplying its diagonals and dividing by 2.

The area of a Rhombus is given by,

$Are{a}_{Rhombus}=\frac{firstdiagonal\times seconddiagonal}{2}=\frac{{d}_{1}\times {d}_{2}}{2}.$The diagonals of a rhombus are 18 cm and 6 cm respectively, find the area of the rhombus.

**Solution:**

The area of a Rhombus is given by,** **

$Are{a}_{Rhombus}=\frac{{d}_{1}\times {d}_{2}}{2}=\frac{18\times 6}{2}=54c{m}^{2}$

#### Parallelogram

A parallelogram has its opposite sides parallel and equal. Its area is calculated by finding the product between the base and height of the parallelogram.

The area of a Parallelogram is given by,

$Are{a}_{paralle\mathrm{log}ram}=b\times h$

Find the area of a parallelogram with a base of 12 cm and a height of 30 cm.

**Solution:**

$Are{a}_{paralle\mathrm{log}ram}=b\times h=12\times 30=360c{m}^{2}$

#### Kite

A kite has its adjacent sides equal the opposite interior angles are equal. The area of a kite is calculated by finding the product between its two diagonals and dividing them by 2.

The area of a Kite is given by,

$Are{a}_{Kite}=\frac{{d}_{1}\times {d}_{2}}{2}$#### Trapezium

A trapezium is a two-dimensional shape, which has two of its sides parallel usually called the bases. Its area is derived by getting the sum of bases and multiplying by half of its height.

The area of a Trapezium is given by,

$Are{a}_{trapezium}=\frac{h(a+b)}{2}$

A trapezoidal-based cabinet with a base height of 15 cm and bases of 13 cm and 27 cm respectively, is placed in Finicky's room. Find the size of the floor it occupies.

**Solution:**

We have to find the area of the trapezoidal-based cabinet using the formula,

${A}_{paralle\mathrm{log}ram}=\frac{h(a+b)}{2}=\frac{15(13+27)}{2}=\frac{15\left(40\right)}{2}=300c{m}^{2}$

### Pentagon

A pentagon is a plane figure with 5 sides. It possesses an apothem which is the perpendicular distance from the midpoint of any of its sides to the center of the pentagon. The sum of all interior angles of a pentagon is 540 degrees while each interior angle equals 108 degrees for a regular pentagon.

We denote by **b** is the side length and **a** is the apothem,

$Are{a}_{Pentagon}=\frac{1}{4}\left(\sqrt{5(5+2\sqrt{5})}\right)\times {b}^{2}$

However, if the apothem is given, the area is thus

$Are{a}_{Pentagon}=\frac{5}{2}\times b\times a$

Find the area of a pentagon with one of its sides 6 cm.

**Solution:**

The area of a pentagon is calculated as,

${A}_{pentagon}=\frac{1}{4}\left(\sqrt{5(5+2\sqrt{5})}\right)\times {b}^{2}$

But b is 6 cm, thus

${A}_{pentagon}=\frac{1}{4}\left(\sqrt{5(5+2\sqrt{5})}\right)\times {6}^{2}=61.937c{m}^{2}$

### Hexagon

A hexagon is a 6-sided polygon. The sum of all interior angles in a hexagon equals 720 degrees and each interior angle of a regular hexagon equals 120 degrees.

Let b be the length of each side and a is the apothem, we have

$Are{a}_{Hexagone}=\frac{3\sqrt{3}}{2}\times {b}^{2}$

In the event that the apothem is given, the area becomes,

${\mathrm{Area}}_{Hexagon}=\frac{1}{2}\times \mathrm{a}\times \mathrm{perimeter}\mathrm{of}\mathrm{hexagon}\phantom{\rule{0ex}{0ex}}$

A regular hexagon, whose sides are all of equal length, we have

$Are{a}_{Regularhexagon}=\frac{1}{2}\times a\times 6b=3ab$

Find the area of a regular hexagon with each side 3.17 m and apothem 16.5 m.

**Solution:**

We use the formula for finding the area of a hexagon with a given apothem,

$A=3ab$

where a is 16.5 m and b is 3.17 m, then,

$Are{a}_{regukarhexagon}=3\times 16.5\times 3.17=156.915{m}^{2}$

### Circle

A circle is a round plane figure whose boundary (circumference) consists of points equidistant from a fixed point called the center. A line that passes the circle from one end of the circle to another through its center is the diameter. A line drawn from any part of the circle's circumference to the center of the circle is the radius.

The area of a circle is given by the formula,

$Are{a}_{circle}=\pi {r}^{2}$Find the area of a circle with a radius of 7 cm. Take id="2942085" role="math" $\pi =\frac{22}{7}.$

**Solution:**

Here r is 7 cm, and $\pi =\frac{22}{7}$ then

id="2942086" role="math" $Are{a}_{circle}=\pi {r}^{2}=\frac{22}{7}\times {7}^{2}=154c{m}^{2}$

### Some Irregular Plane shapes

It is easy to calculate the area of regular shapes because the formula is applied directly. However, when the area of irregular shapes is to be derived, it takes more than just the application of formulas.

The first thing to do is to find a connection or resemblance between the irregular shape and a regular shape. This will enable determine what formula can be used afterward.

