What is the area shaded in this figure?
I was sent this problem by a friend, they say it is a year 6 question. Using trigonometry I got the shaded area as $4 - frac{pi}{3}$.
Now Year six, I doubt if they do trigonometry. Is there another approach out there?
geometry
edited Dec 9 at 12:05
Anubhab Ghosal
74412
asked Dec 9 at 8:40
Waitara Mburu
364
add a comment |
I was sent this problem by a friend, they say it is a year 6 question. Using trigonometry I got the shaded area as $4 - frac{pi}{3}$.
Now Year six, I doubt if they do trigonometry. Is there another approach out there?
geometry
edited Dec 9 at 12:05
Anubhab Ghosal
74412
asked Dec 9 at 8:40
Waitara Mburu
364
Are you sure about your answer? It looks a little large to me. I would say that using the symmetry of the diagram you should be expecting something like $8-2pi$.
– postmortes
Dec 9 at 8:55
Can you provide details about what you'd expect in sixth year? Visitors here come from a wide variety of contries and educational systems. I assume they wouldn't be doing integrals either?
– MvG
Dec 9 at 11:51
@MvG, it doesn't matter now as the actual answer is $frac{32}{5}-4tan^{-1}(frac{3}{4})$, and so you can't do it without trigonometry.
– Anubhab Ghosal
Dec 9 at 13:11
This answer finds the area of a related region. In the notation used there, the target area here can be written as $u-|text{region}PAT|$.
– Blue
Dec 9 at 14:24
add a comment |
I was sent this problem by a friend, they say it is a year 6 question. Using trigonometry I got the shaded area as $4 - frac{pi}{3}$.
Now Year six, I doubt if they do trigonometry. Is there another approach out there?
geometry
edited Dec 9 at 12:05
Anubhab Ghosal
74412
asked Dec 9 at 8:40
Waitara Mburu
364
I was sent this problem by a friend, they say it is a year 6 question. Using trigonometry I got the shaded area as $4 - frac{pi}{3}$.
Now Year six, I doubt if they do trigonometry. Is there another approach out there?
geometry
geometry
edited Dec 9 at 12:05
Anubhab Ghosal
74412
asked Dec 9 at 8:40
Waitara Mburu
364
edited Dec 9 at 12:05
Anubhab Ghosal
74412
asked Dec 9 at 8:40
Waitara Mburu
364
edited Dec 9 at 12:05
Anubhab Ghosal
74412
edited Dec 9 at 12:05
Anubhab Ghosal
74412
edited Dec 9 at 12:05
Anubhab Ghosal
74412
74412
asked Dec 9 at 8:40
Waitara Mburu
364
asked Dec 9 at 8:40
Waitara Mburu
364
asked Dec 9 at 8:40
Waitara Mburu
364
364
Are you sure about your answer? It looks a little large to me. I would say that using the symmetry of the diagram you should be expecting something like $8-2pi$.
– postmortes
Dec 9 at 8:55
Can you provide details about what you'd expect in sixth year? Visitors here come from a wide variety of contries and educational systems. I assume they wouldn't be doing integrals either?
– MvG
Dec 9 at 11:51
@MvG, it doesn't matter now as the actual answer is $frac{32}{5}-4tan^{-1}(frac{3}{4})$, and so you can't do it without trigonometry.
– Anubhab Ghosal
Dec 9 at 13:11
This answer finds the area of a related region. In the notation used there, the target area here can be written as $u-|text{region}PAT|$.
– Blue
Dec 9 at 14:24
add a comment |
Are you sure about your answer? It looks a little large to me. I would say that using the symmetry of the diagram you should be expecting something like $8-2pi$.
– postmortes
Dec 9 at 8:55
Can you provide details about what you'd expect in sixth year? Visitors here come from a wide variety of contries and educational systems. I assume they wouldn't be doing integrals either?
– MvG
Dec 9 at 11:51
@MvG, it doesn't matter now as the actual answer is $frac{32}{5}-4tan^{-1}(frac{3}{4})$, and so you can't do it without trigonometry.
