A sequence of increasingly large squares is constructed as shown in the diagram.
Each right angled triangle has a hypotenuse that bisects two sides of a square.
The area of the smallest square is 4cm2.
Find the nth term for the area of squares.
Find the area of the 9th right angled triangle.
Each square is equivalent in area to its neighbouring right-angled triangle, and both are 1/8 the area of the next square
Therefore S1 = 4; S2 = 32; S3 = 256; Sn = 8^n ÷ 2
T9 = S9 = 8^9 ÷ 2 = 2^26
that’s the 3rd variation of the nth term so far, but it works 🙂 Great method too.
How do we know each square is equivalent in area to its neighboring right-angled triangle?
Take the right angle triangle and draw it with squares on each side (like in pythag proof). Divide the hyp square into 4 equal squares. Each one is the same size as your yellow square. The shorter sides of the triangle have equal sized squares coming off them, as it’s an isosceles right-angled triangle. Rearrange the pythag equation a bit (using a^2 twice instead of a^2 and b^2) and you’ll end up with 1/2 a^2 = 1/4 c^2
In other words, 1/2 base x height = 1/4 of c^2, and we just showed that 1/4 c^2 is the same size as our yellow square.
Alternatively (and not really a proof but a demonstration)
The small square’s side is 2cm.
The hypotenuse of the triangle is therefore 4cm (see the isosceles lines)
As it’s isosceles, each of the shorter sides must therefore be √8 cm
Therefore its area is √8 × √8 ÷ 2 = 4cm²