# Grade 6 maths is more difficult than you realized!

A friend linked me this from 9GAG this morning. He and his colleagues in the office were trying to figure out how to do the problem and sent out a cry for help: I messaged him back to say that I was pretty sure there was an error in the diagram: since the problem was for Grade 6 learners, the little piece at the bottom left was surely supposed to be coloured in.  But figuring out the area of that piece at the bottom is a nice little problem – not suitable for Grade 6 kids, but it will be good for my Grade 10s or 11s.

### Using Analytical Geometry

Analytical Geometry – which may be called Analytic Geometry or Cartesian Geometry – is a branch of algebra to model geometry on the Cartesian plane. This is one way to solve this problem that fits in neatly with our curriculum; like many other things in maths there are probably quite a few other approaches.

### Placing the figure on the Cartesian plane

I redrew the diagram on the Cartesian plane using the wonderful online graphing software at Desmos. The diagram below is to scale, but this is irrelevant for the calculations that follow. I arbitrarily selected the origin in the middle of the figure. Since the height of the rectangle was given as 10 units and the width as 20 units, each circle has a diameter of 10 units and hence a radius of 5 units.  So the points I have labelled have coordinates as follows:

A (-5;0)

E (-10; -5)

F(-5; -5)

The equation of the circle centred at A is The line EO passes through the origin and has a gradient     So the equation of this line is: ### Determine the equation of the green line CG

I added in a line that connects C to A. The point C is where the line EO intersects the circle centred at A, so we can calculate the coordinates of C by a simultaneous solution of these equations. The simplest way to do this is substitute  in place of y in the circle equation giving us:      We can see from the diagram that x=0 is the solution at the origin, but substituting x=-8 into into the original equation gives us the coordinates of C (-8;-4)

Since I have the coordinates of A and C, I can determine the equation of the line AC: and now the y-intercept using the point A (-5;0)  So for line AC ### The inclination of the line and some basic geometry

The inclination of this line will give us angle OAG Hence ### Area of the blue segment

So the area of the blue segment ACF is proportionally a fraction of the whole circle. Which gives us To get the area of triangle ADF we need the coordinates of D; we know the y-value is -5 so    We want the length as a scalar, we don’t care about the direction. The length of line AF is 5 units. So the area of triangle ADF is ### At last…the area of the yellow piece DCF!

Now we can get the area of the yellow piece DCF:  ### What about the green piece?

The the green piece in the diagram is a simple triangle; the coordinates of all the points have already been discovered, so I shall leave the calculation of this area as an exercise for the reader!

### Other solutions?

Did I do this the difficult way? Does anyone else have an alternative method of doing this? I would love to hear from you if you do!