What is the area shaded in this figure?












1












$begingroup$


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?



enter image description here






geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    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 $8-2pi$.
    $endgroup$
    – postmortes
    Dec 9 '18 at 8:55










  • $begingroup$
    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?
    $endgroup$
    – MvG
    Dec 9 '18 at 11:51










  • $begingroup$
    @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.
    $endgroup$
    – Anubhab Ghosal
    Dec 9 '18 at 13:11










  • $begingroup$
    This answer finds the area of a related region. In the notation used there, the target area here can be written as $u-|text{region}PAT|$.
    $endgroup$
    – Blue
    Dec 9 '18 at 14:24


















1












$begingroup$


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?



enter image description here






geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    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 $8-2pi$.
    $endgroup$
    – postmortes
    Dec 9 '18 at 8:55










  • $begingroup$
    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?
    $endgroup$
    – MvG
    Dec 9 '18 at 11:51










  • $begingroup$
    @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.
    $endgroup$
    – Anubhab Ghosal
    Dec 9 '18 at 13:11










  • $begingroup$
    This answer finds the area of a related region. In the notation used there, the target area here can be written as $u-|text{region}PAT|$.
    $endgroup$
    – Blue
    Dec 9 '18 at 14:24
















1












1








1


2



$begingroup$


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?



enter image description here






geometry







share|cite|improve this question











$endgroup$




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?



enter image description here






geometry




geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 '18 at 12:05









Anubhab Ghosal

1,19619




1,19619










asked Dec 9 '18 at 8:40









Waitara MburuWaitara Mburu

415




415












  • $begingroup$
    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 $8-2pi$.
    $endgroup$
    – postmortes
    Dec 9 '18 at 8:55










  • $begingroup$
    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?
    $endgroup$
    – MvG
    Dec 9 '18 at 11:51










  • $begingroup$
    @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.
    $endgroup$
    – Anubhab Ghosal
    Dec 9 '18 at 13:11










  • $begingroup$
    This answer finds the area of a related region. In the notation used there, the target area here can be written as $u-|text{region}PAT|$.
    $endgroup$
    – Blue
    Dec 9 '18 at 14:24




















  • $begingroup$
    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 $8-2pi$.
    $endgroup$
    – postmortes
    Dec 9 '18 at 8:55










  • $begingroup$
    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?
    $endgroup$
    – MvG
    Dec 9 '18 at 11:51










  • $begingroup$
    @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.
    $endgroup$
    – Anubhab Ghosal
    Dec 9 '18 at 13:11










  • $begingroup$
    This answer finds the area of a related region. In the notation used there, the target area here can be written as $u-|text{region}PAT|$.
    $endgroup$
    – Blue
    Dec 9 '18 at 14:24


















$begingroup$
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 $8-2pi$.
$endgroup$
– postmortes
Dec 9 '18 at 8:55




$begingroup$
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 $8-2pi$.
$endgroup$
– postmortes
Dec 9 '18 at 8:55












$begingroup$
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?
$endgroup$
– MvG
Dec 9 '18 at 11:51




$begingroup$
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?
$endgroup$
– MvG
Dec 9 '18 at 11:51












$begingroup$
@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.
$endgroup$
– Anubhab Ghosal
Dec 9 '18 at 13:11




$begingroup$
@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.
$endgroup$
– Anubhab Ghosal
Dec 9 '18 at 13:11












$begingroup$
This answer finds the area of a related region. In the notation used there, the target area here can be written as $u-|text{region}PAT|$.
$endgroup$
– Blue
Dec 9 '18 at 14:24






$begingroup$
This answer finds the area of a related region. In the notation used there, the target area here can be written as $u-|text{region}PAT|$.
$endgroup$
– Blue
Dec 9 '18 at 14:24












2 Answers
2






active

oldest

votes


















1












$begingroup$

Diagram



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 $frac{32}{5}-4pi+8tan^{-1}(frac{4}{3})=frac{32}{5}-4tan^{-1}(frac{3}{4})approx1.252$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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=$16-32/5 -8tan^{-1}(4/3)$
    $endgroup$
    – user126154
    Dec 10 '18 at 10:12










