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10. Shape and Measurement
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VCE 11 General 2023

10.04 Area

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Worksheet
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Lesson

Introduction

Area, is defined as the space within a 2D shape. All these shapes have the same area as they all contain 5 square units each.

The image shows 3 different composite shape made up of 5 square units. Ask your teacher for more information.

As well as whole unit squares, sometimes shapes might be composed of parts of unit squares. Take this shape for example.

The image shows a composite shape made up of different size of square. Ask your teacher for more information.

So this shape would have a total of 3 whole square units, 1 half square unit and 2 quarter square units resulting in a total area of: 3+\dfrac{1}{2}+\dfrac{2}{4} = 4\text{ square units}

Most shapes will not be made up of square however there are formulas for common shapes that will be useful to recall.

Square

A quadrilateral is a polygon with four sides, four vertices and its interior angles add to 360\degree.

Properties of a square:

  • all angles are 90\degree

  • all side lengths are equal

Area of a square:

\displaystyle \text{Area of a square}\displaystyle =\displaystyle \text{length} \times \text{length}
\displaystyle A\displaystyle =\displaystyle l \times l
\displaystyle =\displaystyle l^{2}

Examples

Example 1

Find the area of the square shown.

A square with side length of 12 centrimetres.
Worked Solution
Create a strategy

Use the formula for the area of a square.

Apply the idea
\displaystyle A\displaystyle =\displaystyle l\times lWrite the formula
\displaystyle =\displaystyle 12\times 12Substitute the values
\displaystyle =\displaystyle 144 \text{ cm}^2Evaluate
Idea summary

The area of the square is given by:

\displaystyle A=l\times l
\bm{A}
is the area of the square
\bm{l}
is the side length of the square

Rectangle

Properties of a rectangle:

  • all angles are 90\degree

  • opposite side lengths are equal

Area of a rectangle:

\displaystyle \text{Area of a rectangle}\displaystyle =\displaystyle \text{length} \times \text{width}
\displaystyle A\displaystyle =\displaystyle l \times w

Examples

Example 2

Find the area of the rectangle shown.

A rectangle with length of 12 centrimetres and a width of 9 centrimetres.
Worked Solution
Create a strategy

Use the area of a rectangle formula.

Apply the idea
\displaystyle A\displaystyle =\displaystyle l\times wWrite the formula
\displaystyle =\displaystyle 12\times 9Substitute the values
\displaystyle =\displaystyle 108 \text{ cm}^2Evaluate
Idea summary

The area of the rectangle is given by:

\displaystyle A=l\times w
\bm{A}
is the area of the rectangle
\bm{l}
is the length of the rectangle
\bm{w}
is the width of the rectangle

Parallelogram

Properties of a parallelogram:

  • opposite side lengths are equal and parallel

Area of parallelogram:

\displaystyle \text{Area of a Parallelogram }\displaystyle =\displaystyle \text{Base} \times \text{Height}
\displaystyle A\displaystyle =\displaystyle b \times h

Examples

Example 3

Find the area of the parallelogram shown.

A parallelogram with a base of 15 centimetres and perpendicular height of 8 centimetres.
Worked Solution
Create a strategy

We can find the area of parallelogram using the formula: A=b \times h.

Apply the idea
\displaystyle A\displaystyle =\displaystyle \text{base} \times \text{height}Use the parallelogram area formula
\displaystyle =\displaystyle 15 \times 8Substitute the dimensions
\displaystyle =\displaystyle 120\text{ cm}^{2}Evaluate the product
Idea summary

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

\displaystyle A=b\times h
\bm{b}
is the base of the parallelogram
\bm{h}
is the perpendicular height

Rhombus

Properties of a rhombus:

  • all side lengths are equal

  • opposite sides are parallel

Area of a rhombus:

\displaystyle \text{Area of a Rhombus}\displaystyle =\displaystyle \dfrac{1}{2} \times \text{diagonal } 1 \times \text{diagonal } 2
\displaystyle A\displaystyle =\displaystyle \dfrac{1}{2} \times x \times y

Examples

Example 4

Find the area of the rhombus shown.

A rhombus with a diagonal lengths of 6 centimetres and 9 centimetres.
Worked Solution
Create a strategy

We can find the area of rhombus using the formula: A=\dfrac{1}{2}\times x \times y.

Apply the idea
\displaystyle A\displaystyle =\displaystyle \dfrac{1}{2}\times x \times yUse the rhombus area formula
\displaystyle =\displaystyle \dfrac{1}{2}\times 9 \times 6Substitute the dimensions
\displaystyle =\displaystyle 27\text{ cm}^{2}Evaluate the product

Example 5

Find the shaded area shown in the figure.