– Anubhab Ghosal
Dec 9 at 13:11
This answer finds the area of a related region. In the notation used there, the target area here can be written as $u-|text{region}PAT|$.
– Blue
Dec 9 at 14:24
Are you sure about your answer? It looks a little large to me. I would say that using the symmetry of the diagram you should be expecting something like $8-2pi$.
– postmortes
Dec 9 at 8:55
Are you sure about your answer? It looks a little large to me. I would say that using the symmetry of the diagram you should be expecting something like $8-2pi$.
– postmortes
Dec 9 at 8:55
Can you provide details about what you'd expect in sixth year? Visitors here come from a wide variety of contries and educational systems. I assume they wouldn't be doing integrals either?
– MvG
Dec 9 at 11:51
Can you provide details about what you'd expect in sixth year? Visitors here come from a wide variety of contries and educational systems. I assume they wouldn't be doing integrals either?
– MvG
Dec 9 at 11:51
@MvG, it doesn't matter now as the actual answer is $frac{32}{5}-4tan^{-1}(frac{3}{4})$, and so you can't do it without trigonometry.
– Anubhab Ghosal
Dec 9 at 13:11
@MvG, it doesn't matter now as the actual answer is $frac{32}{5}-4tan^{-1}(frac{3}{4})$, and so you can't do it without trigonometry.
– Anubhab Ghosal
Dec 9 at 13:11
This answer finds the area of a related region. In the notation used there, the target area here can be written as $u-|text{region}PAT|$.
– Blue
Dec 9 at 14:24
This answer finds the area of a related region. In the notation used there, the target area here can be written as $u-|text{region}PAT|$.
– Blue
Dec 9 at 14:24
add a comment |
2 Answers
2
active
oldest
votes
Label the vertices as shown in the diagram.
$tanalpha=frac{1}{2}implies tan 2alpha=frac{4}{3}implies 2alpha=tan^{-1}(frac{4}{3})$
$therefore$ Area of circular sector$ BAG=8tan^{-1}(frac{4}{3})$
Area of $triangle AGC=16cosalphasinalpha=8sin2alpha=frac{32}{5}$
Thus, area of curved figure $BEG=frac{48}{5}-8tan^{-1}(frac{4}{3})$
Therefore, Area of shaded part$=frac{1}{2}$ (Area of $BEFC$)$-frac{1}{2}$ (Area of semicircle)$-$Area of curved figure $BEG$.
This gives $frac{32}{5}-4pi+8tan^{-1}(frac{4}{3})=frac{32}{5}-4tan^{-1}(frac{3}{4})approx1.252$
answered Dec 9 at 12:46
Anubhab Ghosal
74412
the numerical result is correct, I don't believe other calculation: that of BEG and the last passage where $4pi$ disappear. I think that BEG=$16-32/5 -8tan^{-1}(4/3)$
– user126154
Dec 10 at 10:12
ok, now I understant the last passage: $4/3to 3/4$ but I still think BEG is wrong
– user126154
Dec 10 at 10:43
@user126154 $frac{pi}{2}-tan^{-1}(frac{4}{3})=tan^{-1}(frac{3}{4})$
– Anubhab Ghosal
Dec 10 at 11:11
@user126154, Thanks. Fixed the typo.
– Anubhab Ghosal
Dec 10 at 11:22
add a comment |
Flip the image by vertically.
It is not as simple as you think. Year six is probably high school when they can do geometry and calculus. I have done it using calculus and the area is as shown in the figure
answered Dec 9 at 12:13
Satish Ramanathan
9,41531323
It widened my approaches and a fantastic link; thank you all!
– Waitara Mburu
Dec 10 at 10:34
@WaitaraMburu, while one always appreciates if someone benefits from their efforts, it is always good if you upvote and accept an answer(especially answers of new users like me), if you find the answer correct and useful.
– Anubhab Ghosal
Dec 10 at 11:15
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Label the vertices as shown in the diagram.
