Mixing
Two circles X and Y with centres A and B intersect at C and D. If area of circle X is 4 times area of circle...
$begingroup$
This question is solved by taking angleACB = 90 in my book. How can we say that this angle a right angle triangle? Given answer is √5r.
circle
asked Jan 16 at 7:01
sk8sk8
11
$endgroup$
add a comment |
$begingroup$
This question is solved by taking angleACB = 90 in my book. How can we say that this angle a right angle triangle? Given answer is √5r.
circle
asked Jan 16 at 7:01
sk8sk8
11
$endgroup$
$begingroup$
There must be a drawing that goes with this. The angle between a radius and the tangent at the point the radius hits the circle is a right angle, does that do what you need? Without further information I don't see how we answer this.
$endgroup$
– Ross Millikan
Jan 16 at 8:12
add a comment |
$begingroup$
This question is solved by taking angleACB = 90 in my book. How can we say that this angle a right angle triangle? Given answer is √5r.
circle
asked Jan 16 at 7:01
sk8sk8
11
$endgroup$
This question is solved by taking angleACB = 90 in my book. How can we say that this angle a right angle triangle? Given answer is √5r.
circle
circle
asked Jan 16 at 7:01
sk8sk8
11
asked Jan 16 at 7:01
sk8sk8
11
asked Jan 16 at 7:01
sk8sk8
11
asked Jan 16 at 7:01
sk8sk8
11
asked Jan 16 at 7:01
sk8sk8
11
11
$begingroup$
There must be a drawing that goes with this. The angle between a radius and the tangent at the point the radius hits the circle is a right angle, does that do what you need? Without further information I don't see how we answer this.
$endgroup$
– Ross Millikan
Jan 16 at 8:12
add a comment |
$begingroup$
There must be a drawing that goes with this. The angle between a radius and the tangent at the point the radius hits the circle is a right angle, does that do what you need? Without further information I don't see how we answer this.
$endgroup$
– Ross Millikan
Jan 16 at 8:12
$begingroup$
There must be a drawing that goes with this. The angle between a radius and the tangent at the point the radius hits the circle is a right angle, does that do what you need? Without further information I don't see how we answer this.
$endgroup$
– Ross Millikan
Jan 16 at 8:12
$begingroup$
There must be a drawing that goes with this. The angle between a radius and the tangent at the point the radius hits the circle is a right angle, does that do what you need? Without further information I don't see how we answer this.
$endgroup$
– Ross Millikan
Jan 16 at 8:12
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
You need information because you can put the circles so that it intersect and $AB$ can vary. I suppose that $angle ACB =90^circ$ is hypothesis.
answered Jan 16 at 7:15
guchiheguchihe
1188
$endgroup$
$begingroup$
Question is given with four option and in solution ACB=90 [ angle at the point of intersection to the center of the circles.] is written.
$endgroup$
– sk8
Jan 16 at 7:30
add a comment |
$begingroup$
By the triangle inequality $$AC+CB>AB$$ or
$$2r+r>AB$$ or $$AB<3r.$$
Now, by the triangle inequality again we obtain
$$AB+r>2r,$$ which gives
$$r<AB<3r.$$
answered Jan 16 at 7:35
Michael RozenbergMichael Rozenberg
104k1892197
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
You need information because you can put the circles so that it intersect and $AB$ can vary. I suppose that $angle ACB =90^circ$ is hypothesis.
answered Jan 16 at 7:15
guchiheguchihe
1188
$endgroup$
$begingroup$
Question is given with four option and in solution ACB=90 [ angle at the point of intersection to the center of the circles.] is written.
$endgroup$
– sk8
Jan 16 at 7:30
add a comment |
$begingroup$
You need information because you can put the circles so that it intersect and $AB$ can vary. I suppose that $angle ACB =90^circ$ is hypothesis.
answered Jan 16 at 7:15
guchiheguchihe
1188
$endgroup$
$begingroup$
Question is given with four option and in solution ACB=90 [ angle at the point of intersection to the center of the circles.] is written.
$endgroup$
– sk8
Jan 16 at 7:30
add a comment |
$begingroup$
You need information because you can put the circles so that it intersect and $AB$ can vary. I suppose that $angle ACB =90^circ$ is hypothesis.
answered Jan 16 at 7:15
guchiheguchihe
1188
$endgroup$
You need information because you can put the circles so that it intersect and $AB$ can vary. I suppose that $angle ACB =90^circ$ is hypothesis.
answered Jan 16 at 7:15
guchiheguchihe
1188
answered Jan 16 at 7:15
guchiheguchihe
1188
answered Jan 16 at 7:15
guchiheguchihe
1188
answered Jan 16 at 7:15
guchiheguchihe
1188
1188
$begingroup$
Question is given with four option and in solution ACB=90 [ angle at the point of intersection to the center of the circles.] is written.
$endgroup$
– sk8
Jan 16 at 7:30
add a comment |
$begingroup$
Question is given with four option and in solution ACB=90 [ angle at the point of intersection to the center of the circles.] is written.
$endgroup$
– sk8
Jan 16 at 7:30
$begingroup$
Question is given with four option and in solution ACB=90 [ angle at the point of intersection to the center of the circles.] is written.
$endgroup$
– sk8
Jan 16 at 7:30
$begingroup$
Question is given with four option and in solution ACB=90 [ angle at the point of intersection to the center of the circles.] is written.
$endgroup$
– sk8
Jan 16 at 7:30
add a comment |
$begingroup$
By the triangle inequality $$AC+CB>AB$$ or
$$2r+r>AB$$ or $$AB<3r.$$
Now, by the triangle inequality again we obtain
$$AB+r>2r,$$ which gives
$$r<AB<3r.$$
answered Jan 16 at 7:35
Michael RozenbergMichael Rozenberg
104k1892197
$endgroup$
add a comment |
$begingroup$
By the triangle inequality $$AC+CB>AB$$ or
$$2r+r>AB$$ or $$AB<3r.$$
Now, by the triangle inequality again we obtain
$$AB+r>2r,$$ which gives
$$r<AB<3r.$$
answered Jan 16 at 7:35
Michael RozenbergMichael Rozenberg
104k1892197
$endgroup$
add a comment |
$begingroup$
By the triangle inequality $$AC+CB>AB$$ or
$$2r+r>AB$$ or $$AB<3r.$$
Now, by the triangle inequality again we obtain
$$AB+r>2r,$$ which gives
$$r<AB<3r.$$
answered Jan 16 at 7:35
Michael RozenbergMichael Rozenberg
104k1892197
$endgroup$
By the triangle inequality $$AC+CB>AB$$ or
$$2r+r>AB$$ or $$AB<3r.$$
Now, by the triangle inequality again we obtain
$$AB+r>2r,$$ which gives
$$r<AB<3r.$$
answered Jan 16 at 7:35
Michael RozenbergMichael Rozenberg
104k1892197
answered Jan 16 at 7:35
Michael RozenbergMichael Rozenberg
104k1892197
answered Jan 16 at 7:35
Michael RozenbergMichael Rozenberg
104k1892197
answered Jan 16 at 7:35
Michael RozenbergMichael Rozenberg
104k1892197
104k1892197
add a comment |
add a comment |
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$begingroup$
There must be a drawing that goes with this. The angle between a radius and the tangent at the point the radius hits the circle is a right angle, does that do what you need? Without further information I don't see how we answer this.
$endgroup$
– Ross Millikan
Jan 16 at 8:12