A nice little problem in Analytical Geometry


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:

tricky from 9gag

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.

tricky bit 2

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

CodeCogsEqn (9)

The line EO passes through the origin and has a gradient     So the equation of this line is:

CodeCogsEqn (8)

 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:


CodeCogsEqn (1)

CodeCogsEqn (2)

CodeCogsEqn (3)

CodeCogsEqn (11)

CodeCogsEqn (12)

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:

First the gradient:

CodeCogsEqn (13)

and now the y-intercept using the point A (-5;0)

CodeCogsEqn (14)

CodeCogsEqn (15)

So for line AC

CodeCogsEqn (16)

The inclination of the line and some basic geometry

The inclination of this line will give us angle OAG

CodeCogsEqn (17)


CodeCogsEqn (18)

Area of the blue segment

So the area of the blue segment ACF is proportionally a fraction

CodeCogsEqn (19)

of the whole circle. Which gives us

CodeCogsEqn (20)

Area of triangle ADF

To get the area of triangle ADF we need the coordinates of D; we know the y-value is -5 so

CodeCogsEqn (21)

CodeCogsEqn (22)

CodeCogsEqn (23)

CodeCogsEqn (24)

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

CodeCogsEqn (25)

At last…the area of the yellow piece DCF!

Now we can get the area of the yellow piece DCF:

CodeCogsEqn (27)

CodeCogsEqn (28)

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!




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