$tanalpha=frac{1}{2}implies tan 2alpha=frac{4}{3}implies 2alpha=tan^{-1}(frac{4}{3})$
$therefore$ Area of circular sector$ BAG=8tan^{-1}(frac{4}{3})$
Area of $triangle AGC=16cosalphasinalpha=8sin2alpha=frac{32}{5}$
Thus, area of curved figure $BEG=frac{48}{5}-8tan^{-1}(frac{4}{3})$
Therefore, Area of shaded part$=frac{1}{2}$ (Area of $BEFC$)$-frac{1}{2}$ (Area of semicircle)$-$Area of curved figure $BEG$.
This gives $frac{32}{5}-4pi+8tan^{-1}(frac{4}{3})=frac{32}{5}-4tan^{-1}(frac{3}{4})approx1.252$
answered Dec 9 at 12:46
Anubhab Ghosal
74412
the numerical result is correct, I don't believe other calculation: that of BEG and the last passage where $4pi$ disappear. I think that BEG=$16-32/5 -8tan^{-1}(4/3)$
– user126154
Dec 10 at 10:12
ok, now I understant the last passage: $4/3to 3/4$ but I still think BEG is wrong
– user126154
Dec 10 at 10:43
@user126154 $frac{pi}{2}-tan^{-1}(frac{4}{3})=tan^{-1}(frac{3}{4})$
– Anubhab Ghosal
Dec 10 at 11:11
@user126154, Thanks. Fixed the typo.
– Anubhab Ghosal
Dec 10 at 11:22
add a comment |
Label the vertices as shown in the diagram.
$tanalpha=frac{1}{2}implies tan 2alpha=frac{4}{3}implies 2alpha=tan^{-1}(frac{4}{3})$
$therefore$ Area of circular sector$ BAG=8tan^{-1}(frac{4}{3})$
Area of $triangle AGC=16cosalphasinalpha=8sin2alpha=frac{32}{5}$
Thus, area of curved figure $BEG=frac{48}{5}-8tan^{-1}(frac{4}{3})$
Therefore, Area of shaded part$=frac{1}{2}$ (Area of $BEFC$)$-frac{1}{2}$ (Area of semicircle)$-$Area of curved figure $BEG$.
This gives $frac{32}{5}-4pi+8tan^{-1}(frac{4}{3})=frac{32}{5}-4tan^{-1}(frac{3}{4})approx1.252$
answered Dec 9 at 12:46
Anubhab Ghosal
74412
the numerical result is correct, I don't believe other calculation: that of BEG and the last passage where $4pi$ disappear. I think that BEG=$16-32/5 -8tan^{-1}(4/3)$
– user126154
Dec 10 at 10:12
ok, now I understant the last passage: $4/3to 3/4$ but I still think BEG is wrong
– user126154
Dec 10 at 10:43
@user126154 $frac{pi}{2}-tan^{-1}(frac{4}{3})=tan^{-1}(frac{3}{4})$
– Anubhab Ghosal
Dec 10 at 11:11
@user126154, Thanks. Fixed the typo.
– Anubhab Ghosal
Dec 10 at 11:22
add a comment |
Label the vertices as shown in the diagram.
$tanalpha=frac{1}{2}implies tan 2alpha=frac{4}{3}implies 2alpha=tan^{-1}(frac{4}{3})$
$therefore$ Area of circular sector$ BAG=8tan^{-1}(frac{4}{3})$
Area of $triangle AGC=16cosalphasinalpha=8sin2alpha=frac{32}{5}$
Thus, area of curved figure $BEG=frac{48}{5}-8tan^{-1}(frac{4}{3})$
Therefore, Area of shaded part$=frac{1}{2}$ (Area of $BEFC$)$-frac{1}{2}$ (Area of semicircle)$-$Area of curved figure $BEG$.
