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10. Shape and Measurement
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VCE 11 General 2023

10.04 Area

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Lesson

Measuring space

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

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

So this shape would have a total of $3$3 whole square units, $1$1 half square unit and $2$2 quarter square units resulting in a total area of:

$3+\frac{1}{2}+\frac{2}{4}=4$3+12​+24​=4 square units

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

 

Quadrilaterals

A quadrilateral is a polygon with four sides, four vertices and its interior angles add to $360^\circ$360°.

 

Square

Properties of a square:

  • all angles are $90^\circ$90°
  • all side lengths are equal
Area of a square

$\text{Area of a Square }=\text{length }\times\text{length }$Area of a Square =length ×length

$A=l\times l=l^2$A=l×l=l2

Practice questions

Question 1

Find the area of the square shown.

A square with a side measuring 12 cm, indicating its length.  All sides of the square have a single tick mark suggesting they are of equal length. Each corner of the square has a small square bracket, indicating a right angle.

 

Rectangle

Properties of a rectangle:

  • all angles are $90^\circ$90°
  • opposite side lengths are equal
Area of a rectangle

$\text{Area of a Rectangle }=\text{length }\times\text{width }$Area of a Rectangle =length ×width

$A=l\times w$A=l×w

Practice questions

Question 2

Find the area of the rectangle shown.

A rectangle with a length marked 9 m with double ticks on each opposing side and a width labeled 6 m with single ticks on each opposing side.  Small squares are drawn at each corner, signifying that it is a right angle.

 

Parallelogram

Properties of a parallelogram:

  • opposite side lengths are equal and parallel
Area of parallelogram

$\text{Area of a Parallelogram }=\text{Base }\times\text{Height }$Area of a Parallelogram =Base ×Height

$A=b\times h$A=b×h

Practice questions

Question 3

Find the area of the parallelogram shown.

 

A parallelogram with the horizontal top side labeled 15 cm, suggesting the measurement of its base. A vertical dashed line is drawn beyond the parallelogram, connected to a horizontal dashed line perpendicular to the horizontal top side. This vertical dashed line is labeled 14 cm, suggesting the measurement of the parallelogram's height perpendicular to the base.

 

Rhombus

Properties of a rhombus:

  • all side lengths are equal
  • opposite sides are parallel
Area of a rhombus

$\text{Area of a Rhombus }=\frac{1}{2}\times\text{diagonal 1}\times\text{diagonal 2}$Area of a Rhombus =12​×diagonal 1×diagonal 2

$A=\frac{1}{2}\times x\times y$A=12​×x×y

Practice questions

Question 4

Find the area of the rhombus shown.

A rhombus with its two diagonals, one horizontal and one vertical, is shown. The horizontal diagonal measures 9 cm, and the vertical diagonal measures 6 cm. The sides of the rhombus are marked by a single tick indicating that all four sides are congruent. A small square is placed at the upper-right of the intersection of the diagonals to denote a right angle, indicating that the diagonals are perpendicular to each other. 

 

Trapezium

Properties of a trapezium:

  • two sides are parallel
Area of a trapezium

$\text{Area of a Trapezium }=\frac{1}{2}(a+b)\times height$Area of a Trapezium =12​(a+b)×height

where $a$a and $b$b are the lengths of the two parallel sides.

Practice questions

Question 5

Find the area of the trapezium shown.

 

The trapezium with two parallel bases has a measurement of 8 cm on its top base and 11 cm on the bottom base. The distance between the two parallel bases is 6 cm. These bases connects with the lines on the side called legs and they are not labeled.

 

Question 6

Find the value of $x$x if the area of the trapezium shown is $65$65 cm2.

  1. Start by substituting the given values into the formula for the area of a trapezium.

    $A=\frac{1}{2}\left(a+b\right)h$A=12​(a+b)h

 

Kite

Properties of a kites:

  • adjacent pairs of sides are equal
  • the angles where the pairs meet are equal
Area of a kite

$\text{Area of a Kite}=\frac{1}{2}\times\text{diagonal 1}\times\text{diagonal 2}$Area of a Kite=12​×diagonal 1×diagonal 2

$A=\frac{1}{2}\times x\times y$A=12​×x×y

Practice questions

Question 7

Find the area of the kite shown.

Question 8

The area of a kite is $640$640 cm2 and one of the diagonals is $59$59 cm. If the length of the other diagonal is $y$y cm, what is the value of $y$y rounded to two decimal places?

Question 9

Find the shaded area shown in the figure.

 

Triangles

Properties of a triangle:

  • three sides
  • three angles
  • all interior angles add to $180^\circ$180°
Area of a triangle

$\text{Area of a triangle }=\frac{1}{2}\times\text{base }\times\text{height }$Area of a triangle =12​×base ×height

$A=\frac{1}{2}bh$A=12​bh

Practice questions

Question 10

Find the area of the triangle with base length $10$10 m and perpendicular height $8$8 m shown below.

 

A triangle is depicted on the image. Horizontal broken line, indicating the measurement of the base of the triangle, is labeled with 10 m. Vertical broken line, indicating the measurement of height of the triangle, is labeled with 8 m. 

 

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$3 side lengths are known, (no perpendicular heights or angles necessary).

Heron's formula

A triangle with sides of length $a$a, $b$b, $c$c has area given by

$A=\sqrt{s(s-a)(s-b)(s-c)}$A=√s(s−a)(s−b)(s−c),

where $s$s is the semiperimeter of the triangle (half the perimeter), calculated using $s=\frac{a+b+c}{2}$s=a+b+c2​

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.

Created with Geogebra

Practice questions

Question 11

What is the area of the triangle with sides of length $5$5 cm, $6$6 cm and $5$5 cm?

  1. First find the semi-perimeter $s$s.

  2. Hence find the area using Heron's formula.

Question 12

An isosceles triangle has an area $48$48 cm2 and the length of its unequal side is $16$16 cm.

Let $x$x be the length (in cm) of one of its equal sides.

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

  2. Hence solve for $x$x.

 

Circles

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

Created with Geogebra

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:

Area of a circle

$\text{Area of a circle}=\pi r^2$Area of a circle=πr2

Practice questions

Question 13

If the radius of the circle is $5$5 cm, find its area.

Give your answer as an exact value.

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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