Four equal circles intersect: What is the area of the small shaded portion and its height












3












$begingroup$


In terms of $R$ which is the radius of all four circles, what is the area of the intersection region of these four equal circles and the height of the marked arrow in the figure? The marked arrow is along the line CD, also the midpoint of all the circles are points A, B, C and D. Looking for a very short intuitive solution. I have checked similar questions on this site for example this and this.



enter image description here






geometry circle







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Notice anything special about $triangle ABC$?
    $endgroup$
    – Blue
    1 hour ago










  • $begingroup$
    Hmmn! It's equilateral!
    $endgroup$
    – Abdulhameed
    1 hour ago






  • 1




    $begingroup$
    Then just use one of the previous solutions, since you know the angle.
    $endgroup$
    – Andrei
    1 hour ago










  • $begingroup$
    Thanks! I think its clear now! I need to subtract the area of the sector from the triangle to get the small area and then multiply by two. What about the height any clues?
    $endgroup$
    – Abdulhameed
    1 hour ago












  • $begingroup$
    @Abdulhameed: For height, use the equilateral triangles again. If you know the height of a triangle, and the radius of the circle, then ...
    $endgroup$
    – Blue
    1 hour ago
















3












$begingroup$


In terms of $R$ which is the radius of all four circles, what is the area of the intersection region of these four equal circles and the height of the marked arrow in the figure? The marked arrow is along the line CD, also the midpoint of all the circles are points A, B, C and D. Looking for a very short intuitive solution. I have checked similar questions on this site for example this and this.



enter image description here






geometry circle







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Notice anything special about $triangle ABC$?
    $endgroup$
    – Blue
    1 hour ago










  • $begingroup$
    Hmmn! It's equilateral!
    $endgroup$
    – Abdulhameed
    1 hour ago






  • 1




    $begingroup$
    Then just use one of the previous solutions, since you know the angle.
    $endgroup$
    – Andrei
    1 hour ago










  • $begingroup$
    Thanks! I think its clear now! I need to subtract the area of the sector from the triangle to get the small area and then multiply by two. What about the height any clues?
    $endgroup$
    – Abdulhameed
    1 hour ago












  • $begingroup$
    @Abdulhameed: For height, use the equilateral triangles again. If you know the height of a triangle, and the radius of the circle, then ...
    $endgroup$
    – Blue
    1 hour ago














3












3








3





$begingroup$


In terms of $R$ which is the radius of all four circles, what is the area of the intersection region of these four equal circles and the height of the marked arrow in the figure? The marked arrow is along the line CD, also the midpoint of all the circles are points A, B, C and D. Looking for a very short intuitive solution. I have checked similar questions on this site for example this and this.



enter image description here






geometry circle







share|cite|improve this question











$endgroup$




In terms of $R$ which is the radius of all four circles, what is the area of the intersection region of these four equal circles and the height of the marked arrow in the figure? The marked arrow is along the line CD, also the midpoint of all the circles are points A, B, C and D. Looking for a very short intuitive solution. I have checked similar questions on this site for example this and this.



enter image description here






geometry circle




geometry circle






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 1 hour ago







Abdulhameed

















asked 1 hour ago









AbdulhameedAbdulhameed

122113




122113








  • 1




    $begingroup$
    Notice anything special about $triangle ABC$?
    $endgroup$
    – Blue
    1 hour ago










  • $begingroup$
    Hmmn! It's equilateral!
    $endgroup$
    – Abdulhameed
    1 hour ago






  • 1




    $begingroup$
    Then just use one of the previous solutions, since you know the angle.
    $endgroup$
    – Andrei
    1 hour ago










  • $begingroup$
    Thanks! I think its clear now! I need to subtract the area of the sector from the triangle to get the small area and then multiply by two. What about the height any clues?
    $endgroup$
    – Abdulhameed
    1 hour ago












  • $begingroup$
    @Abdulhameed: For height, use the equilateral triangles again. If you know the height of a triangle, and the radius of the circle, then ...
    $endgroup$
    – Blue
    1 hour ago














  • 1




    $begingroup$
    Notice anything special about $triangle ABC$?
    $endgroup$
    – Blue
    1 hour ago










  • $begingroup$
    Hmmn! It's equilateral!
    $endgroup$
    – Abdulhameed
    1 hour ago






  • 1




    $begingroup$
    Then just use one of the previous solutions, since you know the angle.
    $endgroup$
    – Andrei
    1 hour ago










  • $begingroup$
    Thanks! I think its clear now! I need to subtract the area of the sector from the triangle to get the small area and then multiply by two. What about the height any clues?
    $endgroup$
    – Abdulhameed
    1 hour ago












