Area of the region
$begingroup$
Let $O(0,0), A(3,0), B(3,2)$ and $C(0,2)$ be four vertices of a rectangle. Let
$$d(P,OA)≤min {Bigl(d(P,AB),d(P,BC),d(P,OC)Bigl)}$$
where $d$ denotes the distance of the point $P$ with the line segments given. Let $S$ denote the region consisting of all those points $P$ inside the rectangle $OABC$ which satisfies the above inequality. Find the area of region $S$.
The last time I asked this question(which was almost a week ago), this problem was put on hold and then closed. I am preparing for an entrance exam and I came across this problem in my book called Co-ordinate Geometry for Jee Mains and Advanced by Dr. S.K Goyal. I had put my best effort to solve this problem but I get a different answer $frac{9}{5+sqrt{13}} $. The answer given in my textbook is $4$.Any help for this problem is appreciated. Thanks in advance.
inequality analytic-geometry area
asked Jan 3 at 7:22
sskssk
445
$endgroup$
|
show 10 more comments
$begingroup$
Let $O(0,0), A(3,0), B(3,2)$ and $C(0,2)$ be four vertices of a rectangle. Let
$$d(P,OA)≤min {Bigl(d(P,AB),d(P,BC),d(P,OC)Bigl)}$$
where $d$ denotes the distance of the point $P$ with the line segments given. Let $S$ denote the region consisting of all those points $P$ inside the rectangle $OABC$ which satisfies the above inequality. Find the area of region $S$.
The last time I asked this question(which was almost a week ago), this problem was put on hold and then closed. I am preparing for an entrance exam and I came across this problem in my book called Co-ordinate Geometry for Jee Mains and Advanced by Dr. S.K Goyal. I had put my best effort to solve this problem but I get a different answer $frac{9}{5+sqrt{13}} $. The answer given in my textbook is $4$.Any help for this problem is appreciated. Thanks in advance.
inequality analytic-geometry area
asked Jan 3 at 7:22
sskssk
445
$endgroup$
$begingroup$
Please use Mathjax to format your question, math.meta.stackexchange.com/questions/5020/… and also show us what you have done
$endgroup$
– Prakhar Nagpal
Jan 3 at 7:26
1
$begingroup$
We can't tell you where you went wrong if you only show us your final result and not how you got there. Your question is likely to get closed again, but you should edit it to show your work and it will likely get reopened.
$endgroup$
– Rahul
Jan 3 at 7:29
3
$begingroup$
And also please edit the title - it is not informative nor related to your actual question.
$endgroup$
– TheSimpliFire
Jan 3 at 7:39
1
$begingroup$
Did you draw a picture?
$endgroup$
– Michael Burr
Jan 3 at 8:05
1
$begingroup$
You need angle bisectors, not circles. The set of points equidistant from two segments is a (somewhat) complicated shape.
$endgroup$
– Michael Burr
Jan 3 at 8:30
|
show 10 more comments
$begingroup$
Let $O(0,0), A(3,0), B(3,2)$ and $C(0,2)$ be four vertices of a rectangle. Let
$$d(P,OA)≤min {Bigl(d(P,AB),d(P,BC),d(P,OC)Bigl)}$$
where $d$ denotes the distance of the point $P$ with the line segments given. Let $S$ denote the region consisting of all those points $P$ inside the rectangle $OABC$ which satisfies the above inequality. Find the area of region $S$.
The last time I asked this question(which was almost a week ago), this problem was put on hold and then closed. I am preparing for an entrance exam and I came across this problem in my book called Co-ordinate Geometry for Jee Mains and Advanced by Dr. S.K Goyal. I had put my best effort to solve this problem but I get a different answer $frac{9}{5+sqrt{13}} $. The answer given in my textbook is $4$.Any help for this problem is appreciated. Thanks in advance.
inequality analytic-geometry area
asked Jan 3 at 7:22
sskssk
445
$endgroup$
Let $O(0,0), A(3,0), B(3,2)$ and $C(0,2)$ be four vertices of a rectangle. Let
$$d(P,OA)≤min {Bigl(d(P,AB),d(P,BC),d(P,OC)Bigl)}$$
where $d$ denotes the distance of the point $P$ with the line segments given. Let $S$ denote the region consisting of all those points $P$ inside the rectangle $OABC$ which satisfies the above inequality. Find the area of region $S$.
