Circles within circles question ...

P

pazzy78

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Not looking for the solution, I am currently working on it but someone else mentioned that the radius of the outer circle is the sum of the 2 radii of the inner circles, i don't know if this is a rule ... or he is assuming it's part of the problem ..

is it a rule ?
 
omg ... i apologise for the silly q...

if radius of inner circles are a and b ...
diameter of outer circle = 2a + 2b

radius = a + b.

dohhh...
 
Hint: use the chord formula [imath] \overline{PQ} =2\sqrt{r^2-(r-h)^2}= 2\sqrt{2rh-h^2}[/imath] where [imath] r=a+b [/imath] and [imath] h=2a [/imath] in case [imath] a\leq b. [/imath] The area calculation will eliminate [imath] b. [/imath]

... and post your result. I want to know whether I made a mistake in my scribblings ;-)
 
Last edited:
pazzy78 said:
I get 4, forgot about this ...
workings are messy ...
Click to expand...

There is no need for the work to be messy. Just allow Chord PQ to be a diameter
of the larger circle so that the two smaller circles are congruent to each other.

Let the radius of each smaller circle be r. Then, the radius of the larger circle is 2r.
And, the length of Chord PQ is then 4r. The total area of the two smaller circles
is \(\displaystyle \ 2 \pi r^2, \) and the area of the larger circle is \(\displaystyle \ \pi(2r)^2 \ = 4\pi r^2.\)

The area of the shaded region is the area of the larger circle minus the
area of the two smaller circles:

\(\displaystyle 4 \pi r^2 \ - \ 2 \pi r^2 \ = \ 2\pi \)

\(\displaystyle 2\pi r^2 \ = \ 2 \pi \)

\(\displaystyle r^2 \ = \ 1\)

\(\displaystyle radius \ = \ 1 \ unit \)

The length of Chord PQ\(\displaystyle \ = 4*radius = \ 4 \ units. \)
 
Last edited:
I don't see from that diagram how PQ is the diameter of the inner larger circle.
So what I did was ...
Call the radius of outer circle R.
radius big inner circle b
radius small inner circle a

π R^2 = π b^2 + π a^2 + 2 π
R^2 = b^2 + a^2 + 2

2R = 2b + 2a {the diameter of big outer circle is sum of diameters of 2 inner circles}
R = a + b

(a+b)^2 = b^2 + a^2 + 2

.
.
.
2ab = 2
ab =1

a = (1/b)

Now the red triangle in my diagram is a right angled triangle - this is a law.

I called the PQ/2 h.

So Tan BAQ = h/2b
Tan ABQ = h/2a ... but a = 1/b so Tan ABQ = hb/2

But ABQ = 90 - BAQ

so Tan ABQ = 1/TanBAQ

h/2b = 2/hb

h/2 = 2/h

h^2 = 4

h=2

PQ = 4.


Sorry , It's a bit messy but that's me ... the most beautiful solutions are the simplest yes ... but I'm not that good ! :D
 

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pazzy78 said:
Sorry , It's a bit messy but that's me ... the most beautiful solutions are the simplest yes ... but I'm not that good ! :D
Click to expand...
You can now simplify what you got: if [imath]PQ = 2h[/imath] then by noticing similar triangles you can say that [imath]2a/h = h / 2b \Leftrightarrow h^2 = 4ab =4[/imath], which means that [imath]h=2[/imath].
 
Beer induced reaction follows.
lookagain said:
That source has the wrong units on the answer. The answer is supposed to be
\(\displaystyle \ 4 \ m. \)
Click to expand...

Sh(it) sometimes happen.

pazzy78 said:
I don't see from that diagram how PQ is the diameter of the inner larger circle.
So what I did was ...
Call the radius of outer circle R.
radius big inner circle b
radius small inner circle a

π R^2 = π b^2 + π a^2 + 2 π
R^2 = b^2 + a^2 + 2

2R = 2b + 2a {the diameter of big outer circle is sum of diameters of 2 inner circles}
R = a + b

(a+b)^2 = b^2 + a^2 + 2

.
.
.
2ab = 2
ab =1

a = (1/b)

Now the red triangle in my diagram is a right angled triangle - this is a law.

I called the PQ/2 h.

So Tan BAQ = h/2b
Tan ABQ = h/2a ... but a = 1/b so Tan ABQ = hb/2

But ABQ = 90 - BAQ

so Tan ABQ = 1/TanBAQ

h/2b = 2/hb

h/2 = 2/h

h^2 = 4

h=2

PQ = 4.


Sorry , It's a bit messy but that's me ... the most beautiful solutions are the simplest yes ... but I'm not that good ! :D
Click to expand...

Considering that my good friend lookagain didn't let my source hack incorrect answer unit go unchallenged, it is somewhat surprising that lookagain didn't make any recommendations concerning your literal messy notation regarding the use of grouping symbols to make it all clear(er). I guess lookagain loves me too much to give my source hack a pass. A real good friend that lookagain. I guess that's what friends are for.
 
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