Mixing
Finding two diametres |$AC$| and |$AD$| where $B$ is the center of the larger one and both the circle touch...
$begingroup$
Two circle as shown in the figure, A is the tangent point of both the circle. B is the centre of the large circle. The distance of CD = 90 mm(according to estimation) and EF = 50 mm. What is the value of diametre of both the circle?
I couldn't catch the right process although I tried with some steps. But what I figured out wasn't so usefull at all. I need some help to solve the problem.
Thanks in advance.
geometry circle plane-geometry
asked Jan 20 at 7:29
Anirban NiloyAnirban Niloy
666218
$endgroup$
add a comment |
$begingroup$
Two circle as shown in the figure, A is the tangent point of both the circle. B is the centre of the large circle. The distance of CD = 90 mm(according to estimation) and EF = 50 mm. What is the value of diametre of both the circle?
I couldn't catch the right process although I tried with some steps. But what I figured out wasn't so usefull at all. I need some help to solve the problem.
Thanks in advance.
geometry circle plane-geometry
asked Jan 20 at 7:29
Anirban NiloyAnirban Niloy
666218
$endgroup$
add a comment |
$begingroup$
Two circle as shown in the figure, A is the tangent point of both the circle. B is the centre of the large circle. The distance of CD = 90 mm(according to estimation) and EF = 50 mm. What is the value of diametre of both the circle?
I couldn't catch the right process although I tried with some steps. But what I figured out wasn't so usefull at all. I need some help to solve the problem.
Thanks in advance.
geometry circle plane-geometry
asked Jan 20 at 7:29
Anirban NiloyAnirban Niloy
666218
$endgroup$
Two circle as shown in the figure, A is the tangent point of both the circle. B is the centre of the large circle. The distance of CD = 90 mm(according to estimation) and EF = 50 mm. What is the value of diametre of both the circle?
I couldn't catch the right process although I tried with some steps. But what I figured out wasn't so usefull at all. I need some help to solve the problem.
Thanks in advance.
geometry circle plane-geometry
geometry circle plane-geometry
asked Jan 20 at 7:29
Anirban NiloyAnirban Niloy
666218
asked Jan 20 at 7:29
Anirban NiloyAnirban Niloy
666218
edited Feb 21 at 5:12
Anirban Niloy
asked Jan 20 at 7:29
Anirban NiloyAnirban Niloy
666218
asked Jan 20 at 7:29
Anirban NiloyAnirban Niloy
666218
asked Jan 20 at 7:29
Anirban NiloyAnirban Niloy
666218
666218
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
$AB = r$, $BC = r - 90$, $EB = r - 50$
By power of a point,
$$
ABcdot BC = EB^2
$$
and from this equation $r$ can be found.
edited Jan 20 at 14:46
Daniele Tampieri
2,3272922
answered Jan 20 at 12:46
MickMick
11.9k21641
$endgroup$
$begingroup$
Oops, I didn't know about the 'By power of a point'(Intersecting chords theorem). Whatever! You helped me a lot.
$endgroup$
– Anirban Niloy
Jan 20 at 17:13
$begingroup$
@AnirbanNiloy You are welcome.
$endgroup$
– Mick
Jan 20 at 18:07
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$AB = r$, $BC = r - 90$, $EB = r - 50$
By power of a point,
$$
ABcdot BC = EB^2
$$
and from this equation $r$ can be found.
edited Jan 20 at 14:46
Daniele Tampieri
2,3272922
answered Jan 20 at 12:46
MickMick
11.9k21641
$endgroup$
$begingroup$
Oops, I didn't know about the 'By power of a point'(Intersecting chords theorem). Whatever! You helped me a lot.
$endgroup$
– Anirban Niloy
Jan 20 at 17:13
$begingroup$
@AnirbanNiloy You are welcome.
$endgroup$
– Mick
Jan 20 at 18:07
add a comment |
$begingroup$
$AB = r$, $BC = r - 90$, $EB = r - 50$
By power of a point,
$$
ABcdot BC = EB^2
$$
and from this equation $r$ can be found.
edited Jan 20 at 14:46
Daniele Tampieri
2,3272922
answered Jan 20 at 12:46
MickMick
11.9k21641
$endgroup$
$begingroup$
Oops, I didn't know about the 'By power of a point'(Intersecting chords theorem). Whatever! You helped me a lot.
$endgroup$
– Anirban Niloy
Jan 20 at 17:13
$begingroup$
@AnirbanNiloy You are welcome.
$endgroup$
– Mick
Jan 20 at 18:07
add a comment |
$begingroup$
$AB = r$, $BC = r - 90$, $EB = r - 50$
By power of a point,
$$
ABcdot BC = EB^2
$$
and from this equation $r$ can be found.
edited Jan 20 at 14:46
Daniele Tampieri
2,3272922
answered Jan 20 at 12:46
MickMick
11.9k21641
$endgroup$
$AB = r$, $BC = r - 90$, $EB = r - 50$
By power of a point,
$$
ABcdot BC = EB^2
$$
and from this equation $r$ can be found.
edited Jan 20 at 14:46
Daniele Tampieri
2,3272922
answered Jan 20 at 12:46
MickMick
11.9k21641
edited Jan 20 at 14:46
Daniele Tampieri
2,3272922
edited Jan 20 at 14:46
Daniele Tampieri
2,3272922
edited Jan 20 at 14:46
Daniele Tampieri
2,3272922
2,3272922
answered Jan 20 at 12:46
MickMick
11.9k21641
answered Jan 20 at 12:46
MickMick
11.9k21641
answered Jan 20 at 12:46
MickMick
11.9k21641
11.9k21641
$begingroup$
Oops, I didn't know about the 'By power of a point'(Intersecting chords theorem). Whatever! You helped me a lot.
$endgroup$
– Anirban Niloy
Jan 20 at 17:13
$begingroup$
@AnirbanNiloy You are welcome.
$endgroup$
– Mick
Jan 20 at 18:07
add a comment |
$begingroup$
Oops, I didn't know about the 'By power of a point'(Intersecting chords theorem). Whatever! You helped me a lot.
$endgroup$
– Anirban Niloy
Jan 20 at 17:13
$begingroup$
@AnirbanNiloy You are welcome.
$endgroup$
– Mick
Jan 20 at 18:07
$begingroup$
Oops, I didn't know about the 'By power of a point'(Intersecting chords theorem). Whatever! You helped me a lot.
$endgroup$
– Anirban Niloy
Jan 20 at 17:13
$begingroup$
Oops, I didn't know about the 'By power of a point'(Intersecting chords theorem). Whatever! You helped me a lot.
$endgroup$
– Anirban Niloy
Jan 20 at 17:13
$begingroup$
@AnirbanNiloy You are welcome.
$endgroup$
– Mick
Jan 20 at 18:07
$begingroup$
@AnirbanNiloy You are welcome.
$endgroup$
– Mick
Jan 20 at 18:07
add a comment |
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