A rhombus with another rhombus inside. The rhombus has a diagonal lengths of 12 centimetres and 16 centimetres.
Worked Solution
Create a strategy

Subtract the area of the inner rhombus from the outer rhombus.

Apply the idea
\displaystyle \text{Area of Outer Rhombus}\displaystyle =\displaystyle \dfrac{1}{2}\times 16 \times 12Substitute x=16,\,y=12
\displaystyle =\displaystyle 96\text{ cm}^2Evaluate
\displaystyle \text{Area of Inner Rhombus}\displaystyle =\displaystyle \dfrac{1}{2}\times 12 \times 12Substitute x=12,\,y=12
\displaystyle =\displaystyle 72\text{ cm}^2Evaluate

Subtract the area of the outer rhombus from the area of the inner rhombus:

\displaystyle \text{Shaded area}\displaystyle =\displaystyle 96-72Subtract 72 from 96
\displaystyle =\displaystyle 24\text{ cm}^2Evaluate
Idea summary

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

\displaystyle A=\dfrac{1}{2}\times x \times y
\bm{x}
is the length of one diagonal
\bm{y}
is the length of the other diagonal

Trapezium

Properties of a trapezium:

  • two sides are parallel

\text{Area of a Trapezium }= \dfrac{1}{2}(a+b)\times \text{height}, where a and b are the lengths of the two parallel sides.

Examples

Example 6

Find the area of the trapezium shown.

A trapezium with parallel side lengths of 8 centimetres and 11 centimetres, and a height of 6 centimetres.
Worked Solution
Create a strategy

We can find the area of trapezium using the formula:A= \dfrac{1}{2}(a+b)h

Apply the idea

The side lengths are a=8\ \text{cm} and b=11\ \text{cm}, and the height is h=6 cm.

\displaystyle A\displaystyle =\displaystyle \dfrac{1}{2}(a+b)hUse the formula
\displaystyle =\displaystyle \frac{1}{2}\times \left(8+11\right)\times 6Substitute the values
\displaystyle =\displaystyle 57\ \text{cm}^2Evaluate the product

Example 7

Find the value of x if the area of the trapezium shown is 52.5\ \text{cm}^{2} .

A trapezium with parallel side lengths of 8 centimetres and 13 centimetres, and a height of x centimetres.
Worked Solution
Create a strategy

We can find the area of trapezium using the formula:A= \dfrac{1}{2}(a+b)h

Apply the idea

We are given: A = 52.5,\, a= 8,\,b=13,\, and h=x.

\displaystyle A\displaystyle =\displaystyle \dfrac{1}{2}(a+b)hUse the formula
\displaystyle 52.5\displaystyle =\displaystyle \dfrac{1}{2}(8 + 13)xSubstitute the values
\displaystyle 52.5\displaystyle =\displaystyle \dfrac{1}{2}\times 21 xEvaluate the sum inside the brackets
\displaystyle 52.5\displaystyle =\displaystyle 10.5xEvaluate the product
\displaystyle x\displaystyle =\displaystyle \dfrac{52.5}{10.5}Divide both sides by 10.5
\displaystyle =\displaystyle 5\text{ cm}Evaluate
Idea summary

The area of a trapezium is given by:

\displaystyle A=\dfrac{1}{2}(a+b)h
\bm{a}
is a parallel side length
\bm{b}
is the other parallel side length
\bm{h}
is the perpendicular height

Kite

Properties of a kites:

  • adjacent pairs of sides are equal

  • the angles where the pairs meet are equal

Area of a kite:

\displaystyle \text{Area of a Kite}\displaystyle =\displaystyle \dfrac{1}{2}\times \text{diagonal } 1 \times \text{diagonal } 2
\displaystyle A\displaystyle =\displaystyle \dfrac{1}{2} \times x \times y

Examples

Example 8

Find the area of the kite shown.

A kite with diagonal lengths of 10 centimetres and 6 centimetres.
Worked Solution
Create a strategy

We can use the formula: A=\dfrac{1}{2}\times x \times y.

Apply the idea
\displaystyle \text{Area }\displaystyle =\displaystyle \frac{1}{2}\times x \times yUse the formula
\displaystyle =\displaystyle \frac{1}{2}\times 6\times 10Substitute the diagonal lengths
\displaystyle =\displaystyle 30\ \text{cm}^2Evaluate

Example 9

The area of a kite is 308 \text{ cm}^{2} and one of the diagonals is 47 cm. If the length of the other diagonal is y cm, find the value of y, rounded to two decimal places.

