Question

Answer this Math Question!

Answer 1

let me draw the diagram

Answer 2

Diagram?

Answer 3

The radius of the larger circle is 10 cm. Find the area of the largest circle that will fit in the middle.

Answer 4

this is toughest of all the problems i have solved

Answer 5

hint(it's simple algebra of grade 9 0r 10)

Answer 6

The radius of the LARGEST circle is 10cm?

Answer 7

find the area of the blue circle

Answer 8

The radius of the LARGEST circle?
The orange circle radii is 10cm or what?

Answer 9

10 cm

Answer 10

largest circle radius is 10 cm

Answer 11

Anything else given?

Answer 12

nop!

Answer 13

my internet connection is frequently disconnecting so i will get back on this question tommorow solve it till tommorow!!!!

Answer 14

Radius of the largest circle(orange) is 10 cm. Find the area of the blue circle??

Answer 15

Let R be radius of white circle
Let r be radius of blue circle

\[2R+r=10\]
\[R+r = \sqrt{2}R\]
solving for R in 2nd equation
\[R = \frac{r}{\sqrt{2}-1} = r(\sqrt{2} +1)\]
sub into 1st equation
\[r(2\sqrt{2}+2) +r = 10\]
\[r = \frac{10}{2\sqrt{2}+3} = 10(3-2\sqrt{2})\]
\[Area = \pi r^{2} = 100(3-2\sqrt{2})^{2} \pi \approx 9.25\]

Answer 16

How is R+r = sqrt2.R
@dumbcow? :/

Answer 17

Using this section of the circles:

we get:\[\begin{align}
r+2R&=10\\
\therefore R&=\frac{10-r}{2}\tag{1}\\
2R&=\sqrt{2}(r+R)\tag{2}\\
\therefore 10-r&=\sqrt{2}(r+\frac{10-r}{2})\qquad\text{(using (1))}\\
\therefore 20-2r&=\sqrt{2}(2r+10-r)=\sqrt{2}(r+10)\\
\therefore r(2+\sqrt{2})&=20-10\sqrt{2}=10(2-\sqrt{2})\\
\therefore r&=\frac{10(2-\sqrt{2})}{2+\sqrt{2}}=\frac{10(2-\sqrt{2})^2}{4-2}=5(4-2\sqrt{2}+2)\\
&=5(6-2\sqrt{2})\\
&=10(3-\sqrt{2})
\end{align}\]
Use this to calculate the area of the blue circle as \(\pi r^2\)