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Circle with equilateral triangle [on hold]
Geometry - Equilateral triangle covered with five circlesEquilateral triangle coveringArea of equilateral triangle from circumcircleCommon area between Circle and Equilateral triangleequilateral triangle and inscribed circle6 Circle in Equilateral TriangleHomework: Is the triangle an equilateral triangle?Area of enclosed overlapping circles within an equilateral trianglecircumcentre of equilateral triangleInscribe an equilateral triangle inside a triangle
$begingroup$
How to use property of equilateral triangle which is given in the question .
geometry triangle circle
edited 2 days ago
Dr. Mathva
2,296526
asked 2 days ago
Abhinov SinghAbhinov Singh
1354
$endgroup$
put on hold as off-topic by John Omielan, steven gregory, Eevee Trainer, user21820, José Carlos Santos 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – John Omielan, steven gregory, Eevee Trainer, user21820, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
How to use property of equilateral triangle which is given in the question .
geometry triangle circle
edited 2 days ago
Dr. Mathva
2,296526
asked 2 days ago
Abhinov SinghAbhinov Singh
1354
$endgroup$
put on hold as off-topic by John Omielan, steven gregory, Eevee Trainer, user21820, José Carlos Santos 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – John Omielan, steven gregory, Eevee Trainer, user21820, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
How to use property of equilateral triangle which is given in the question .
geometry triangle circle
edited 2 days ago
Dr. Mathva
2,296526
asked 2 days ago
Abhinov SinghAbhinov Singh
1354
$endgroup$
How to use property of equilateral triangle which is given in the question .
geometry triangle circle
geometry triangle circle
edited 2 days ago
Dr. Mathva
2,296526
asked 2 days ago
Abhinov SinghAbhinov Singh
1354
edited 2 days ago
Dr. Mathva
2,296526
asked 2 days ago
Abhinov SinghAbhinov Singh
1354
edited 2 days ago
Dr. Mathva
2,296526
edited 2 days ago
Dr. Mathva
2,296526
edited 2 days ago
Dr. Mathva
2,296526
2,296526
asked 2 days ago
Abhinov SinghAbhinov Singh
1354
asked 2 days ago
Abhinov SinghAbhinov Singh
1354
asked 2 days ago
Abhinov SinghAbhinov Singh
1354
1354
put on hold as off-topic by John Omielan, steven gregory, Eevee Trainer, user21820, José Carlos Santos 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – John Omielan, steven gregory, Eevee Trainer, user21820, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by John Omielan, steven gregory, Eevee Trainer, user21820, José Carlos Santos 2 days ago
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – John Omielan, steven gregory, Eevee Trainer, user21820, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Here's a picture, with a bunch of useful things drawn in and labeled:
The center of the circles are marked with dots; $O$ is the center of the yellow circle and $E$ is the center of the green circle.
So now, what's the ratio of $ED$ to $AE$?
answered 2 days ago
jmerryjmerry
13.4k1629
$endgroup$
add a comment |
$begingroup$
According to Jmerry's diagram,
$$2R = AE + ED = r+ frac{r}{cos60°} = r+ frac{r}{1/2} = r+ 2r$$
Thus,
$$frac{R}{r}=frac{3}{2}$$ or
$$frac{r}{R}=frac{2}{3}$$
edited 2 days ago
Oscar Lanzi
13.1k12136
answered 2 days ago
Mathew MahindaratneMathew Mahindaratne
562212
$endgroup$
$begingroup$
Therefore $$frac{text{yellow}}{text{green}}=frac{R^2pi-r^2pi}{r^2pi}=frac{R^2pi}{r^2pi}-1=big(frac{R}{r}big)^2-1=big(frac{3}{2}big)^2-1=frac{5}{4}$$ Thus $$frac{text{green}}{text{yellow}}=frac{4}{5}$$
$endgroup$
– Dr. Mathva
2 days ago
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here's a picture, with a bunch of useful things drawn in and labeled:
The center of the circles are marked with dots; $O$ is the center of the yellow circle and $E$ is the center of the green circle.
So now, what's the ratio of $ED$ to $AE$?
answered 2 days ago
jmerryjmerry
13.4k1629
$endgroup$
add a comment |
$begingroup$
Here's a picture, with a bunch of useful things drawn in and labeled:
The center of the circles are marked with dots; $O$ is the center of the yellow circle and $E$ is the center of the green circle.
So now, what's the ratio of $ED$ to $AE$?
answered 2 days ago
jmerryjmerry
13.4k1629
$endgroup$
add a comment |
$begingroup$
Here's a picture, with a bunch of useful things drawn in and labeled:
The center of the circles are marked with dots; $O$ is the center of the yellow circle and $E$ is the center of the green circle.
So now, what's the ratio of $ED$ to $AE$?
answered 2 days ago
jmerryjmerry
13.4k1629
$endgroup$
Here's a picture, with a bunch of useful things drawn in and labeled:
The center of the circles are marked with dots; $O$ is the center of the yellow circle and $E$ is the center of the green circle.