Worked Solution
Create a strategy

We can use the formula: A=\dfrac{1}{2}\times x \times y.

Apply the idea

We are given: A=308 and x=47.

\displaystyle A\displaystyle =\displaystyle \dfrac{1}{2}\times x \times yUse the formula
\displaystyle 308\displaystyle =\displaystyle \dfrac{1}{2}\times 47 \times ySubstitute the values
\displaystyle 616\displaystyle =\displaystyle 47yMultiply both sides by 2
\displaystyle y\displaystyle =\displaystyle \dfrac{616}{47}Divide both sides by 47
\displaystyle =\displaystyle 13.11\text{ cm}^{2}Evaluate
Idea summary

The area of a kite can be found using the same formula in finding the area of a rhombus.

Triangles

Properties of a triangle:

  • three sides

  • three angles

  • all interior angles add to 180\degree

Area of a triangle:

\displaystyle \text{Area of a triangle}\displaystyle =\displaystyle \dfrac{1}{2}\times \text{base} \times \text{height}
\displaystyle A\displaystyle =\displaystyle \dfrac{1}{2}bh
Triangle with a height of h and base of b.

Examples

Example 10

Find the area of the triangle with base length 10\text{ m} and perpendicular height 8\text{ m} shown below.

Triangle with a height of 8 metres and base of 10 metres.
Worked Solution
Create a strategy

We can find the area of triangle using the formula: \text{A}=\dfrac{1}{2}bh.

Apply the idea
\displaystyle \text{Area of triangle}\displaystyle =\displaystyle \dfrac{1}{2} \times 10 \times 8Susbstitute b=10,\,h=8
\displaystyle =\displaystyle 40\text{ m}^2Evaluate
Idea summary

The area of a triangle is given by:

\displaystyle A=\dfrac{1}{2}bh
\bm{b}
is the base
\bm{h}
is the perpendicular height

Heron's formula

There is another method for finding the area of a triangle, it's called Heron's Formula. Heron's Formula is used when all 3 side lengths are known, (no perpendicular heights or angles necessary).

A triangle with angles A, B and C and opposite sides of lower case A, B, and C.

A triangle with sides of length a,\, b,\, c has area given by A=\sqrt{s\left(s-a \right) \left(s-b \right)\left(s-c \right)} where s is the semiperimeter of the triangle (half the perimeter), calculated using s= \dfrac{a+b+c}{2}.

Exploration

This applet will allow you manipulate a triangle. Heron's formula calculation can be seen and compared with the more common formula using base and height.

Loading interactive...

The applet shows a triangle with the side lengths, semiperimeter and area using Heron's formula. When we click the box next to Compare, it will show the perpendiular height and the Area of the triangle using its formula. It will also show in the triangle the perpendicular height in dashed line.

Examples

Example 11

Consider a triangle with side lengths 5\text{ cm}, \, 6\text{ cm} and 5\text{ cm}.

a

First find the semi-perimeter s.

Worked Solution
Create a strategy

We can use the formula: s= \dfrac{a+b+c}{2}, where a,\,b and c are the side lengths.

Apply the idea
\displaystyle s\displaystyle =\displaystyle \dfrac{5+6+5}{2}Substitute a=5,\,b=6 and c=5
\displaystyle =\displaystyle \dfrac{16}{2}Add the numbers in the numerator
\displaystyle =\displaystyle 8\text{ cm}^2Evaluate
b

Hence find the area using Heron's formula.

Worked Solution
Create a strategy

We can use the Heron's formula: A=\sqrt{s\left(s-a \right) \left(s-b \right)\left(s-c \right)}, where a,\,b and c are the side lengths, and s is the semi-perimeter.

Apply the idea
\displaystyle A\displaystyle =\displaystyle \sqrt{8\left(8-5 \right) \left(8-6 \right)\left(8-5 \right)}Substitute a=5,\,b=6,\,c=5 and s=8
\displaystyle =\displaystyle \sqrt{8\times 3 \times 2 \times 3}Evaluate the subtraction inside the brackets
\displaystyle =\displaystyle \sqrt{144}Evaluate the product
\displaystyle =\displaystyle 12\text{ cm}^2Simplify

Example 12

An isosceles triangle has an area 48\text{ cm}^{2} and the length of its unequal side is 16\text{ cm}. Let x be the length of the equal sides.

a

Find an expression for the semi-perimeter of the triangle.

