(a) Cut the rectangle into two identical quadrilaterals.
(b) Cut the rectangle into two identical hexagons.
(c) Cut the rectangle into two identical heptagons.
a) You have a \(10\times20\) chocolate bar and 19 friends. Since you are good at maths they ask you to split this bar into 19 pieces (always breaking along the lines between squares). All the pieces have to be of a rectangular shape. Your friends don’t really care how much they will get, they just want to be special, so you need to split the bar in such way that no two pieces are the same.
(b) The friends are quite impressed by your problem solving skills. But one of them is not that happy with the fact you didn’t get a single piece of the chocolate bar. He thinks you might feel that you are too special, therefore he convinces the others that you should get another \(10\times20\) chocolate bar and now split it into 20 different pieces, all of rectangular shapes (and still you need to break along the lines between squares). Can you do it now?
Cut an equilateral triangle into 4 smaller equilateral triangles. Then can another equilateral triangle be cut into 7 smaller equilateral triangles (triangles do not necessarily have to be identical)?
Consider another equilateral triangle. Is it possible to cut it into (a) 9; (b) 16; (c) 28; (d) 2; (e) 42 smaller equilateral triangles (which are not necessarily identical)?
(f) Kyle claims he can cut an equilateral triangle into any number of smaller (not necessarily identical) equilateral triangles if this number is either greater than 8 and divisible by 3, or greater than 3 and has remainder 1 when divided by 3. Prove or disprove Kyle’s statement.
(g)* Let \(n\) be a natural number greater than 5. Is it true one can cut an equilateral triangle into \(n\) smaller equilateral triangles?