Five rectangles are arranged in a row. Each rectangle is half as tall as the previous one. Also, each rectangle's width is half its height. The first rectangle is

32

cm. tall. What is the sum of the areas of the five rectangles? (Do not include cm. in your answer)

The first rectangle's area is 32cm*16cm = 512cm^2

Each subsequent rectangle is half the width and height, meaning it's 1/4 the area.

You can express this as a geometric sequence with first term 512, common ratio 1/4, and number of terms 5.

The sum of the geometric sequence is S = 512 (1 - 0.25^5) / (1 - 0.25) = 682cm^2.

I have no idea what it means by "do not include cm in your answer", is cm^2 considered including cm? There is no way to express the area without using units given that the areas are in units.

If you don't want it in cm^2, 682cm^2 = 0.0682m^2

Answer: 682

Also, each rectangle’s width is half its height. It can be expressed as

w = 1/2h

The first rectangle is 32cm tall. Then the measurement would be

h=32

w = 1/2h = 16

The first rectangle is 32cm tall and e ach rectangle is half as tall as the previous one. That can be expressed as

h (n) = 32 (1/2) ^n-1

The area of rectangle is

area (n) = h*w

area (n) = h * (1/2h) = 1/2h ^2

area (n) = 1/2 {32 (1/2) ^n-1}^2

area (n) = 512 (1/2) ^2n-2

Then the sum of the 5 rectangle would be:

area (1) + area (2) + area (3) + area (4) + area (5)

512 (1/2) ^2 (1) - 2 + 512 (1/2) ^2 (2) - 2 + 512 (1/2) ^2 (3) - 2 + 512 (1/2) ^2 (4) - 2 + 512 (1/2) ^2 (5) - 2 =

512 + 512/4 + 512/16 + 512 / 64 + 512 / 256=

512 + 128 + 32 + 8 + 2 = 682