The sides of a square are 16 cm in length.

Question

5a. The sides of a square are 16 cm in length. The midpoints of the sides of this square are joined to form a new square and four triangles. The process is repeated twice.
Let Xn denote the length of one of the equal sides of each new triangle.
Let An denote the area of each new triangle.
The following table gives the values of Xn and An, for 1 less than or equal to n less than or equal to 3.

N    1 2 3
Xn  8 __ 4
An 32 16 __


5b. The process described above is repeated. Find A6.


5c. Consider an initial square of side length k cm. The process describes above is repeated indefinitely. The total area of the shaded regions is k cm^2. Find the value of k.

Answer 1

You can dither with the Pythagorean Theorem or you can draw a diagram!

If you do you shall notice the each created square is one-half the area of the originator.

Using a geometric series

A₆ = A₁(1/2)⁵

A₆ = 16²(1/2)⁵

A₆ = 2⁸(2^-5)

A₆ = 2³

A₆ = 8

Answer 2

Hi Megan from Toledo...hope you had a great day at school

 

This is an terrific and interesting problem...I'm going to give you a hint in beginning to answer this question...but I won't do it for you because once you figure it out...you will see how simple it is. If you still have problems...contact me on Wyzant I will be glad to help.

 

Okay...imagine...better yet draw the first square. With that given, we know each side is equal And the midpoints are 8 cm in length...Right???  Now, if we draw another square inside the first square, with each corner intersecting at the midpoints it tell/shows us what??  Another hint...we now have the new square, with sides of 8 cm and a side of the new square which would be what?! ANOTHER Hint...an isosceles right triangle. So how do we find the measure of that side? How about the Pythagorean Theorem!! Ring a bell.

Okay...I've got you going...try to solve it from here.

Like I said...if you still have issues...contact me on Wyzant...and I will certainly be glad to help you...but I'm not answering it for you.

Okay...go to it