The last time I asked this question(which was almost a week ago), this problem was put on hold and then closed. I am preparing for an entrance exam and I came across this problem in my book called Co-ordinate Geometry for Jee Mains and Advanced by Dr. S.K Goyal. I had put my best effort to solve this problem but I get a different answer $frac{9}{5+sqrt{13}} $. The answer given in my textbook is $4$.Any help for this problem is appreciated. Thanks in advance.
inequality analytic-geometry area
inequality analytic-geometry area
asked Jan 3 at 7:22
sskssk
445
asked Jan 3 at 7:22
sskssk
445
edited Jan 3 at 8:44
ssk
asked Jan 3 at 7:22
sskssk
445
asked Jan 3 at 7:22
sskssk
445
asked Jan 3 at 7:22
sskssk
445
445
$begingroup$
Please use Mathjax to format your question, math.meta.stackexchange.com/questions/5020/… and also show us what you have done
$endgroup$
– Prakhar Nagpal
Jan 3 at 7:26
1
$begingroup$
We can't tell you where you went wrong if you only show us your final result and not how you got there. Your question is likely to get closed again, but you should edit it to show your work and it will likely get reopened.
$endgroup$
– Rahul
Jan 3 at 7:29
3
$begingroup$
And also please edit the title - it is not informative nor related to your actual question.
$endgroup$
– TheSimpliFire
Jan 3 at 7:39
1
$begingroup$
Did you draw a picture?
$endgroup$
– Michael Burr
Jan 3 at 8:05
1
$begingroup$
You need angle bisectors, not circles. The set of points equidistant from two segments is a (somewhat) complicated shape.
$endgroup$
– Michael Burr
Jan 3 at 8:30
|
show 10 more comments
$begingroup$
Please use Mathjax to format your question, math.meta.stackexchange.com/questions/5020/… and also show us what you have done
$endgroup$
– Prakhar Nagpal
Jan 3 at 7:26
1
$begingroup$
We can't tell you where you went wrong if you only show us your final result and not how you got there. Your question is likely to get closed again, but you should edit it to show your work and it will likely get reopened.
$endgroup$
– Rahul
Jan 3 at 7:29
3
$begingroup$
And also please edit the title - it is not informative nor related to your actual question.
$endgroup$
– TheSimpliFire
Jan 3 at 7:39
1
$begingroup$
Did you draw a picture?
$endgroup$
– Michael Burr
Jan 3 at 8:05
1
$begingroup$
You need angle bisectors, not circles. The set of points equidistant from two segments is a (somewhat) complicated shape.
$endgroup$
– Michael Burr
Jan 3 at 8:30
$begingroup$
Please use Mathjax to format your question, math.meta.stackexchange.com/questions/5020/… and also show us what you have done
$endgroup$
– Prakhar Nagpal
Jan 3 at 7:26
$begingroup$
Please use Mathjax to format your question, math.meta.stackexchange.com/questions/5020/… and also show us what you have done
$endgroup$
– Prakhar Nagpal
Jan 3 at 7:26
1
1
$begingroup$
We can't tell you where you went wrong if you only show us your final result and not how you got there. Your question is likely to get closed again, but you should edit it to show your work and it will likely get reopened.
$endgroup$
– Rahul
Jan 3 at 7:29
$begingroup$
We can't tell you where you went wrong if you only show us your final result and not how you got there. Your question is likely to get closed again, but you should edit it to show your work and it will likely get reopened.
$endgroup$
– Rahul
Jan 3 at 7:29
3
3
$begingroup$
And also please edit the title - it is not informative nor related to your actual question.
$endgroup$
– TheSimpliFire
Jan 3 at 7:39
$begingroup$
And also please edit the title - it is not informative nor related to your actual question.
$endgroup$
– TheSimpliFire
Jan 3 at 7:39
1
1
$begingroup$
Did you draw a picture?
$endgroup$
– Michael Burr
Jan 3 at 8:05
$begingroup$
Did you draw a picture?
$endgroup$
– Michael Burr
Jan 3 at 8:05
1
1
$begingroup$
You need angle bisectors, not circles. The set of points equidistant from two segments is a (somewhat) complicated shape.
$endgroup$
– Michael Burr
Jan 3 at 8:30
$begingroup$
You need angle bisectors, not circles. The set of points equidistant from two segments is a (somewhat) complicated shape.