  • $begingroup$
    @Abdulhameed: For height, use the equilateral triangles again. If you know the height of a triangle, and the radius of the circle, then ...
    $endgroup$
    – Blue
    1 hour ago








1




1




$begingroup$
Notice anything special about $triangle ABC$?
$endgroup$
– Blue
1 hour ago




$begingroup$
Notice anything special about $triangle ABC$?
$endgroup$
– Blue
1 hour ago












$begingroup$
Hmmn! It's equilateral!
$endgroup$
– Abdulhameed
1 hour ago




$begingroup$
Hmmn! It's equilateral!
$endgroup$
– Abdulhameed
1 hour ago




1




1




$begingroup$
Then just use one of the previous solutions, since you know the angle.
$endgroup$
– Andrei
1 hour ago




$begingroup$
Then just use one of the previous solutions, since you know the angle.
$endgroup$
– Andrei
1 hour ago












$begingroup$
Thanks! I think its clear now! I need to subtract the area of the sector from the triangle to get the small area and then multiply by two. What about the height any clues?
$endgroup$
– Abdulhameed
1 hour ago






$begingroup$
Thanks! I think its clear now! I need to subtract the area of the sector from the triangle to get the small area and then multiply by two. What about the height any clues?
$endgroup$
– Abdulhameed
1 hour ago














$begingroup$
@Abdulhameed: For height, use the equilateral triangles again. If you know the height of a triangle, and the radius of the circle, then ...
$endgroup$
– Blue
1 hour ago




$begingroup$
@Abdulhameed: For height, use the equilateral triangles again. If you know the height of a triangle, and the radius of the circle, then ...
$endgroup$
– Blue
1 hour ago










2 Answers
2






active

oldest

votes


















4












$begingroup$

enter image description hereIf you repeat this pattern infinitely, you'll find that each circle consists of $12$ of these football shaped areas, together with $6$ 'triangular' areas in between (see red circle in figure). So, setting the area of the footballs to $A$, and that of the triangles to $B$, we have:



$pi^2=12A +6B$ (I am setting radius to $1$ ... you can easily adjust for $R$)



OK, now consider the green rectangle formed by four of the points on a circle. We see that the height of such a rectangle is equal to the radius, so that is $1$, and the width is easily found to be $sqrt{3}$, and this area includes $4$ whole footballs, $2$ half footballs, $2$ whole 'triangles', and $4$ half triangles. So:



$sqrt{3}=5A+4B$



Now you can easily solve for $A$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I appreciate this effort but could you make this clearer.
    $endgroup$
    – Abdulhameed
    1 hour ago










  • $begingroup$
    @Abdulhameed Added a picture
    $endgroup$
    – Bram28
    42 mins ago










  • $begingroup$
    @Bram28 really nice picture, +1
    $endgroup$
    – Zubin Mukerjee
    39 mins ago










  • $begingroup$
    This is so beautiful! Its well appreciated
    $endgroup$
    – Abdulhameed
    35 mins ago










  • $begingroup$
    @Abdulhameed Another solution is to take $1/6$ of the area of the area of the circle of radius $1$, subtract the area of an equilateral triangle with side length $1$, then multiply by $2$: $$2left(frac{1}{6}pi^2 - frac{sqrt{3}}{4}right)$$
    $endgroup$
    – Zubin Mukerjee
    29 mins ago



















1












$begingroup$

enter image description here



Notice that the area of the equilateral triangle with edge $R$ plus $frac12$ the area of the shaded region is $frac16$ the area of the circle. The height of the triangle is $frac{Rsqrt3}{2}$ and the area is $frac12cdot Rcdotfrac{Rsqrt3}{2}=frac{R^2sqrt3}{4}$. The area of the shaded region is:
$$2left(frac{pi R^2}{6}-frac{R^2sqrt3}{4}right)=frac{2pi R^2}{6}-frac{3R^2sqrt3}{6}=frac{R^2(2pi-3sqrt3)}{6}$$
Also notice that the height of the triangle plus $frac12$ the height of the shaded area is equal to the radius of the circle. The height of the shaded area is:
$$2left(R-frac{Rsqrt3}{2}right)=2R-Rsqrt3=R(2-sqrt3)$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The area of the triangle should be $$frac{R^2sqrt{3}}{4}$$
    $endgroup$
    – Zubin Mukerjee
    24 mins ago










  • $begingroup$
    @ZubinMukerjee oh, yeah... mental slip
    $endgroup$
    – Daniel Mathias
    23 mins ago











