# Sequences Problem #2 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.

## 5 thoughts on “Sequences Problem #2”

1. Neil Pickup (@npickup) says:

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

2. solvemymaths says:

that’s the 3rd variation of the nth term so far, but it works 🙂 Great method too.

3. Tyler says:

How do we know each square is equivalent in area to its neighboring right-angled triangle?

4. solvemymaths says:

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.

5. Neil Pickup (@npickup) says:

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