$endgroup$
– Michael Burr
Jan 3 at 8:30
|
show 10 more comments
1 Answer
1
active
oldest
votes
$begingroup$
An answer of $4$ seems wrong. The entire rectangle has area $6$ and the set of points closer to the bottom edge than the top edge is half the rectangle. These are contradictory statements.
A quick calculation in my head gives an area of $2$ (a trapezoid of bases $1$ and $3$ and of height $1$).
answered Jan 3 at 8:51
Michael BurrMichael Burr
27k23262
$endgroup$
$begingroup$
Mr. Michael Burr ,I guess the answer given in my text book might have a printing mistake.
$endgroup$
– ssk
Jan 3 at 8:53
$begingroup$
Mr. Michael Burr, can you explain of how you considered a trapezoid with the dimensions that you had mentioned .
$endgroup$
– ssk
Jan 3 at 8:57
2
$begingroup$
Draw the angle bisectors of the two lower corners along with the horizontal mid-line.
$endgroup$
– Michael Burr
Jan 3 at 9:21
$begingroup$
The typo could be in the inequality symbol.
$endgroup$
– Macavity
Jan 3 at 11:10
$begingroup$
@Macavity If that's the typo, it would work out nicely. Well spotted.
$endgroup$
– Michael Burr
Jan 3 at 14:02
|
show 3 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
An answer of $4$ seems wrong. The entire rectangle has area $6$ and the set of points closer to the bottom edge than the top edge is half the rectangle. These are contradictory statements.
A quick calculation in my head gives an area of $2$ (a trapezoid of bases $1$ and $3$ and of height $1$).
answered Jan 3 at 8:51
Michael BurrMichael Burr
27k23262
$endgroup$
$begingroup$
Mr. Michael Burr ,I guess the answer given in my text book might have a printing mistake.
$endgroup$
– ssk
Jan 3 at 8:53
$begingroup$
Mr. Michael Burr, can you explain of how you considered a trapezoid with the dimensions that you had mentioned .
$endgroup$
– ssk
Jan 3 at 8:57
2
$begingroup$
Draw the angle bisectors of the two lower corners along with the horizontal mid-line.
$endgroup$
– Michael Burr
Jan 3 at 9:21
$begingroup$
The typo could be in the inequality symbol.
$endgroup$
– Macavity
Jan 3 at 11:10
$begingroup$
@Macavity If that's the typo, it would work out nicely. Well spotted.
$endgroup$
– Michael Burr
Jan 3 at 14:02
|
show 3 more comments
$begingroup$
An answer of $4$ seems wrong. The entire rectangle has area $6$ and the set of points closer to the bottom edge than the top edge is half the rectangle. These are contradictory statements.
A quick calculation in my head gives an area of $2$ (a trapezoid of bases $1$ and $3$ and of height $1$).
answered Jan 3 at 8:51
Michael BurrMichael Burr
27k23262
$endgroup$
$begingroup$
Mr. Michael Burr ,I guess the answer given in my text book might have a printing mistake.
$endgroup$
– ssk
Jan 3 at 8:53
$begingroup$
Mr. Michael Burr, can you explain of how you considered a trapezoid with the dimensions that you had mentioned .
$endgroup$
– ssk
Jan 3 at 8:57
2
$begingroup$
Draw the angle bisectors of the two lower corners along with the horizontal mid-line.
$endgroup$
– Michael Burr
Jan 3 at 9:21
$begingroup$
The typo could be in the inequality symbol.
$endgroup$
– Macavity
Jan 3 at 11:10
$begingroup$
@Macavity If that's the typo, it would work out nicely. Well spotted.
$endgroup$
– Michael Burr
Jan 3 at 14:02
|
show 3 more comments
$begingroup$
An answer of $4$ seems wrong. The entire rectangle has area $6$ and the set of points closer to the bottom edge than the top edge is half the rectangle. These are contradictory statements.
A quick calculation in my head gives an area of $2$ (a trapezoid of bases $1$ and $3$ and of height $1$).
answered Jan 3 at 8:51
Michael BurrMichael Burr
27k23262
$endgroup$
An answer of $4$ seems wrong. The entire rectangle has area $6$ and the set of points closer to the bottom edge than the top edge is half the rectangle. These are contradictory statements.
