Circle inside 1/4-circle w/ r=50: find inner circle's radius
- Thread starter Kenshin
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Draw the diagonal through the center of the small inner circle. You already have the horizontal line from the center to the straight side of the quarter-circle; now draw in the vertical line down to the other straight side.
Logically, the four sides (the two lines you've drawn, plus portions of the straight sides of the quarter-circle) must form a square with side length "r" (using your labelling). The diagonal line splits this square into two halves. What sorts of right triangles are the two halves?
The diagonal line obviously has length 50. What then is the length of the line from the corner of the quarter-circle to the center of the inner circle? That is, what is the length of the hypotenuse of the two triangles formed above?
Now use what you know about that sort of triangle and the ratios of its sides. :wink:
Eliz.
Logically, the four sides (the two lines you've drawn, plus portions of the straight sides of the quarter-circle) must form a square with side length "r" (using your labelling). The diagonal line splits this square into two halves. What sorts of right triangles are the two halves?
The diagonal line obviously has length 50. What then is the length of the line from the corner of the quarter-circle to the center of the inner circle? That is, what is the length of the hypotenuse of the two triangles formed above?
Now use what you know about that sort of triangle and the ratios of its sides. :wink:
Eliz.
Re: Circle inside 1/4-circle w/ r=50: find inner circle's ra
Hello, Kenshin!
It's even simpler than you think . . .
\(\displaystyle C\) is the center of the small circle.
Its radius is: \(\displaystyle \,r \,=\,CA\,=\,CB\,=\,CD\)
Diameter \(\displaystyle AB\) passes through the origin \(\displaystyle O,\)
. . (though it doesn't look that way in my diagram).
Consider right triangle CDO.
It has sides: \(\displaystyle \,CD\,=\,DO \,=\,r\)
Hence: \(\displaystyle OC \:=\:\sqrt{2}r\)
OA is a radius of the quarter-circle: \(\displaystyle \,OA \:=\:50\)
We have: .\(\displaystyle \underbrace{OC}\,+\,\underbrace{CA}\:=\:50\)
. . . . . . . . . ↓. . . ↓
. . . . . . . . \(\displaystyle \overbrace{\sqrt{2}r}\:+\:\overbrace{r}\;\;=\:50\)
Hence: \(\displaystyle \
\sqrt{2}\,+\,1)r \;=\;50\)
Therefore: \(\displaystyle \:r \;=\;\frac{50}{1\,+\,\sqrt{2}} \;\approx\;20.71\) mm.
Hello, Kenshin!
It's even simpler than you think . . .
Code:
|
50 o o
| o
| * * * o A
| * *o
| * / o
|* r / *
| / o
* / *o
* *C *
* / : * o
| r / :
|* / :r *
| * / : * o
| B* : *
- + - - - * * * - - - - -o- -
O | D 50\(\displaystyle C\) is the center of the small circle.
Its radius is: \(\displaystyle \,r \,=\,CA\,=\,CB\,=\,CD\)
Diameter \(\displaystyle AB\) passes through the origin \(\displaystyle O,\)
. . (though it doesn't look that way in my diagram).
Consider right triangle CDO.
It has sides: \(\displaystyle \,CD\,=\,DO \,=\,r\)
Hence: \(\displaystyle OC \:=\:\sqrt{2}r\)
OA is a radius of the quarter-circle: \(\displaystyle \,OA \:=\:50\)
We have: .\(\displaystyle \underbrace{OC}\,+\,\underbrace{CA}\:=\:50\)
. . . . . . . . . ↓. . . ↓
. . . . . . . . \(\displaystyle \overbrace{\sqrt{2}r}\:+\:\overbrace{r}\;\;=\:50\)
Hence: \(\displaystyle \
\sqrt{2}\,+\,1)r \;=\;50\)Therefore: \(\displaystyle \:r \;=\;\frac{50}{1\,+\,\sqrt{2}} \;\approx\;20.71\) mm.
There is a circle inside of a quarter of a circle.. The quarter-circle's radius is 50mm. Can anyone help me find out the radius of the inside circle?
Alternativelsy:
The center of the 50mm quarter circle is O.
The center of the inner circle in P.
From P, drop a line to, and perpendicular ro, the lower horizontal line of the quarter circle at point A.
PD = r, the radius of the inner circle.
OP = 50 - r.
Since triangle OPD is a right triangle, r^2 + r^2 = (50 - r)^2 leading to r^2 + 100r -2500 = 0 which results in r = 20.71.
Alternativelsy:
The center of the 50mm quarter circle is O.
The center of the inner circle in P.
From P, drop a line to, and perpendicular ro, the lower horizontal line of the quarter circle at point A.
PD = r, the radius of the inner circle.
OP = 50 - r.
Since triangle OPD is a right triangle, r^2 + r^2 = (50 - r)^2 leading to r^2 + 100r -2500 = 0 which results in r = 20.71.
- Joined
- Feb 4, 2004
- Messages
- 16,582
At what point did you get confused? You know how to draw straight lines, and you know what "diagonal", "corner" and "center of the inner circle" mean, so you could draw the diagonal line, right? :wink:Kenshin said:
You already draw one line, so you can certainly draw the second (vertical) line, and you can recognize a square, so you could do that bit, too, right? :?:
I'm guessing you haven't yet learned about 45-45-90 triangles...? :shock:
If not, then I'm not sure how you were supposed to proceed, and soroban's complete worked solution probably should not be copied into your homework, since you haven't covered that stuff yet, and thus your teacher would know that you copied.
Try the other suggestion instead, but you might want to use the Quadratic Formula solve the equation you're the tutor gave you, so you can get the exact solution, which is probably what your teacher wants.
Eliz.