The figure shows a sequence of squares inscribed in the first-quadrant angle formed by the line y = (1/2)x and the positive x-axis. Each square has two vertices on the x-axis and one on the line y = (1/2)x, and neighboring squares share a vertex. The first (smallest) square is 8 cm tall. How tall are the next four squares in the sequence? How tall is the nth square in the sequence?
To find the height of the next square in the sequence, we can observe that each square is similar to the previous one, with a scale factor of 2. This means that the height of each square is twice the height of the previous square.
Starting with a height of 8 cm for the first square, the heights of the next four squares are:
1st square: 8 cm
2nd square: 2 * 8 cm = 16 cm
3rd square: 2 * 16 cm = 32 cm
4th square: 2 * 32 cm = 64 cm
5th square: 2 * 64 cm = 128 cm
The nth square in the sequence will have a height of 8 * 2^(n-1) cm, where n is the position of the square in the sequence.