Brian was repairing some fencing in the garden. He was replacing feather board in a section of the fence. The gap between the posts was 2.34m and each piece of feather board was 14.5cm wide. He wanted to use 20 pieces of feather board in the gap and wanted to know what overlap to allow. He needed to know the overlap to help him nail up neatly and consistently.
After some false starts, my thinking eventually went something like this.
20 pieces with 19 overlaps. Let an overlap be x cm so …
19(14.5-x) + 14.5 = 234
But it seemed rather like using a hammer to crack a nut – resorting to algebra. Of course, as an alternative, I could have iterated!
Then it occurred to me that I was looking at the problem the wrong way round – I should be thinking of the 20 pieces of feather board first… 20 x 14.5cm = 290cm. This gives me how much there was available for the overlaps.
290cm – 234cm = 56cm
So the overlap needed to be 56/19 cm.
The interesting thing was that I was much happier with this non-algebraic answer. Also my original approach was so focused on the gap to start with that I could not see the alternative and I think, neater approach. But isn’t this typical?
I wonder what your pupils would do if given this problem. Of course they could use strips of paper and model the problem using trial and improvement.
I think this might make a good short problem to lead to discussion of different methods.
Seeing Things Differently – Hexagonal Numbers
A problem for secondary students familiar with triangle numbers. Try to give time for them to come up with their own structures that help them see ways of extending the sequence.
Seeing hexagonal numbers
This task and related ideas have been inspired by the “Twist and Lock” problem below.
Twist and Lock
I was visiting the Tate Modern in London and found these in the shop. A string of cubes linked together with some elastic and which allow you to twist each cube onto different sides of adjacent cubes. Find some related mathematical ideas here.
A version for Farhan
Just seen a reference to the NRICH problem ‘Farhan’s Poor Square’. Though I am not excited by this problem – it made me think of this:
- AB is the diameter of the circle on ABC and part of a diagonal of the large square.
- The large square has a side length of 2x.
- The triangle is isosceles and right angled.
What is the area of the large square that is left white?
Is it possible to make the grey shape larger and reduce the white area? What is the best you can do? Does a constraint that the grey shape has to remain similar matter? For example, you don’t have to stick to a semicircle, just a segment of a circle.
Diameter, circumference and Pi
Three tasks that are designed to help develop a firm understanding of the relationship between the diameter and the circumference of a circle and multiples of Pi.
Patterns and Coordinates
A series of whole class and group activities to explore, conjecture and generalise (using algebra if appropriate) in the context of coordinates in all four quadrants.
Inferring and Extending
Can you describe what you see in the images which introduce this problem and can you extend the idea if there were three circles and not two?