Puzzles

Let \(S\) be the area of the large square, \(T\) be the area of one of the large triangles, \(U\) be one of the red overlaps and V be the uncovered blue square. We can write $$S=4T-4U+V$$ as the area of the square is the total of the four triangles, take away the overlaps as they have been double counted, add the blue square as it has been missed.

We know that 4U=V, so

Therefore one of the triangles covers one quarter of the square.

Five congruent triangles are drawn in a regular pentagon. The total area which the triangles overlap (red) is equal to the area they don't cover (blue). What proportion of the area of the large pentagon does each triangle take up?

\(n\) congruent triangles are drawn in a regular \(n\) sided polygon. The total area which the triangles overlap is equal to the area they don't cover. What proportion of the area of the large \(n\) sided polygon does each triangle take up?

Name the regions as follows:

\(E\) is a 1×1 square. Placed together, \(A\), \(C\), \(G\) and \(I\) also make a 1×1 square. \(B\) is equal to \(H\) and \(D\) is equal to \(F\).

Therefore \(B+E+F=A+C+D+G+H+I\). Therefore the hatched region is \(C\) larger than the shaded region. The area of \(C\) (and therefore the difference) is \(\frac{1}{4}\).

What is the difference between the shaded and the hatched regions in this dodecagon?

As our shape is a triangle, the 4cm and 5cm sides must be adjacent. Call the angle between them be \(\theta\).

The area of the triangle is \(\frac{1}{2}\times 4\times 5 \times \sin{\theta}\) or \(10\sin{\theta}\). This has a maximum value when \(\theta=90^\circ\), so the largest triangle has and area of 10cm² and looks like:

What is the largest area triangle with a perimeter of 12cm?

Let \(4x\) be the side length of the square. This means that the radius of the red circle is \(2x\) and the radius of a blue circle is \(x\). Therefore the area of the red circle is \(4\pi x^2\).

The area of one of the blue squares is \(\pi x^2\) so the blue area is \(4\pi x^2\). Therefore the two areas are the same.

Is the red or blue area larger?

Let \(A\) be the area of the square (and the triangle).

The length of a side of the square is \(\sqrt{A}\), so the perimeter of the square is \(4\sqrt{A}\).

Let \(l\) be the length of a side the triangle. Then \(\frac{1}{2}l^2\sin{60}=A\), so \(l^2=\frac{4A}{\sqrt{3}}\). Therefore \(l=\frac{2\sqrt{A}}{3^\frac{1}{4}}\) and the perimeter of the triangle is \(\frac{6\sqrt{A}}{3^\frac{1}{4}}\).

Hence the ratio of the perimeters is \(\frac{6\sqrt{A}}{3^\frac{1}{4}} : 4\sqrt{A}\) which simplifies to 3^3/4:2

If an \(n\) sided regular polygon has the area \(A\), what is the length of one of its sides?