This gives $frac{32}{5}-4pi+8tan^{-1}(frac{4}{3})=frac{32}{5}-4tan^{-1}(frac{3}{4})approx1.252$
answered Dec 9 at 12:46
Anubhab Ghosal
74412
Label the vertices as shown in the diagram.
$tanalpha=frac{1}{2}implies tan 2alpha=frac{4}{3}implies 2alpha=tan^{-1}(frac{4}{3})$
$therefore$ Area of circular sector$ BAG=8tan^{-1}(frac{4}{3})$
Area of $triangle AGC=16cosalphasinalpha=8sin2alpha=frac{32}{5}$
Thus, area of curved figure $BEG=frac{48}{5}-8tan^{-1}(frac{4}{3})$
Therefore, Area of shaded part$=frac{1}{2}$ (Area of $BEFC$)$-frac{1}{2}$ (Area of semicircle)$-$Area of curved figure $BEG$.
This gives $frac{32}{5}-4pi+8tan^{-1}(frac{4}{3})=frac{32}{5}-4tan^{-1}(frac{3}{4})approx1.252$
answered Dec 9 at 12:46
Anubhab Ghosal
74412
edited Dec 10 at 11:22
answered Dec 9 at 12:46
Anubhab Ghosal
74412
answered Dec 9 at 12:46
Anubhab Ghosal
74412
answered Dec 9 at 12:46
Anubhab Ghosal
74412
74412
the numerical result is correct, I don't believe other calculation: that of BEG and the last passage where $4pi$ disappear. I think that BEG=$16-32/5 -8tan^{-1}(4/3)$
– user126154
Dec 10 at 10:12
ok, now I understant the last passage: $4/3to 3/4$ but I still think BEG is wrong
– user126154
Dec 10 at 10:43
@user126154 $frac{pi}{2}-tan^{-1}(frac{4}{3})=tan^{-1}(frac{3}{4})$
– Anubhab Ghosal
Dec 10 at 11:11
@user126154, Thanks. Fixed the typo.
– Anubhab Ghosal
Dec 10 at 11:22
add a comment |
the numerical result is correct, I don't believe other calculation: that of BEG and the last passage where $4pi$ disappear. I think that BEG=$16-32/5 -8tan^{-1}(4/3)$
– user126154
Dec 10 at 10:12
ok, now I understant the last passage: $4/3to 3/4$ but I still think BEG is wrong
– user126154
Dec 10 at 10:43
@user126154 $frac{pi}{2}-tan^{-1}(frac{4}{3})=tan^{-1}(frac{3}{4})$
– Anubhab Ghosal
Dec 10 at 11:11
@user126154, Thanks. Fixed the typo.
– Anubhab Ghosal
Dec 10 at 11:22
the numerical result is correct, I don't believe other calculation: that of BEG and the last passage where $4pi$ disappear. I think that BEG=$16-32/5 -8tan^{-1}(4/3)$
– user126154
Dec 10 at 10:12
the numerical result is correct, I don't believe other calculation: that of BEG and the last passage where $4pi$ disappear. I think that BEG=$16-32/5 -8tan^{-1}(4/3)$
– user126154
Dec 10 at 10:12
ok, now I understant the last passage: $4/3to 3/4$ but I still think BEG is wrong
– user126154
Dec 10 at 10:43
ok, now I understant the last passage: $4/3to 3/4$ but I still think BEG is wrong
– user126154
Dec 10 at 10:43
@user126154 $frac{pi}{2}-tan^{-1}(frac{4}{3})=tan^{-1}(frac{3}{4})$
– Anubhab Ghosal
Dec 10 at 11:11
@user126154 $frac{pi}{2}-tan^{-1}(frac{4}{3})=tan^{-1}(frac{3}{4})$
– Anubhab Ghosal
Dec 10 at 11:11
@user126154, Thanks. Fixed the typo.
– Anubhab Ghosal
Dec 10 at 11:22
@user126154, Thanks. Fixed the typo.
– Anubhab Ghosal
Dec 10 at 11:22
add a comment |
Flip the image by vertically.