So now, what's the ratio of $ED$ to $AE$?
answered 2 days ago
jmerryjmerry
13.4k1629
answered 2 days ago
jmerryjmerry
13.4k1629
answered 2 days ago
jmerryjmerry
13.4k1629
answered 2 days ago
jmerryjmerry
13.4k1629
13.4k1629
add a comment |
add a comment |
$begingroup$
According to Jmerry's diagram,
$$2R = AE + ED = r+ frac{r}{cos60°} = r+ frac{r}{1/2} = r+ 2r$$
Thus,
$$frac{R}{r}=frac{3}{2}$$ or
$$frac{r}{R}=frac{2}{3}$$
edited 2 days ago
Oscar Lanzi
13.1k12136
answered 2 days ago
Mathew MahindaratneMathew Mahindaratne
562212
$endgroup$
$begingroup$
Therefore $$frac{text{yellow}}{text{green}}=frac{R^2pi-r^2pi}{r^2pi}=frac{R^2pi}{r^2pi}-1=big(frac{R}{r}big)^2-1=big(frac{3}{2}big)^2-1=frac{5}{4}$$ Thus $$frac{text{green}}{text{yellow}}=frac{4}{5}$$
$endgroup$
– Dr. Mathva
2 days ago
add a comment |
$begingroup$
According to Jmerry's diagram,
$$2R = AE + ED = r+ frac{r}{cos60°} = r+ frac{r}{1/2} = r+ 2r$$
Thus,
$$frac{R}{r}=frac{3}{2}$$ or
$$frac{r}{R}=frac{2}{3}$$
edited 2 days ago
Oscar Lanzi
13.1k12136
answered 2 days ago
Mathew MahindaratneMathew Mahindaratne
562212
$endgroup$
$begingroup$
Therefore $$frac{text{yellow}}{text{green}}=frac{R^2pi-r^2pi}{r^2pi}=frac{R^2pi}{r^2pi}-1=big(frac{R}{r}big)^2-1=big(frac{3}{2}big)^2-1=frac{5}{4}$$ Thus $$frac{text{green}}{text{yellow}}=frac{4}{5}$$
$endgroup$
– Dr. Mathva
2 days ago
add a comment |
$begingroup$
According to Jmerry's diagram,
$$2R = AE + ED = r+ frac{r}{cos60°} = r+ frac{r}{1/2} = r+ 2r$$
Thus,
$$frac{R}{r}=frac{3}{2}$$ or
$$frac{r}{R}=frac{2}{3}$$
edited 2 days ago
Oscar Lanzi
13.1k12136
answered 2 days ago
Mathew MahindaratneMathew Mahindaratne
562212
$endgroup$
According to Jmerry's diagram,
$$2R = AE + ED = r+ frac{r}{cos60°} = r+ frac{r}{1/2} = r+ 2r$$
Thus,
$$frac{R}{r}=frac{3}{2}$$ or
$$frac{r}{R}=frac{2}{3}$$
edited 2 days ago
Oscar Lanzi
13.1k12136
answered 2 days ago
Mathew MahindaratneMathew Mahindaratne
562212
edited 2 days ago
Oscar Lanzi
13.1k12136
edited 2 days ago
Oscar Lanzi
13.1k12136
edited 2 days ago
Oscar Lanzi
13.1k12136
13.1k12136
answered 2 days ago
Mathew MahindaratneMathew Mahindaratne
562212
answered 2 days ago
Mathew MahindaratneMathew Mahindaratne
562212
answered 2 days ago
Mathew MahindaratneMathew Mahindaratne
562212
562212
$begingroup$
Therefore $$frac{text{yellow}}{text{green}}=frac{R^2pi-r^2pi}{r^2pi}=frac{R^2pi}{r^2pi}-1=big(frac{R}{r}big)^2-1=big(frac{3}{2}big)^2-1=frac{5}{4}$$ Thus $$frac{text{green}}{text{yellow}}=frac{4}{5}$$
$endgroup$
– Dr. Mathva
2 days ago
add a comment |
$begingroup$
Therefore $$frac{text{yellow}}{text{green}}=frac{R^2pi-r^2pi}{r^2pi}=frac{R^2pi}{r^2pi}-1=big(frac{R}{r}big)^2-1=big(frac{3}{2}big)^2-1=frac{5}{4}$$ Thus $$frac{text{green}}{text{yellow}}=frac{4}{5}$$
$endgroup$
– Dr. Mathva
2 days ago
$begingroup$
Therefore $$frac{text{yellow}}{text{green}}=frac{R^2pi-r^2pi}{r^2pi}=frac{R^2pi}{r^2pi}-1=big(frac{R}{r}big)^2-1=big(frac{3}{2}big)^2-1=frac{5}{4}$$ Thus $$frac{text{green}}{text{yellow}}=frac{4}{5}$$
$endgroup$
– Dr. Mathva
2 days ago
$begingroup$
Therefore $$frac{text{yellow}}{text{green}}=frac{R^2pi-r^2pi}{r^2pi}=frac{R^2pi}{r^2pi}-1=big(frac{R}{r}big)^2-1=big(frac{3}{2}big)^2-1=frac{5}{4}$$ Thus $$frac{text{green}}{text{yellow}}=frac{4}{5}$$
$endgroup$
– Dr. Mathva
2 days ago
add a comment |
