can any one plz do questions #1 , 3 , 7 and 8 thx
i already done the rest.....however i dont know how to solve the above questions.....so plz help me....by the way.....plz show ur steps thnx...
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 $\displaystyle O_1.$
The first iteration step is to inscribe square $\displaystyle S_1$ in $\displaystyle O_1.$
The second step is to inscribe circle $\displaystyle O_2$ in $\displaystyle S_1$ and another square $\displaystyle S_2$ in $\displaystyle O_2.$
Repeat this process to the $\displaystyle n^{th}$step. .Find the area of $\displaystyle S_1,\;S_2,\;S_3.$Code:* * * * * A*---------------*B *| \ / |* | \ / | * | \ / | * * | * | * * | / O \ | * | / \ | *| / \ |* D*---------------*C * * * * *
Since $\displaystyle O_1$ is a unit circle: $\displaystyle OA = OB = OC = OD = 1.$
. . $\displaystyle \Delta AOB$ is an isosceles right triangle.
Since $\displaystyle AO = BO = 1$, then $\displaystyle AB = \sqrt{2}.$
. . The side of the inscribed square $\displaystyle S_1$ is $\displaystyle \sqrt{2}.$
In general, if $\displaystyle d$ is the diameter of a circle,
. . the side of the inscribed square is: $\displaystyle s = \frac{d}{\sqrt{2}}$
The diameter of $\displaystyle O_2$ is$\displaystyle \sqrt{2}$ . . . Hence, the side of $\displaystyle S_2$ is: .$\displaystyle \frac{\sqrt{2}}{\sqrt{2}}\,=\,1$
The diameter of $\displaystyle O_3$ is $\displaystyle 1$ . . . Hence, the side of $\displaystyle S_3$ is: .$\displaystyle \frac{1}{\sqrt{2}}$
Therefore, the areas of $\displaystyle S_1,\;S_2,\;S_3$ are: .$\displaystyle (\sqrt{2})^2,\;(1)^2,\;\left(\frac{1}{\sqrt{2}} \right)^2\;=\;2,\;1,\;\frac{1}{2}$
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)
. . $\displaystyle \begin{array}{ccccc}A:\;2 \\ B:\;3 \\ C:\;3 \\ D:\:4 \\ E:\;2 \end{array}$
(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 $\displaystyle S_1$ isn't equal to $\displaystyle r_2^2$ but rather $\displaystyle d_2^2$
and the method he used to figure out $\displaystyle r_n$ is completely pointless, because you need to know $\displaystyle r_{n-1}$ and to do that you need to know $\displaystyle r_n$
I'm thinking about emailing the academy and telling them about this...
The diagnol of the square, $\displaystyle d$ is equal to the diameter of the circle $\displaystyle d$: (forgive my poor art skills)Originally Posted by sonymd23
and $\displaystyle d$ is the hypotenuse of a right triangleCode:* * * * * *---------------* *| \ |* | \ | * | \ | * * | \d | * * | \ | * | \ | *| \ |* *---------------* * * * * *
the area of the square is equal to $\displaystyle s^2$Code:* * * * * E*---------------* *| \ |* | \ | * | \ | * * | \d | * * | \ | * | \ | *| \ |* D*---------------*F * * * * *
therefore you can use the pythagoreum theorum to solve for $\displaystyle s^2$Code:* * * * * E*---------------* *| \ |* | \ | * | \ | * * | s \d | * * | \ | * | \ | *| s \ |* D*---------------*F * * * * *
$\displaystyle d^2=s^2+s^2\quad\rightarrow\quad d^2=2s^2\quad\rightarrow\quad\frac{d^2}{2}=s^2$
so the area of any square inscribed in a circle is $\displaystyle \frac{d^2}{2}$
now, let's find the area of square 1
$\displaystyle \frac{d^2}{2}=s_1^2$
$\displaystyle \frac{2^2}{2}=s_1^2$
$\displaystyle 2=s_1^2$ 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...
$\displaystyle \frac{d^2}{2}=s_2^2$
$\displaystyle \frac{2}{2}=s_2^2$
$\displaystyle 1=s_2^2$
therefore the area of the second square is 1, lets go onto three...
$\displaystyle \frac{d^2}{2}=s_3^2$
$\displaystyle \frac{1}{2}=s_3^2$
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...
$\displaystyle \frac{2^2}{2^1}=s_1^2$
$\displaystyle \frac{2^2}{2^2}=s_2^2$
$\displaystyle \frac{2^2}{2^3}=s_3^2$
therefore...
$\displaystyle \frac{2^2}{2^n}=s_n^2$
which can be simplified...
$\displaystyle 2^2\times 2^{-n}=s_n^2$ use the multiplicative rule of powers...
$\displaystyle 2^{(2-n)}=s_n^2$
~ $\displaystyle Q\!u\!i\!c\!k$
(I must say that this equation is ten times easier than your teacher's)