  • $begingroup$
    ok, now I understant the last passage: $4/3to 3/4$ but I still think BEG is wrong
    $endgroup$
    – user126154
    Dec 10 '18 at 10:43










  • $begingroup$
    @user126154 $frac{pi}{2}-tan^{-1}(frac{4}{3})=tan^{-1}(frac{3}{4})$
    $endgroup$
    – Anubhab Ghosal
    Dec 10 '18 at 11:11












  • $begingroup$
    @user126154, Thanks. Fixed the typo.
    $endgroup$
    – Anubhab Ghosal
    Dec 10 '18 at 11:22



















1












$begingroup$

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
enter image description here






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It widened my approaches and a fantastic link; thank you all!
    $endgroup$
    – Waitara Mburu
    Dec 10 '18 at 10:34










  • $begingroup$
    @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.
    $endgroup$
    – Anubhab Ghosal
    Dec 10 '18 at 11:15











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

Diagram



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 $frac{32}{5}-4pi+8tan^{-1}(frac{4}{3})=frac{32}{5}-4tan^{-1}(frac{3}{4})approx1.252$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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=$16-32/5 -8tan^{-1}(4/3)$
    $endgroup$
    – user126154
    Dec 10 '18 at 10:12










  • $begingroup$
    ok, now I understant the last passage: $4/3to 3/4$ but I still think BEG is wrong
    $endgroup$
    – user126154
    Dec 10 '18 at 10:43










  • $begingroup$
    @user126154 $frac{pi}{2}-tan^{-1}(frac{4}{3})=tan^{-1}(frac{3}{4})$
    $endgroup$
    – Anubhab Ghosal
    Dec 10 '18 at 11:11












  • $begingroup$
    @user126154, Thanks. Fixed the typo.
    $endgroup$
    – Anubhab Ghosal
    Dec 10 '18 at 11:22
















1












$begingroup$

Diagram



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 $frac{32}{5}-4pi+8tan^{-1}(frac{4}{3})=frac{32}{5}-4tan^{-1}(frac{3}{4})approx1.252$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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=$16-32/5 -8tan^{-1}(4/3)$
    $endgroup$
    – user126154
    Dec 10 '18 at 10:12










  • $begingroup$
    ok, now I understant the last passage: $4/3to 3/4$ but I still think BEG is wrong
    $endgroup$
    – user126154
    Dec 10 '18 at 10:43










  • $begingroup$
    @user126154 $frac{pi}{2}-tan^{-1}(frac{4}{3})=tan^{-1}(frac{3}{4})$
    $endgroup$
    – Anubhab Ghosal
    Dec 10 '18 at 11:11












  • $begingroup$
    @user126154, Thanks. Fixed the typo.
    $endgroup$
    – Anubhab Ghosal
    Dec 10 '18 at 11:22














1












1








1





$begingroup$

Diagram



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 $frac{32}{5}-4pi+8tan^{-1}(frac{4}{3})=frac{32}{5}-4tan^{-1}(frac{3}{4})approx1.252$






share|cite|improve this answer











$endgroup$



Diagram



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 $frac{32}{5}-4pi+8tan^{-1}(frac{4}{3})=frac{32}{5}-4tan^{-1}(frac{3}{4})approx1.252$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 10 '18 at 11:22

























answered Dec 9 '18 at 12:46









Anubhab GhosalAnubhab Ghosal

1,19619




1,19619












  • $begingroup$
    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=$16-32/5 -8tan^{-1}(4/3)$
    $endgroup$
    – user126154
    Dec 10 '18 at 10:12










  • $begingroup$
    ok, now I understant the last passage: $4/3to 3/4$ but I still think BEG is wrong
    $endgroup$
    – user126154
    Dec 10 '18 at 10:43










  • $begingroup$
    @user126154 $frac{pi}{2}-tan^{-1}(frac{4}{3})=tan^{-1}(frac{3}{4})$
    $endgroup$
    – Anubhab Ghosal
    Dec 10 '18 at 11:11












  • $begingroup$
    @user126154, Thanks. Fixed the typo.
    $endgroup$
    – Anubhab Ghosal
    Dec 10 '18 at 11:22


















  • $begingroup$
    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=$16-32/5 -8tan^{-1}(4/3)$
    $endgroup$
    – user126154
    Dec 10 '18 at 10:12