Find the area of the figure below,

**Solution:**

The area of the figure above is best derived if the figure is broken into regular shapes.

So, the figure has been broken into rectangles A, B and C.

Let us find the area of each. We recall that the area of a rectangle is given by,

$Are{a}_{rec\mathrm{tan}gle}=length\times breadth$

Thus

$Are{a}_{A}=9\times 3=27c{m}^{2}$.

Now, to find the area of rectangle B, we need to find its length first.

$Lengt{h}_{rec\mathrm{tan}gleB}=3+9+3=15cm.$

Thus, the area of rectangle B is given by,

$Are{a}_{recatngleB}=lengt{h}_{rec\mathrm{tan}gleB}\times breadt{h}_{rec\mathrm{tan}gleB}=15\times 8=120c{m}^{2}$

But

$Are{a}_{A}=Are{a}_{C}=27c{m}^{2}.$

Hence, the area of the initial shape is the sum of the areas of the three rectangles.

$Are{a}_{irregularshape}=Are{a}_{A}+Are{a}_{B}+Are{a}_{C}=27+120+27=174c{m}^{2}$

Find the area of the unshaded portion in the diagram below. Take $\pi =\frac{22}{7}.$

**Solution:**

We define the known values in the above figure, the base of the triangle is of length 16 cm, the height of the triangle is of length 20 cm, and the radius r is 7 cm.

To find the area of the unshaded region which is indeed irregular, we note that the shape consists of a big triangle with an inscribed circle. The area of the unshaded region is the part not affected by the circle. Therefore,

$Are{a}_{unshadedregion}=Are{a}_{triangle}-Are{a}_{circle}$

But we recall that the area of the triangle can be calculated while using the formula,

$Are{a}_{triangle}=\frac{1}{2}base\times height=\frac{1}{2}\times 16\times 20=160c{m}^{2}.$

The area of the circle can be calculated using the formula,

$Are{a}_{Circle}=\pi {r}^{2}=\frac{22}{7}\times {7}^{2}=154c{m}^{2}.\phantom{\rule{0ex}{0ex}}$

Finally, the area of the unshaded region is,

$Are{a}_{unshadedregion}=Are{a}_{traingle}-Are{a}_{circle=}160-154=6c{m}^{2}.$

## Examples on area of plane figures

To improve your competence on problems regarding the area of plane figures, you are advised to practice more problems. Here are a few to strengthen your competence.

If A cuboidal block has a top surface dimension of 8 m by 5 m and a height 12 cm. If a trampoline cover is used to prevent birds' droppings from staining its top. What should be the minimum area of the trampoline such the top is top surface is totally covered.

**Solution:**

Be careful not to confuse the question for the total surface area of a cuboid as you have only been asked to find the surface area of the top of the block. Essentially, the trampoline should have at a minimum, the area of the top. Hence, the minimum area of trampoline needed is

$Area=8m\times 5m\phantom{\rule{0ex}{0ex}}=40{m}^{2}$

A trapezoidal frame with top and base distance of 4 m and 8 m respectively with a height of 6 m is to be made from a 2 m side square tile. How many square tiles are needed?

**Solution:**

The first step is to find the area of the frame which is a trapezium.

$Are{a}_{trapezium}=\frac{h(a+b)}{2}\phantom{\rule{0ex}{0ex}}Are{a}_{trapezium}=\frac{6m(4m+8m)}{2}\phantom{\rule{0ex}{0ex}}Are{a}_{trapezium}=\frac{6m(12m)}{2}\phantom{\rule{0ex}{0ex}}Are{a}_{trapezium}=\frac{72{m}^{2}}{2}\phantom{\rule{0ex}{0ex}}Are{a}_{trapezium}=36{m}^{2}\phantom{\rule{0ex}{0ex}}$

Next, find the area of the tile which is a square.

$Are{a}_{square}=length\times breadth={l}^{2}\phantom{\rule{0ex}{0ex}}Are{a}_{square}={\left(2m\right)}^{2}Are{a}_{square}=4{m}^{2}\phantom{\rule{0ex}{0ex}}$

Now we have the area of both the frame (trapezium) and the tile (square), let's find how many of the tiles can be used to make the trapezium.

$No.oftilesneeded=\frac{36{m}^{2}}{4{m}^{2}}\phantom{\rule{0ex}{0ex}}No.oftilesneeded=9\phantom{\rule{0ex}{0ex}}$

## Area of Plane Figures - Key takeaways

- Plane figures are two-dimensional surfaces that have no height or thickness.
- Plane figures are
Plane figures are irregular

The area is measured in square units.

The area of a plane figure is calculated depending on the type of its shape.

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##### Frequently Asked Questions about Area of Plane Figures

How to find the area of plane figures?

Area of plane figures are found by determining the shape and applying the right formula.

What is area of plane figures?

The area of plane figures is the space covered by the figure when placed on a surface.

What is an example of area of plane figures?

An example of an area of a plane figure would be to find the area of a rectangle which is the product of the length and breadth of the rectangle.

What are the different types of area of plane figures?

The different types of area of plane figures is dependent on the type of the plane shape.

What are the uses of area of plane figures?

Area of plane figures helps us know the space occupied by a plane shape. It also helps us derive the total surface area of solid shapes since the components of solids are plane surfaces.

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