6..2.1 Understand the concept of perimeter (4)
6..2.2 Understand the concept of area (5)
This is our final set of perimeter and area tasks. This time the shapes involve circles.
Monday: The first part of this task is fairly straightforward. It involves four 3-by-3 squares and two quarter circles. The second part touches on something we have met before (Week 10): the smaller J's area is one ninth rather than one third of the larger J's area. The diagram is visually quite appealing and might prompt students to imagine other Js belonging to this family.
You might want to encourage students to write expressions such as this for the area of the larger J, rather than simply evaluate it:
4×3² + (𝛑×6² – 𝛑×3²)/4 ≈ (4 + 9/4)3² = 6.25×3².
In turn, the area of the smaller J is approximately 6.25×3² ÷ 3² = 6.25.
4×3² + (𝛑×6² – 𝛑×3²)/4 ≈ (4 + 9/4)3² = 6.25×3².
In turn, the area of the smaller J is approximately 6.25×3² ÷ 3² = 6.25.
Tuesday: In this task, the area of the sets of circles clearly increases. However, it turns out that the perimeter doesn't - which may well surprise students at first.
We can summarise the situation like this:
for n circles, each circle has a diameter of 6/n, so their total perimeter (in cm) is n × 𝛑 × 6/n = 6𝛑.
for n circles, each circle has a diameter of 6/n, so their total perimeter (in cm) is n × 𝛑 × 6/n = 6𝛑.
Wednesday: This time the area stays the same - which might be less surprising, though it is still a nice result.
We can summarise the situation like this:
for n×n circles, each circle has a radius of 6/2n = 3/n, so their total area (in cm²) is n² × 𝛑 × (3/n)² = 9𝛑.
for n×n circles, each circle has a radius of 6/2n = 3/n, so their total area (in cm²) is n² × 𝛑 × (3/n)² = 9𝛑.
Thursday: This is a variant of the classic ropes-around-the-earth task. If we start with a circle, however large, and increase its radius by an amount e, we increase its circumference by 2𝛑e. Here the cycle tracks are equally wide, so the difference in radius of the middle and inner circle is the same as for the outer and middle circle. So the differences in the circumferences is also the same, namely 6 m.
Some students might approach the task by assigning a specific value to the radius of the inner circle (and using 𝛑 ≈ 3, to keep things simple). However, it is a moot point whether this will make the task easier, unless they carefully track the numbers, in this kind of way:
Let the inner circle have a radius (in m) of 40, say. Then its circumference (in m) will be 2×3×40. So the middle circle has a circumference of 2×3×40 + 6 and so its radius is (2×3×40 + 6)/2×3 = 40+1. So the radius of the outer circle will be 40+1+1, and its circumference will be 2×3×(40+1+1), which is 6 more than the middle circle.
Let the inner circle have a radius (in m) of 40, say. Then its circumference (in m) will be 2×3×40. So the middle circle has a circumference of 2×3×40 + 6 and so its radius is (2×3×40 + 6)/2×3 = 40+1. So the radius of the outer circle will be 40+1+1, and its circumference will be 2×3×(40+1+1), which is 6 more than the middle circle.
Friday: This is based on a task I came across on Twitter and which I wrote about in the March 2016 issue of Mathematics in School (Visualising situations that can vary). Students may well surmise that the area stays the same - but can they prove it?!
