look hereOriginally Posted by sonymd23
ur answer is different from the teacher
he wrote the first one......
r1 = 1 x1=2^1/2
s1 = 2^1/2 x 2^1/2 = 2
r2= [r1^1/2 - (x1/2)^2]^1/2
= [1 - (2^1/2/2)^2]^1/2
2
r n= [r n-1 - (xn - 1/2)^2 ]^1/2
= ...............r 1
however i dont understand what the teacher wrote......
I can't answer questions 7 and 8, question 3's has been answered in the hyperlink, and I don't really feel like redoing my circle-square equation to match your teachers work (maybe you could turn in my old equation and tell him you found a different way )Originally Posted by sonymd23
Hello, sonymd23!
I have no idea what you teacher did for #1 . . .
Given a unit circle
The first iteration step is to inscribe square in
The second step is to inscribe circle in and another square in
Repeat this process to the step. .Find the area ofCode:* * * * * A*---------------*B *| \ / |* | \ / | * | \ / | * * | * | * * | / O \ | * | / \ | *| / \ |* D*---------------*C * * * * *
Since is a unit circle:
. . is an isosceles right triangle.
Since , then
. . The side of the inscribed square is
In general, if is the diameter of a circle,
. . the side of the inscribed square is:
The diameter of is . . . Hence, the side of is: .
The diameter of is . . . Hence, the side of is: .
Therefore, the areas of are: .
Hello again, sonymd23!
Here's #3 . . .
For the map given at the right:
(a) Represent the map with a network.
(b) Find the degree of each vertex.
(c) State whether the network is traceable and explain the reason.Code:(a) D o * * * * * * * * A o * * o E * * * * * * * * o * * * o B C
(b)
. .
(c) The network is traceable; there are exactly two odd vertices.
I would like to point out that your teacher is incorrect, the area of isn't equal to but rather
and the method he used to figure out is completely pointless, because you need to know and to do that you need to know
I'm thinking about emailing the academy and telling them about this...
The diagnol of the square, is equal to the diameter of the circle : (forgive my poor art skills)Originally Posted by sonymd23
and is the hypotenuse of a right triangleCode:* * * * * *---------------* *| \ |* | \ | * | \ | * * | \d | * * | \ | * | \ | *| \ |* *---------------* * * * * *
the area of the square is equal toCode:* * * * * E*---------------* *| \ |* | \ | * | \ | * * | \d | * * | \ | * | \ | *| \ |* D*---------------*F * * * * *
therefore you can use the pythagoreum theorum to solve forCode:* * * * * E*---------------* *| \ |* | \ | * | \ | * * | s \d | * * | \ | * | \ | *| s \ |* D*---------------*F * * * * *
so the area of any square inscribed in a circle is
now, let's find the area of square 1
so that's the area of the first square
The diameter of the second circle is the same as the length of one of the sides of the first square, that means that the area of the square is equal to the diameter squared, so substitute the previous numbers into our equation...
therefore the area of the second square is 1, lets go onto three...
therefore the area of the third square is 1/2, lets figure out the area of the "n"th square.
notice that each time you solve for the area of the square you divide the last diameter^2 (area) by 2 therfore...
therefore...
which can be simplified...
use the multiplicative rule of powers...
~
(I must say that this equation is ten times easier than your teacher's)