Mixing
What proportion of the quarter circle is shaded?
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1
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Interesting yet challenging quiz I found on a website. My answer is a $frac{1}{ sqrt{2}}$.
After I assumed the semicircle has radius $rsin{45}$, where $r$ is the radius of the quarter circular part.
Any objections or comment?
geometry area
edited 42 mins ago
Robert Z
89.7k1056128
asked 2 hours ago
Waitara Mburu
261
add a comment |
up vote
1
down vote
favorite
Interesting yet challenging quiz I found on a website. My answer is a $frac{1}{ sqrt{2}}$.
After I assumed the semicircle has radius $rsin{45}$, where $r$ is the radius of the quarter circular part.
Any objections or comment?
geometry area
edited 42 mins ago
Robert Z
89.7k1056128
asked 2 hours ago
Waitara Mburu
261
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Interesting yet challenging quiz I found on a website. My answer is a $frac{1}{ sqrt{2}}$.
After I assumed the semicircle has radius $rsin{45}$, where $r$ is the radius of the quarter circular part.
Any objections or comment?
geometry area
edited 42 mins ago
Robert Z
89.7k1056128
asked 2 hours ago
Waitara Mburu
261
Interesting yet challenging quiz I found on a website. My answer is a $frac{1}{ sqrt{2}}$.
After I assumed the semicircle has radius $rsin{45}$, where $r$ is the radius of the quarter circular part.
Any objections or comment?
geometry area
geometry area
edited 42 mins ago
Robert Z
89.7k1056128
asked 2 hours ago
Waitara Mburu
261
edited 42 mins ago
Robert Z
89.7k1056128
asked 2 hours ago
Waitara Mburu
261
edited 42 mins ago
Robert Z
89.7k1056128
edited 42 mins ago
Robert Z
89.7k1056128
edited 42 mins ago
Robert Z
89.7k1056128
89.7k1056128
asked 2 hours ago
Waitara Mburu
261
asked 2 hours ago
Waitara Mburu
261
asked 2 hours ago
Waitara Mburu
261
261
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
Your result is not correct. Let $R$ be the radius of the quarter circle and $r$ be the radius of the half-circle. Then, by Pythagoras theorem applied to $triangle OO'C$,
$$R^2=r^2+d^2$$
where $d=sqrt{2}r$ is the distance of the two centers $O$ and $O'$ (note that $angle O'OH=45^{circ}$).
Can you take it from here?
answered 1 hour ago
Robert Z
89.7k1056128
Looks good to me. Of course a good picture is necessary to draw. The picture in the post is not very clear on what the actual geometrical constraints are.
– Matti P.
1 hour ago
Sorry my omission, the semicircle is tangent to the quarter circle. I can take it from here which gives 2/3 as the probable answer, no solution was given though. Thanks.
– Waitara Mburu
1 hour ago
@MattiP. Now we have a better picture.
– Robert Z
43 mins ago
@WaitaraMburu Yes, the required ratio is $2/3$. If my answer was useful please consider "accepting" it. See math.stackexchange.com/tour
– Robert Z
26 mins ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Your result is not correct. Let $R$ be the radius of the quarter circle and $r$ be the radius of the half-circle. Then, by Pythagoras theorem applied to $triangle OO'C$,
$$R^2=r^2+d^2$$
where $d=sqrt{2}r$ is the distance of the two centers $O$ and $O'$ (note that $angle O'OH=45^{circ}$).
Can you take it from here?
answered 1 hour ago
Robert Z
89.7k1056128
Looks good to me. Of course a good picture is necessary to draw. The picture in the post is not very clear on what the actual geometrical constraints are.
– Matti P.
1 hour ago
Sorry my omission, the semicircle is tangent to the quarter circle. I can take it from here which gives 2/3 as the probable answer, no solution was given though. Thanks.
– Waitara Mburu
1 hour ago
@MattiP. Now we have a better picture.
– Robert Z
43 mins ago
@WaitaraMburu Yes, the required ratio is $2/3$. If my answer was useful please consider "accepting" it. See math.stackexchange.com/tour
– Robert Z
26 mins ago
add a comment |
up vote
1
down vote
Your result is not correct. Let $R$ be the radius of the quarter circle and $r$ be the radius of the half-circle. Then, by Pythagoras theorem applied to $triangle OO'C$,
$$R^2=r^2+d^2$$
where $d=sqrt{2}r$ is the distance of the two centers $O$ and $O'$ (note that $angle O'OH=45^{circ}$).
