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4.04 Areas of rectangles and triangles

Lesson

Areas of rectangles

The area of a rectangle is the size of the space within its boundary. This is the number of unit squares it takes to cover the shape. For a rectangle with length $5$5 centimetres and width $3$3 centimetres, how many squares of size $1$1 cm2 would it take to cover it?

If we divide the shape into a grid of $1$1 cm2, we can quickly see that it would take $3$3 rows of $5$5 squares to cover the shape, and so the area is $15$15 cm2. We can always take this approach of dividing a rectangle into a grid - and in doing so, we obtain the formula for the area of a rectangle:

Area of a rectangle

For rectangles:

$Area=\text{length}\times\text{width}$Area=length×width, this can be abbreviated to $A=l\times w$A=l×w

 

A square is a special rectangle where the length and width are equal, so we can use the following formula:

$A=l\times l$A=l×l, which simplifies to $A=l^2$A=l2

 

Note that units for area are length units squared. Don't forget to include units in your answer!

 

Practice questions

Question 1

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.

Question 2

Find the area of the attached figure.

A quadrilateral with all of its sides marked with a single tick mark, indicating that they are congruent. Each corners have a small square indicating right angles. One of the sides is labeled $11$11 cm, indicating its length.

Question 3

A kitchen floor is tiled with the tiles shown in the picture. If $30$30 tiles are needed to tile the floor, what is the total area of the floor? Give your answer in square centimetres.

 

Finding a side length

Sometimes we might know the area of a rectangle, and either the length or width. Using division, we can work out the missing value. In the case of a square, since it has equal length and width, if we know its area, we can work out its side length by taking the square root of the area.

 

Worked examples

Example 1

A rectangle has an area of $42$42 cm2 and a length of $7$7 cm, how wide is the rectangle?

Think: The area is $\text{length}\times\text{width}$length×width. To find the length we need to know what number multiplied by $7$7 is $42$42. We can write this formally in an equation and reverse the multiplication by dividing.

Do:

$A$A $=$= $L\times W$L×W

Write the formula.

$42$42 $=$= $7\times W$7×W

Substitute in known values.

$42\div7$42÷7 $=$= $W$W

Divide $42$42 by $7$7.

$\therefore W$W $=$= $6$6 cm

 

Example 2

A square has an area of $64$64 cm2, what is the side length of the square?

Think: The area is length multiplied by itself, to find the length we need to know what number multiplied by itself is $64$64. We can write this formally in an equation and use the square root to find the length.

Do:

$A$A $=$= $L^2$L2

Write the formula.

$64$64 $=$= $L^2$L2

Substitute in known values.

$\sqrt{64}$64 $=$= $L$L

Take the positive square root of $64$64.

$\therefore L$L $=$= $8$8 cm

 

 

Practice questions

Question 4

Find the width of a rectangle that has an area of $27$27 mm2 and a length of $9$9 mm.

Question 5

Find the perimeter of a square whose area is $49$49cm2.

 

Area of triangles

The area of a triangle is given by the formula $Area=\frac{1}{2}\text{base}\times\text{height}$Area=12base×height, which can be abbreviated to $A=\frac{1}{2}bh$A=12bh.

The height in this formula is referring to the height that is perpendicular to the base length. This may be one of the sides of the triangle (like in the right-angled triangle shown above), it may be within the triangle (like the second triangle above), or it may be outside of the triangle (like in the third triangle above).

Use the following applet to see a range of triangles and their related rectangles by moving the points and slider.

 

Worked example

Example 3

Find the area of the triangle pictured below.

Think: Identify the base and perpendicular height and use the formula. Here the base is $7$7 cm and the height perpendicular to this is $8$8 cm.

Do:

 

$A$A $=$= $\frac{1}{2}bh$12bh

Write the formula.

  $=$= $\frac{1}{2}\times7\times8$12×7×8 cm2

Substitute in known values and write down units.

  $=$= $\frac{1}{2}\times56$12×56 cm2

 

  $=$= $28$28 cm2

 

 

So the area of this triangle is $28$28 cm2.

 

Remember!

The rule for the area of a triangle is:

$Area=\frac{1}{2}\text{base}\times\text{height}$Area=12base×height

This can be abbreviated to:

$A=\frac{1}{2}bh$A=12bh

The base and height must be perpendicular, as shown in the diagrams below:

   

 

 

Practice question

Question 6

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. 

 

Finding an unknown measurement

Just as with rectangles we could be given the area and asked to find the base or height. To do so, we use the area formula and work backwards.

Worked example

Example 4

A triangle has an area of $24$24 cm2 and a base of $6$6 cm. What is the height of the triangle?

Think: The area is half the base multiplied by the height. To undo this, we can double the area and then divide by the given length. We can write this formally in an equation and rearrange the equation to find the height.

Do:

$A$A $=$= $\frac{1}{2}bh$12bh

Write the formula.

$24$24 $=$= $\frac{1}{2}\times b\times6$12×b×6

Substitute in known values.

$48$48 $=$= $6b$6b

Multiply both sides by $2$2.

$\frac{48}{6}$486 $=$= $\frac{6b}{6}$6b6

Divide $48$48 by $6$6.

$\therefore b$b $=$= $8$8 cm

 

 

Practice questions

Question 7

Find the value of $h$h in the triangle with base length $6$6 cm if its area is $54$54 cm2.

A triangle is has a base positioned vertically. The altitude of a triangle is drawn perpendicularly from the vertex of the triangle to the base, creating two right angled triangles. The altitude is the height of the triangle labeled h cm, and the base is labeled 6 cm.

Question 8

A gutter running along the roof of a house has a cross-section in the shape of a triangle. If the area of the cross-section is $50$50 cm2, and the length of the base of the gutter is $10$10 cm, find the perpendicular height $h$h of the gutter.

A right-angled triangle is drawn with a dashed line indicating its height 'h cm' dropping from the vertex opposite the right angle to the base. The base is labeled as 10cm. There are arrows on both ends of the base line and the height line indicating that they are measurements. The hypotenuse extends from the endpoint of the base opposite the right angle to the peak of the triangle.

Outcomes

1.3.8

calculate areas of rectangles and triangles, and composites of these shapes

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