Puzzles

The largest rectangle will be 11 by 12 and have area 132.

The largest dodecagon will be regular. It's area will be 172.

Let the radius of the small circles be \(r\). The are of half of one of these circles is \(\frac{1}{2}\pi r^2\).

The side of the square is \(2r\) and so the area of the square is \(4r^2\). Therefore the area of the whole shape is \((4+2\pi)r^2\).

By Pythagoras' Theorem, the radius of the large circle is \(r\sqrt{2}\). Therefore the area of the circle is \(2\pi r^2\). This means that the shaded area is \((4+2\pi)r^2 - 2\pi r^2\) or \(4r^2\).

This is the same as the area of the square, so the ratio is 1:1.

Draw on the following lines parallel to those which were added in the question.

Then a grid of copies of the smaller blue triangle has been created. Now consider the three triangles which are coloured green, purple and orange in the following diagram:

Each of these traingles covers half a parallelogram made from four blue triangles. Therefore the area of each of these triangles is twice the area of the small blue triangle.

And so the blue triangle covers one seventh of the large triangle.

If the sides of the triangle were split into \(n\) pieces the the lines added, what would the area of the smaller blue triangle be as a fraction of the area of the original large triangle?

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?