Worked Solution
Create a strategy

We can use the formula: s= \dfrac{a+b+c}{2}, where a,\,b and c are the side lengths.

Apply the idea
\displaystyle s\displaystyle =\displaystyle \dfrac{x+x+16}{2}Substitute a=x,\,b=x and c=16
\displaystyle s\displaystyle =\displaystyle \dfrac{2x+16}{2}Add the numbers in the numerator
\displaystyle s\displaystyle =\displaystyle x+8Evaluate
b

Hence solve for x.

Worked Solution
Create a strategy

We can use the Heron's formula: A=\sqrt{s\left(s-a \right) \left(s-b \right)\left(s-c \right)}, where a,\,b and c are the side lengths, and s is the semi-perimeter.

Apply the idea
\displaystyle \sqrt{(x+8)(x+8-x)(x+8-x)(x+8-16)}\displaystyle =\displaystyle 48Substitute a=x,\,b=x,\,c=16,\\s=x+8 and A=48
\displaystyle \sqrt{(x+8)\times 8 \times 8(x-8)}\displaystyle =\displaystyle 48Simplify the expressions in the brackets
\displaystyle 8\sqrt{(x+8)(x-8)}\displaystyle =\displaystyle 48Use the surd law \sqrt{A^{2}B}=A\sqrt{B}
\displaystyle \sqrt{(x+8)(x-8)}\displaystyle =\displaystyle 6Divide both side by 8
\displaystyle (x+8)(x-8)\displaystyle =\displaystyle 36Square both sides
\displaystyle x^{2}-64\displaystyle =\displaystyle 36Expand (x+8)(x-8)
\displaystyle x^{2}-64+64\displaystyle =\displaystyle 64+36Add 36 to both sides
\displaystyle x^{2}\displaystyle =\displaystyle 100Evaluate
\displaystyle x\displaystyle =\displaystyle 10Take the square root of each side
Idea summary

To find the area of a triangle when we know the lengths of all 3 sides, we can use Heron's Formula:

\displaystyle A=\sqrt{s\left(s-a \right) \left(s-b \right)\left(s-c \right)}
\bm{s}
is the semi-perimeter of the triangle
\bm{a, \,b, \,c}
side lengths of the triangle

To find the semi-perimeter we can use the equation:

\displaystyle s= \dfrac{a+b+c}{2}
\bm{a, \,b, \,c}
side lengths of the triangle

Circles

To find the area of a circle we know there is a rule involving \pi. The following investigation will demonstrate what happens when a circle is cut into segments and unraveled to approximate the area.

Exploration

Take a look at the following applet by moving the sliders:

Loading interactive...

The more triangles we use, the closer this area gets to the area of the circle, and the closer the base of the resulting parallelogram gets to being half of the circumference of the circle. So we can see that the area of a circle is given by \pi r^2.

When the segments are realigned, an approximation of a parallelogram is formed. In a circle, the more segments that are cut make a shape where the base is half the circumference and the height is the radius. This leads to the following area formula: \text{Area of a circle}=\pi r^{2}.

Examples

Example 13

If the radius of the circle is 5\text{ cm}, find its area correct to 1 decimal place.

Worked Solution
Create a strategy

We can use the formula: A=\pi r^{2}.

Apply the idea
\displaystyle A\displaystyle =\displaystyle \pi \times (5)^{2}Substitute r=5
\displaystyle =\displaystyle 78.5 \text{ cm}^2Evaluate
Idea summary

We can calculate the area of a circle using the formula:

\displaystyle A=\pi r^2
\bm{A}
is the area
\bm{r}
is the radius of the circle

This formula can also be used to find the radius if we know the area of the circle.

Outcomes

U2.AoS4.5

the perimeter and areas of triangles (using several methods based on information available), quadrilaterals, circles and composite shapes, including arcs

U2.AoS4.11

calculate the perimeter and areas of triangles (calculating the areas of triangles in practical situations using the rules A=1/2 bh, A=1/2 ab sin(c) or A=\sqrt{s(s-a)(s-b)(s-c)} where s=(a+b+c)/2

U2.AoS4.12

use quadrilaterals, circles and composite shapes including arcs and sectors in practical situations

U2.AoS4.13

calculate the perimeter, areas, volumes and surface areas of solids (spheres, cylinders, pyramids and prisms and composite objects) in practical situations, including simple uses of Pythagoras’ in three dimensions

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