A quick calculation in my head gives an area of $2$ (a trapezoid of bases $1$ and $3$ and of height $1$).
answered Jan 3 at 8:51
Michael BurrMichael Burr
27k23262
edited Jan 3 at 8:54
answered Jan 3 at 8:51
Michael BurrMichael Burr
27k23262
answered Jan 3 at 8:51
Michael BurrMichael Burr
27k23262
answered Jan 3 at 8:51
Michael BurrMichael Burr
27k23262
27k23262
$begingroup$
Mr. Michael Burr ,I guess the answer given in my text book might have a printing mistake.
$endgroup$
– ssk
Jan 3 at 8:53
$begingroup$
Mr. Michael Burr, can you explain of how you considered a trapezoid with the dimensions that you had mentioned .
$endgroup$
– ssk
Jan 3 at 8:57
2
$begingroup$
Draw the angle bisectors of the two lower corners along with the horizontal mid-line.
$endgroup$
– Michael Burr
Jan 3 at 9:21
$begingroup$
The typo could be in the inequality symbol.
$endgroup$
– Macavity
Jan 3 at 11:10
$begingroup$
@Macavity If that's the typo, it would work out nicely. Well spotted.
$endgroup$
– Michael Burr
Jan 3 at 14:02
|
show 3 more comments
$begingroup$
Mr. Michael Burr ,I guess the answer given in my text book might have a printing mistake.
$endgroup$
– ssk
Jan 3 at 8:53
$begingroup$
Mr. Michael Burr, can you explain of how you considered a trapezoid with the dimensions that you had mentioned .
$endgroup$
– ssk
Jan 3 at 8:57
2
$begingroup$
Draw the angle bisectors of the two lower corners along with the horizontal mid-line.
$endgroup$
– Michael Burr
Jan 3 at 9:21
$begingroup$
The typo could be in the inequality symbol.
$endgroup$
– Macavity
Jan 3 at 11:10
$begingroup$
@Macavity If that's the typo, it would work out nicely. Well spotted.
$endgroup$
– Michael Burr
Jan 3 at 14:02
$begingroup$
Mr. Michael Burr ,I guess the answer given in my text book might have a printing mistake.
$endgroup$
– ssk
Jan 3 at 8:53
$begingroup$
Mr. Michael Burr ,I guess the answer given in my text book might have a printing mistake.
$endgroup$
– ssk
Jan 3 at 8:53
$begingroup$
Mr. Michael Burr, can you explain of how you considered a trapezoid with the dimensions that you had mentioned .
$endgroup$
– ssk
Jan 3 at 8:57
$begingroup$
Mr. Michael Burr, can you explain of how you considered a trapezoid with the dimensions that you had mentioned .
$endgroup$
– ssk
Jan 3 at 8:57
2
2
$begingroup$
Draw the angle bisectors of the two lower corners along with the horizontal mid-line.
$endgroup$
– Michael Burr
Jan 3 at 9:21
$begingroup$
Draw the angle bisectors of the two lower corners along with the horizontal mid-line.
$endgroup$
– Michael Burr
Jan 3 at 9:21
$begingroup$
The typo could be in the inequality symbol.
$endgroup$
– Macavity
Jan 3 at 11:10
$begingroup$
The typo could be in the inequality symbol.
$endgroup$
– Macavity
Jan 3 at 11:10
$begingroup$
@Macavity If that's the typo, it would work out nicely. Well spotted.
$endgroup$
– Michael Burr
Jan 3 at 14:02
$begingroup$
@Macavity If that's the typo, it would work out nicely. Well spotted.
$endgroup$
– Michael Burr
Jan 3 at 14:02
|
show 3 more comments
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$begingroup$
Please use Mathjax to format your question, math.meta.stackexchange.com/questions/5020/… and also show us what you have done
$endgroup$
– Prakhar Nagpal
Jan 3 at 7:26
1
$begingroup$
We can't tell you where you went wrong if you only show us your final result and not how you got there. Your question is likely to get closed again, but you should edit it to show your work and it will likely get reopened.
$endgroup$
– Rahul
Jan 3 at 7:29
3
$begingroup$
And also please edit the title - it is not informative nor related to your actual question.
$endgroup$
– TheSimpliFire
Jan 3 at 7:39
1
$begingroup$
Did you draw a picture?
$endgroup$
– Michael Burr
Jan 3 at 8:05
1
$begingroup$
You need angle bisectors, not circles. The set of points equidistant from two segments is a (somewhat) complicated shape.
$endgroup$
– Michael Burr
Jan 3 at 8:30