# Thread: how do u solve these? help!!!

1. ## how do u solve these? help!!!

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...

2. Originally Posted by sonymd23
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...

look here

3. 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......

4. Originally 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 am having the worst time trying to figure out what you just wrote, is your teachers solving for r? what is r? the radius of the circle?

5. here is what my teacher wrote........by the way can u do question #3,7,8? plz

6. Originally Posted by sonymd23
here is what my teacher wrote........by the way can u do question #3,7,8? plz
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 )

7. Hello, sonymd23!

I have no idea what you teacher did for #1 . . .

Given a unit circle $O_1.$
The first iteration step is to inscribe square $S_1$ in $O_1.$
The second step is to inscribe circle $O_2$ in $S_1$ and another square $S_2$ in $O_2.$
Repeat this process to the $n^{th}$step. .Find the area of $S_1,\;S_2,\;S_3.$
Code:
              * * *
*           *
A*---------------*B
*| \           / |*
|   \       /   |
* |     \   /     | *
* |       *       | *
* |     / O \     | *
|   /       \   |
*| /           \ |*
D*---------------*C
*           *
* * *

Since $O_1$ is a unit circle: $OA = OB = OC = OD = 1.$
. . $\Delta AOB$ is an isosceles right triangle.
Since $AO = BO = 1$, then $AB = \sqrt{2}.$
. . The side of the inscribed square $S_1$ is $\sqrt{2}.$

In general, if $d$ is the diameter of a circle,
. . the side of the inscribed square is: $s = \frac{d}{\sqrt{2}}$

The diameter of $O_2$ is $\sqrt{2}$ . . . Hence, the side of $S_2$ is: . $\frac{\sqrt{2}}{\sqrt{2}}\,=\,1$

The diameter of $O_3$ is $1$ . . . Hence, the side of $S_3$ is: . $\frac{1}{\sqrt{2}}$

Therefore, the areas of $S_1,\;S_2,\;S_3$ are: . $(\sqrt{2})^2,\;(1)^2,\;\left(\frac{1}{\sqrt{2}} \right)^2\;=\;2,\;1,\;\frac{1}{2}$

8. can any one do #3,7,8 ? plz

9. 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)
. . $\begin{array}{ccccc}A:\;2 \\ B:\;3 \\ C:\;3 \\ D:\:4 \\ E:\;2 \end{array}$

(c) The network is traceable; there are exactly two odd vertices.

10. ^thnx every one..........but i really need help on 7 and 8 .......

11. I would like to point out that your teacher is incorrect, the area of $S_1$ isn't equal to $r_2^2$ but rather $d_2^2$
and the method he used to figure out $r_n$ is completely pointless, because you need to know $r_{n-1}$ and to do that you need to know $r_n$

12. Originally Posted by Quick
I would like to point out that your teacher is incorrect, the area of $S_1$ isn't equal to $r_2^2$ but rather $d_2^2$
and the method he used to figure out $r_n$ is completely pointless, because you need to know $r_{n-1}$ and to do that you need to know $r_n$
so wahts the proper way to do this? and would u mind show me every steps....thnx

13. Originally Posted by sonymd23
so wahts the proper way to do this? and would u mind show me every steps....thnx
The diagnol of the square, $d$ is equal to the diameter of the circle $d$: (forgive my poor art skills)
Code:
              * * *
*           *
*---------------*
*| \             |*
|   \           |
* |     \         | *
* |       \d      | *
* |         \     | *
|           \   |
*|             \ |*
*---------------*
*           *
* * *
and $d$ is the hypotenuse of a right triangle
Code:
              * * *
*           *
E*---------------*
*| \             |*
|   \           |
* |     \         | *
* |       \d      | *
* |         \     | *
|           \   |
*|             \ |*
D*---------------*F
*           *
* * *
the area of the square is equal to $s^2$
Code:
              * * *
*           *
E*---------------*
*| \             |*
|   \           |
* |     \         | *
* | s     \d      | *
* |         \     | *
|           \   |
*|        s    \ |*
D*---------------*F
*           *
* * *
therefore you can use the pythagoreum theorum to solve for $s^2$

$d^2=s^2+s^2\quad\rightarrow\quad d^2=2s^2\quad\rightarrow\quad\frac{d^2}{2}=s^2$

so the area of any square inscribed in a circle is $\frac{d^2}{2}$
now, let's find the area of square 1

$\frac{d^2}{2}=s_1^2$

$\frac{2^2}{2}=s_1^2$

$2=s_1^2$ 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...

$\frac{d^2}{2}=s_2^2$

$\frac{2}{2}=s_2^2$

$1=s_2^2$

therefore the area of the second square is 1, lets go onto three...

$\frac{d^2}{2}=s_3^2$

$\frac{1}{2}=s_3^2$

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...

$\frac{2^2}{2^1}=s_1^2$

$\frac{2^2}{2^2}=s_2^2$

$\frac{2^2}{2^3}=s_3^2$

therefore...

$\frac{2^2}{2^n}=s_n^2$

which can be simplified...

$2^2\times 2^{-n}=s_n^2$ use the multiplicative rule of powers...

$2^{(2-n)}=s_n^2$

~ $Q\!u\!i\!c\!k$

(I must say that this equation is ten times easier than your teacher's)

14. questiuon 7 and8 plz.......thnx