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3089033%2ffour-equal-circles-intersect-what-is-the-area-of-the-small-shaded-portion-and-i%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

enter image description hereIf you repeat this pattern infinitely, you'll find that each circle consists of $12$ of these football shaped areas, together with $6$ 'triangular' areas in between (see red circle in figure). So, setting the area of the footballs to $A$, and that of the triangles to $B$, we have:



$pi^2=12A +6B$ (I am setting radius to $1$ ... you can easily adjust for $R$)



OK, now consider the green rectangle formed by four of the points on a circle. We see that the height of such a rectangle is equal to the radius, so that is $1$, and the width is easily found to be $sqrt{3}$, and this area includes $4$ whole footballs, $2$ half footballs, $2$ whole 'triangles', and $4$ half triangles. So:



$sqrt{3}=5A+4B$



Now you can easily solve for $A$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I appreciate this effort but could you make this clearer.
    $endgroup$
    – Abdulhameed
    1 hour ago










  • $begingroup$
    @Abdulhameed Added a picture
    $endgroup$
    – Bram28
    42 mins ago










  • $begingroup$
    @Bram28 really nice picture, +1
    $endgroup$
    – Zubin Mukerjee
    39 mins ago










  • $begingroup$
    This is so beautiful! Its well appreciated
    $endgroup$
    – Abdulhameed
    35 mins ago










  • $begingroup$
    @Abdulhameed Another solution is to take $1/6$ of the area of the area of the circle of radius $1$, subtract the area of an equilateral triangle with side length $1$, then multiply by $2$: $$2left(frac{1}{6}pi^2 - frac{sqrt{3}}{4}right)$$
    $endgroup$
    – Zubin Mukerjee
    29 mins ago
















4












$begingroup$

enter image description hereIf you repeat this pattern infinitely, you'll find that each circle consists of $12$ of these football shaped areas, together with $6$ 'triangular' areas in between (see red circle in figure). So, setting the area of the footballs to $A$, and that of the triangles to $B$, we have:



$pi^2=12A +6B$ (I am setting radius to $1$ ... you can easily adjust for $R$)



OK, now consider the green rectangle formed by four of the points on a circle. We see that the height of such a rectangle is equal to the radius, so that is $1$, and the width is easily found to be $sqrt{3}$, and this area includes $4$ whole footballs, $2$ half footballs, $2$ whole 'triangles', and $4$ half triangles. So:



$sqrt{3}=5A+4B$



Now you can easily solve for $A$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    I appreciate this effort but could you make this clearer.
    $endgroup$
    – Abdulhameed
    1 hour ago










  • $begingroup$
    @Abdulhameed Added a picture
    $endgroup$
    – Bram28
    42 mins ago










  • $begingroup$
    @Bram28 really nice picture, +1
    $endgroup$
    – Zubin Mukerjee
    39 mins ago










  • $begingroup$
    This is so beautiful! Its well appreciated
    $endgroup$
    – Abdulhameed
    35 mins ago










  • $begingroup$
    @Abdulhameed Another solution is to take $1/6$ of the area of the area of the circle of radius $1$, subtract the area of an equilateral triangle with side length $1$, then multiply by $2$: $$2left(frac{1}{6}pi^2 - frac{sqrt{3}}{4}right)$$
    $endgroup$
    – Zubin Mukerjee
    29 mins ago














4












4








4





$begingroup$

enter image description hereIf you repeat this pattern infinitely, you'll find that each circle consists of $12$ of these football shaped areas, together with $6$ 'triangular' areas in between (see red circle in figure). So, setting the area of the footballs to $A$, and that of the triangles to $B$, we have:



$pi^2=12A +6B$ (I am setting radius to $1$ ... you can easily adjust for $R$)



OK, now consider the green rectangle formed by four of the points on a circle. We see that the height of such a rectangle is equal to the radius, so that is $1$, and the width is easily found to be $sqrt{3}$, and this area includes $4$ whole footballs, $2$ half footballs, $2$ whole 'triangles', and $4$ half triangles. So:



$sqrt{3}=5A+4B$



Now you can easily solve for $A$






share|cite|improve this answer











$endgroup$



enter image description hereIf you repeat this pattern infinitely, you'll find that each circle consists of $12$ of these football shaped areas, together with $6$ 'triangular' areas in between (see red circle in figure). So, setting the area of the footballs to $A$, and that of the triangles to $B$, we have:



$pi^2=12A +6B$ (I am setting radius to $1$ ... you can easily adjust for $R$)



OK, now consider the green rectangle formed by four of the points on a circle. We see that the height of such a rectangle is equal to the radius, so that is $1$, and the width is easily found to be $sqrt{3}$, and this area includes $4$ whole footballs, $2$ half footballs, $2$ whole 'triangles', and $4$ half triangles. So:



$sqrt{3}=5A+4B$



Now you can easily solve for $A$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 42 mins ago

























answered 1 hour ago









Bram28Bram28

61.2k44591




61.2k44591












  • $begingroup$
    I appreciate this effort but could you make this clearer.
    $endgroup$
    – Abdulhameed
    1 hour ago










  • $begingroup$
    @Abdulhameed Added a picture
    $endgroup$
    – Bram28
    42 mins ago










  • $begingroup$
    @Bram28 really nice picture, +1
    $endgroup$
    – Zubin Mukerjee
    39 mins ago










  • $begingroup$
    This is so beautiful! Its well appreciated
    $endgroup$
    – Abdulhameed
    35 mins ago










  • $begingroup$
    @Abdulhameed Another solution is to take $1/6$ of the area of the area of the circle of radius $1$, subtract the area of an equilateral triangle with side length $1$, then multiply by $2$: $$2left(frac{1}{6}pi^2 - frac{sqrt{3}}{4}right)$$
    $endgroup$
    – Zubin Mukerjee
    29 mins ago


















  • $begingroup$
    I appreciate this effort but could you make this clearer.
    $endgroup$
    – Abdulhameed
    1 hour ago










  • $begingroup$
    @Abdulhameed Added a picture
    $endgroup$
    – Bram28
    42 mins ago










  • $begingroup$
    @Bram28 really nice picture, +1
    $endgroup$
    – Zubin Mukerjee
    39 mins ago










  • $begingroup$
    This is so beautiful! Its well appreciated
    $endgroup$
    – Abdulhameed
    35 mins ago










  • $begingroup$
    @Abdulhameed Another solution is to take $1/6$ of the area of the area of the circle of radius $1$, subtract the area of an equilateral triangle with side length $1$, then multiply by $2$: $$2left(frac{1}{6}pi^2 - frac{sqrt{3}}{4}right)$$
    $endgroup$
    – Zubin Mukerjee
    29 mins ago
















$begingroup$
I appreciate this effort but could you make this clearer.
$endgroup$
– Abdulhameed
1 hour ago




$begingroup$
I appreciate this effort but could you make this clearer.
$endgroup$
– Abdulhameed
1 hour ago












$begingroup$
@Abdulhameed Added a picture
$endgroup$
– Bram28
42 mins ago




$begingroup$
@Abdulhameed Added a picture
$endgroup$
– Bram28
42 mins ago












$begingroup$
@Bram28 really nice picture, +1
$endgroup$
– Zubin Mukerjee
39 mins ago




$begingroup$
@Bram28 really nice picture, +1
$endgroup$
– Zubin Mukerjee
39 mins ago












$begingroup$
This is so beautiful! Its well appreciated
$endgroup$
– Abdulhameed
35 mins ago




$begingroup$
This is so beautiful! Its well appreciated
$endgroup$
– Abdulhameed
35 mins ago












$begingroup$
@Abdulhameed Another solution is to take $1/6$ of the area of the area of the circle of radius $1$, subtract the area of an equilateral triangle with side length $1$, then multiply by $2$: $$2left(frac{1}{6}pi^2 - frac{sqrt{3}}{4}right)$$
$endgroup$
– Zubin Mukerjee
29 mins ago




$begingroup$
@Abdulhameed Another solution is to take $1/6$ of the area of the area of the circle of radius $1$, subtract the area of an equilateral triangle with side length $1$, then multiply by $2$: $$2left(frac{1}{6}pi^2 - frac{sqrt{3}}{4}right)$$
$endgroup$
– Zubin Mukerjee
29 mins ago











1












$begingroup$

enter image description here



Notice that the area of the equilateral triangle with edge $R$ plus $frac12$ the area of the shaded region is $frac16$ the area of the circle. The height of the triangle is $frac{Rsqrt3}{2}$ and the area is $frac12cdot Rcdotfrac{Rsqrt3}{2}=frac{R^2sqrt3}{4}$. The area of the shaded region is:
$$2left(frac{pi R^2}{6}-frac{R^2sqrt3}{4}right)=frac{2pi R^2}{6}-frac{3R^2sqrt3}{6}=frac{R^2(2pi-3sqrt3)}{6}$$
Also notice that the height of the triangle plus $frac12$ the height of the shaded area is equal to the radius of the circle. The height of the shaded area is:
$$2left(R-frac{Rsqrt3}{2}right)=2R-Rsqrt3=R(2-sqrt3)$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The area of the triangle should be $$frac{R^2sqrt{3}}{4}$$
    $endgroup$
    – Zubin Mukerjee
    24 mins ago