  • $begingroup$
    ok, now I understant the last passage: $4/3to 3/4$ but I still think BEG is wrong
    $endgroup$
    – user126154
    Dec 10 '18 at 10:43










  • $begingroup$
    @user126154 $frac{pi}{2}-tan^{-1}(frac{4}{3})=tan^{-1}(frac{3}{4})$
    $endgroup$
    – Anubhab Ghosal
    Dec 10 '18 at 11:11












  • $begingroup$
    @user126154, Thanks. Fixed the typo.
    $endgroup$
    – Anubhab Ghosal
    Dec 10 '18 at 11:22
















$begingroup$
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=$16-32/5 -8tan^{-1}(4/3)$
$endgroup$
– user126154
Dec 10 '18 at 10:12




$begingroup$
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=$16-32/5 -8tan^{-1}(4/3)$
$endgroup$
– user126154
Dec 10 '18 at 10:12












$begingroup$
ok, now I understant the last passage: $4/3to 3/4$ but I still think BEG is wrong
$endgroup$
– user126154
Dec 10 '18 at 10:43




$begingroup$
ok, now I understant the last passage: $4/3to 3/4$ but I still think BEG is wrong
$endgroup$
– user126154
Dec 10 '18 at 10:43












$begingroup$
@user126154 $frac{pi}{2}-tan^{-1}(frac{4}{3})=tan^{-1}(frac{3}{4})$
$endgroup$
– Anubhab Ghosal
Dec 10 '18 at 11:11






$begingroup$
@user126154 $frac{pi}{2}-tan^{-1}(frac{4}{3})=tan^{-1}(frac{3}{4})$
$endgroup$
– Anubhab Ghosal
Dec 10 '18 at 11:11














$begingroup$
@user126154, Thanks. Fixed the typo.
$endgroup$
– Anubhab Ghosal
Dec 10 '18 at 11:22




$begingroup$
@user126154, Thanks. Fixed the typo.
$endgroup$
– Anubhab Ghosal
Dec 10 '18 at 11:22











1












$begingroup$

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
enter image description here






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It widened my approaches and a fantastic link; thank you all!
    $endgroup$
    – Waitara Mburu
    Dec 10 '18 at 10:34










  • $begingroup$
    @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.
    $endgroup$
    – Anubhab Ghosal
    Dec 10 '18 at 11:15
















1












$begingroup$

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
enter image description here






share|cite|improve this answer









$endgroup$













  • $begingroup$
    It widened my approaches and a fantastic link; thank you all!
    $endgroup$
    – Waitara Mburu
    Dec 10 '18 at 10:34










  • $begingroup$
    @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.
    $endgroup$
    – Anubhab Ghosal
    Dec 10 '18 at 11:15














1












1








1





$begingroup$

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
enter image description here






share|cite|improve this answer









$endgroup$



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
enter image description here







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 9 '18 at 12:13









Satish RamanathanSatish Ramanathan

10k31323




10k31323












  • $begingroup$
    It widened my approaches and a fantastic link; thank you all!
    $endgroup$
    – Waitara Mburu
    Dec 10 '18 at 10:34










  • $begingroup$
    @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.
    $endgroup$
    – Anubhab Ghosal
    Dec 10 '18 at 11:15


















  • $begingroup$
    It widened my approaches and a fantastic link; thank you all!
    $endgroup$
    – Waitara Mburu
    Dec 10 '18 at 10:34










  • $begingroup$
    @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.
    $endgroup$
    – Anubhab Ghosal
    Dec 10 '18 at 11:15
















$begingroup$
It widened my approaches and a fantastic link; thank you all!
$endgroup$
– Waitara Mburu
Dec 10 '18 at 10:34




$begingroup$
It widened my approaches and a fantastic link; thank you all!
$endgroup$
– Waitara Mburu
Dec 10 '18 at 10:34












$begingroup$
@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.
$endgroup$
– Anubhab Ghosal
Dec 10 '18 at 11:15




$begingroup$
@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.
$endgroup$
– Anubhab Ghosal
Dec 10 '18 at 11:15


















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How to change which sound is reproduced for terminal bell?

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3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
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Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

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up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
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Can I use Tabulator js library in my java Spring + Thymeleaf project?

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0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
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