Can you take it from here?
answered 1 hour ago
Robert Z
89.7k1056128
Looks good to me. Of course a good picture is necessary to draw. The picture in the post is not very clear on what the actual geometrical constraints are.
– Matti P.
1 hour ago
Sorry my omission, the semicircle is tangent to the quarter circle. I can take it from here which gives 2/3 as the probable answer, no solution was given though. Thanks.
– Waitara Mburu
1 hour ago
@MattiP. Now we have a better picture.
– Robert Z
43 mins ago
@WaitaraMburu Yes, the required ratio is $2/3$. If my answer was useful please consider "accepting" it. See math.stackexchange.com/tour
– Robert Z
26 mins ago
add a comment |
up vote
1
down vote
up vote
1
down vote
Your result is not correct. Let $R$ be the radius of the quarter circle and $r$ be the radius of the half-circle. Then, by Pythagoras theorem applied to $triangle OO'C$,
$$R^2=r^2+d^2$$
where $d=sqrt{2}r$ is the distance of the two centers $O$ and $O'$ (note that $angle O'OH=45^{circ}$).
Can you take it from here?
answered 1 hour ago
Robert Z
89.7k1056128
Your result is not correct. Let $R$ be the radius of the quarter circle and $r$ be the radius of the half-circle. Then, by Pythagoras theorem applied to $triangle OO'C$,
$$R^2=r^2+d^2$$
where $d=sqrt{2}r$ is the distance of the two centers $O$ and $O'$ (note that $angle O'OH=45^{circ}$).
Can you take it from here?
answered 1 hour ago
Robert Z
89.7k1056128
edited 31 mins ago
answered 1 hour ago
Robert Z
89.7k1056128
answered 1 hour ago
Robert Z
89.7k1056128
answered 1 hour ago
Robert Z
89.7k1056128
89.7k1056128
Looks good to me. Of course a good picture is necessary to draw. The picture in the post is not very clear on what the actual geometrical constraints are.
– Matti P.
1 hour ago
Sorry my omission, the semicircle is tangent to the quarter circle. I can take it from here which gives 2/3 as the probable answer, no solution was given though. Thanks.
– Waitara Mburu
1 hour ago
@MattiP. Now we have a better picture.
– Robert Z
43 mins ago
@WaitaraMburu Yes, the required ratio is $2/3$. If my answer was useful please consider "accepting" it. See math.stackexchange.com/tour
– Robert Z
26 mins ago
add a comment |
Looks good to me. Of course a good picture is necessary to draw. The picture in the post is not very clear on what the actual geometrical constraints are.
– Matti P.
1 hour ago
Sorry my omission, the semicircle is tangent to the quarter circle. I can take it from here which gives 2/3 as the probable answer, no solution was given though. Thanks.
– Waitara Mburu
1 hour ago
@MattiP. Now we have a better picture.
– Robert Z
43 mins ago
@WaitaraMburu Yes, the required ratio is $2/3$. If my answer was useful please consider "accepting" it. See math.stackexchange.com/tour
– Robert Z
26 mins ago
Looks good to me. Of course a good picture is necessary to draw. The picture in the post is not very clear on what the actual geometrical constraints are.
– Matti P.
1 hour ago
Looks good to me. Of course a good picture is necessary to draw. The picture in the post is not very clear on what the actual geometrical constraints are.
– Matti P.
1 hour ago
Sorry my omission, the semicircle is tangent to the quarter circle. I can take it from here which gives 2/3 as the probable answer, no solution was given though. Thanks.
– Waitara Mburu
1 hour ago
Sorry my omission, the semicircle is tangent to the quarter circle. I can take it from here which gives 2/3 as the probable answer, no solution was given though. Thanks.
– Waitara Mburu
1 hour ago
@MattiP. Now we have a better picture.
– Robert Z
43 mins ago
@MattiP. Now we have a better picture.
– Robert Z
43 mins ago
@WaitaraMburu Yes, the required ratio is $2/3$. If my answer was useful please consider "accepting" it. See math.stackexchange.com/tour
– Robert Z
26 mins ago
@WaitaraMburu Yes, the required ratio is $2/3$. If my answer was useful please consider "accepting" it. See math.stackexchange.com/tour
– Robert Z
26 mins ago
add a comment |
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