It is not as simple as you think. Year six is probably high school when they can do geometry and calculus. I have done it using calculus and the area is as shown in the figure
answered Dec 9 at 12:13
Satish Ramanathan
9,41531323
It widened my approaches and a fantastic link; thank you all!
– Waitara Mburu
Dec 10 at 10:34
@WaitaraMburu, while one always appreciates if someone benefits from their efforts, it is always good if you upvote and accept an answer(especially answers of new users like me), if you find the answer correct and useful.
– Anubhab Ghosal
Dec 10 at 11:15
add a comment |
Flip the image by vertically.
It is not as simple as you think. Year six is probably high school when they can do geometry and calculus. I have done it using calculus and the area is as shown in the figure
answered Dec 9 at 12:13
Satish Ramanathan
9,41531323
It widened my approaches and a fantastic link; thank you all!
– Waitara Mburu
Dec 10 at 10:34
@WaitaraMburu, while one always appreciates if someone benefits from their efforts, it is always good if you upvote and accept an answer(especially answers of new users like me), if you find the answer correct and useful.
– Anubhab Ghosal
Dec 10 at 11:15
add a comment |
Flip the image by vertically.
It is not as simple as you think. Year six is probably high school when they can do geometry and calculus. I have done it using calculus and the area is as shown in the figure
answered Dec 9 at 12:13
Satish Ramanathan
9,41531323
Flip the image by vertically.
It is not as simple as you think. Year six is probably high school when they can do geometry and calculus. I have done it using calculus and the area is as shown in the figure
answered Dec 9 at 12:13
Satish Ramanathan
9,41531323
answered Dec 9 at 12:13
Satish Ramanathan
9,41531323
answered Dec 9 at 12:13
Satish Ramanathan
9,41531323
answered Dec 9 at 12:13
Satish Ramanathan
9,41531323
9,41531323
It widened my approaches and a fantastic link; thank you all!
– Waitara Mburu
Dec 10 at 10:34
@WaitaraMburu, while one always appreciates if someone benefits from their efforts, it is always good if you upvote and accept an answer(especially answers of new users like me), if you find the answer correct and useful.
– Anubhab Ghosal
Dec 10 at 11:15
add a comment |
It widened my approaches and a fantastic link; thank you all!
– Waitara Mburu
Dec 10 at 10:34
@WaitaraMburu, while one always appreciates if someone benefits from their efforts, it is always good if you upvote and accept an answer(especially answers of new users like me), if you find the answer correct and useful.
– Anubhab Ghosal
Dec 10 at 11:15
It widened my approaches and a fantastic link; thank you all!
– Waitara Mburu
Dec 10 at 10:34
It widened my approaches and a fantastic link; thank you all!
– Waitara Mburu
Dec 10 at 10:34
@WaitaraMburu, while one always appreciates if someone benefits from their efforts, it is always good if you upvote and accept an answer(especially answers of new users like me), if you find the answer correct and useful.
– Anubhab Ghosal
Dec 10 at 11:15
@WaitaraMburu, while one always appreciates if someone benefits from their efforts, it is always good if you upvote and accept an answer(especially answers of new users like me), if you find the answer correct and useful.
– Anubhab Ghosal
Dec 10 at 11:15
add a comment |
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Are you sure about your answer? It looks a little large to me. I would say that using the symmetry of the diagram you should be expecting something like $8-2pi$.
– postmortes
Dec 9 at 8:55
Can you provide details about what you'd expect in sixth year? Visitors here come from a wide variety of contries and educational systems. I assume they wouldn't be doing integrals either?
– MvG
Dec 9 at 11:51
@MvG, it doesn't matter now as the actual answer is $frac{32}{5}-4tan^{-1}(frac{3}{4})$, and so you can't do it without trigonometry.
– Anubhab Ghosal
Dec 9 at 13:11
This answer finds the area of a related region. In the notation used there, the target area here can be written as $u-|text{region}PAT|$.
– Blue
Dec 9 at 14:24