One of the great areas of confusion for students in the
measurement strand is Area and Perimeter, in fact it sometimes
seems that there is another term out there, “Arimeter”.
The confusion probably stems from this concept:
6 cm
The Perimeter is 2×(6 + 2) cm
2 cm
The Area is 6 × 2 cm2
While this is correct students who learn the process simply
remember that they have to do something with the 6 and the 2
but they don’t quite know what.
Introductory Activity 2
• Find all the perimeters of rectangles that have an area of 12 cm2.
• Students draw all the rectangles they can think of that have an area of 12 cm2.
• Is a 6 by 2 rectangle the same as a 2 by 6 rectangle?
Answer: Maybe, mathematically
speaking the answer is yes, it is, but
what happens if you have a long
narrow back yard you might want a
2 m by 6 m garden not a 6 m by 2 m
garden.
• Students now calculate each
perimeter using the TI-15
ExplorerTM calculator then they
scroll through their calculations to
find the greatest and the least
perimeter.
Introductory Activity 2
Area = 12 cm2
Length in cm
Breadth in cm
Perimeter in cm
Significant Questions:
• What did you notice about the perimeter as the area changed?
• Can you see a pattern?
• Can you see a quick way to calculate the perimeter?
• Can you describe the shape with the smallest perimeter?
• What happens to the perimeter if the shape is irregular, for example a
“T” or an “L” or an “H” shape?
Using the Scroll Feature
Calculate the perimeters of the rectangles on the
previous slide:
2
×
(
12
+
1
)
Enter
=
Or you could just key in 12 + 1 + 12 + 1 =
Your screen for the first one should look like this:
2 x ( 12 + 1 ) =
26
Repeat this for all the other rectangles.
Then use the up and down scroll keys to review
all of your answers.
Screen Shots from the Scroll Process
2 x ( 12 + 1 ) = 26
2 x ( 6 + 2 ) = 16
2 x ( 4 + 3 ) = 14
2 x ( 3 + 4 ) = 14
2 x ( 2 + 6 ) = 16
2 x ( 1 + 12 ) = 26
The scroll button only scrolls back one entry at a time so in this instance
there would be 6 screen shots.
From this it is an easy step to make the deduction that the rectangle with
the smallest perimeter but still with an area of 12 cm2 is the 4 by 3
rectangle (or the 3 by 4 rectangle).
Perimeter and Area
Students now repeat this process but with the
perimeter fixed at 12 cm and different areas.
Using graph paper students should produce
something like this:
NB use integer values only at this stage.
• A rectangle has a perimeter of 12 cm.
Find all the areas of rectangles that have an
perimeter of 12 cm.
• Students draw all the rectangles they can think
of that have a perimeter of 12 cm.
• Students now calculate each area using the
TI-15 ExplorerTM calculator then they scroll
through their calculations to find the greatest
and the least area.
Students repeat the earlier activities using
different areas and perimeters.
See Session 2b.
Session 2(a)
Exploring Area with a Fixed Perimeter
Perimeter = 12 cm
Length in cm
Breadth in cm
Area in cm2
5
1
5
4
2
8
3
3
9
2
4
8
1
5
5
Significant Questions:
• Can you see a pattern?
• What can you say about the area as the shape changes?
• Can you describe the shape with the greatest area?
Worksheet 1
Worked Example for Part 1
Sample Question:
Use your TI-15 ExplorerTM calculator to find the smallest possible and the largest possible
perimeter for the following rectangles:
Rectangle with an area of 24 cm2
Smallest perimeter______20 cm______ (using only integer dimensions)
Largest perimeter______50 cm______
Working:
If the area is 24 cm2 then possible rectangles are: 24x1, 12x2, 8x3 and 6x4 using only integers.
The perimeters of these rectangles are: 24 + 1 + 24 + 1 or 2 x (24+1) etc.
If we use the calculator the screens should look something like this:
2 x (24 + 1) = 50
2 x (12 + 2) = 28
2 x (8 + 3) = 22
2 x (6 + 4) = 20
It is now a simple matter to scroll back through your answers to find the smallest and the largest.
Worksheet 1
Worked Example for Part 2
Sample Question:
Use your TI-15 ExplorerTM calculator to find the smallest possible and the largest possible area for the following
rectangles:
Rectangle with a perimeter of 18 cm
Smallest area_______8 cm2_______
Largest area_______20 cm2_______
Working:
If the a perimeter is 18 cm then possible rectangles are: 8x1, 7x2, 6x3, and 5x4 using only integers. Hint: to
figure out your rectangles first halve the perimeter, this gives you the sum of 2 sides, now you can work out all
your different rectangles.
8 cm
Eg 1 cm
1 cm Perimeter = 8 + 1 + 8 + 1 = 18
8 cm
so the area is 8 x 1 = 8 cm 2
If we use the calculator the screens should look something like this:
8x1=8
7 x 2 = 14
6 x 3 = 18
5 x 4 = 20
It is now a simple matter to scroll back through your answers to find the smallest and the largest.
As the perimeters get larger the number of possible rectangles increases for example the family of rectangles
with a perimeter of 40 cm consists of 10 rectangles if we stick to integers, if we use rational numbers (fractions
etc) the number becomes infinite.
Assessment Task
The Calf Paddock (and Calf Pen)
The maximum area without the hayshed wall would be a square with 9 m side lengths.
There are many possible pens with 3 sides measuring a total of 36 m for the paddocks that use the
hayshed wall, 2 are shown here. Many students assume that a square pen (blue)will have the greatest
area based on the earlier answers but in this case the rectangular pen (red) has the greatest area. Why?
Hay Shed Wall
Red Paddock Area = 162 m2
9 + 18 + 9 = 36 m
Green Paddock Area = 144 m2
12 + 12 + 12 = 36 m
Circle Facts You Need to Know
Remember π “the circle number”.
Circumference (the perimeter of the circle)
Radius
Diameter (2 x Radius); cuts the circle in half.
Area: the pale blue shaded bit.
The circumference = 2 x π x the radius written as 2 x π x r and abbreviated
to 2πr or π x diameter (πd).
The area of a circle = π x the (radius)2 or π x r2 (abbreviated to πr2).
Moving Away from Rectangles
Is there another shape, other than a square, that will maximise area and
minimise perimeter.
Students test other shapes using cm grid paper and measurement to
determine perimeter and area.
Students estimate the area by counting the squares and develop an idea
that maybe the circle has the greatest area for the least perimeter.
Session 3: More Calf Pens
If the farmer could build whatever shape he wanted to, what would his calf
pen look like if he wanted to get the maximum area from his 36 m of fencing.
Again most students will say that a circle would give him the greatest area.
The calculation would be something like this:
If the circumference is 36 m and the circumference is also
2 × π × the radius then:
2 × π × the radius = 36 m
So the radius is 36 m ÷ (2 × π ) = 5.73
Students should know from the “finding pi” unit that:
The Area of a Circle = π × (radius)2 = π × 5.732 = 103.1 m2
Remember the maximum area for the rectangle was 81 m2
But: What happens with the Hay Shed?
What Happens with the Hayshed?
A semicircle with a radius of 11.5 m has a curved side of 36.1 m, pretty
close to the farmers 36 m of allowable fencing.
The Area is ½ (π × 11.52 ) = 207.7 m2 That’s 51 calves he can
accommodate.
Hay Shed Wall
Give students time
to figure this out for
themselves.
Each square represents
1 square metre