Sequences Problem #2

square sequence copy

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.

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5 thoughts on “Sequences Problem #2

  1. 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.

  2. 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

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