  • $begingroup$
    @ZubinMukerjee oh, yeah... mental slip
    $endgroup$
    – Daniel Mathias
    23 mins ago
















1












$begingroup$

enter image description here



Notice that the area of the equilateral triangle with edge $R$ plus $frac12$ the area of the shaded region is $frac16$ the area of the circle. The height of the triangle is $frac{Rsqrt3}{2}$ and the area is $frac12cdot Rcdotfrac{Rsqrt3}{2}=frac{R^2sqrt3}{4}$. The area of the shaded region is:
$$2left(frac{pi R^2}{6}-frac{R^2sqrt3}{4}right)=frac{2pi R^2}{6}-frac{3R^2sqrt3}{6}=frac{R^2(2pi-3sqrt3)}{6}$$
Also notice that the height of the triangle plus $frac12$ the height of the shaded area is equal to the radius of the circle. The height of the shaded area is:
$$2left(R-frac{Rsqrt3}{2}right)=2R-Rsqrt3=R(2-sqrt3)$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    The area of the triangle should be $$frac{R^2sqrt{3}}{4}$$
    $endgroup$
    – Zubin Mukerjee
    24 mins ago










  • $begingroup$
    @ZubinMukerjee oh, yeah... mental slip
    $endgroup$
    – Daniel Mathias
    23 mins ago














1












1








1





$begingroup$

enter image description here



Notice that the area of the equilateral triangle with edge $R$ plus $frac12$ the area of the shaded region is $frac16$ the area of the circle. The height of the triangle is $frac{Rsqrt3}{2}$ and the area is $frac12cdot Rcdotfrac{Rsqrt3}{2}=frac{R^2sqrt3}{4}$. The area of the shaded region is:
$$2left(frac{pi R^2}{6}-frac{R^2sqrt3}{4}right)=frac{2pi R^2}{6}-frac{3R^2sqrt3}{6}=frac{R^2(2pi-3sqrt3)}{6}$$
Also notice that the height of the triangle plus $frac12$ the height of the shaded area is equal to the radius of the circle. The height of the shaded area is:
$$2left(R-frac{Rsqrt3}{2}right)=2R-Rsqrt3=R(2-sqrt3)$$






share|cite|improve this answer











$endgroup$



enter image description here



Notice that the area of the equilateral triangle with edge $R$ plus $frac12$ the area of the shaded region is $frac16$ the area of the circle. The height of the triangle is $frac{Rsqrt3}{2}$ and the area is $frac12cdot Rcdotfrac{Rsqrt3}{2}=frac{R^2sqrt3}{4}$. The area of the shaded region is:
$$2left(frac{pi R^2}{6}-frac{R^2sqrt3}{4}right)=frac{2pi R^2}{6}-frac{3R^2sqrt3}{6}=frac{R^2(2pi-3sqrt3)}{6}$$
Also notice that the height of the triangle plus $frac12$ the height of the shaded area is equal to the radius of the circle. The height of the shaded area is:
$$2left(R-frac{Rsqrt3}{2}right)=2R-Rsqrt3=R(2-sqrt3)$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 20 mins ago

























answered 27 mins ago









Daniel MathiasDaniel Mathias

81917




81917












  • $begingroup$
    The area of the triangle should be $$frac{R^2sqrt{3}}{4}$$
    $endgroup$
    – Zubin Mukerjee
    24 mins ago










  • $begingroup$
    @ZubinMukerjee oh, yeah... mental slip
    $endgroup$
    – Daniel Mathias
    23 mins ago


















  • $begingroup$
    The area of the triangle should be $$frac{R^2sqrt{3}}{4}$$
    $endgroup$
    – Zubin Mukerjee
    24 mins ago










  • $begingroup$
    @ZubinMukerjee oh, yeah... mental slip
    $endgroup$
    – Daniel Mathias
    23 mins ago
















$begingroup$
The area of the triangle should be $$frac{R^2sqrt{3}}{4}$$
$endgroup$
– Zubin Mukerjee
24 mins ago




$begingroup$
The area of the triangle should be $$frac{R^2sqrt{3}}{4}$$
$endgroup$
– Zubin Mukerjee
24 mins ago












$begingroup$
@ZubinMukerjee oh, yeah... mental slip
$endgroup$
– Daniel Mathias
23 mins ago




$begingroup$
@ZubinMukerjee oh, yeah... mental slip
$endgroup$
– Daniel Mathias
23 mins ago


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid …



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3089033%2ffour-equal-circles-intersect-what-is-the-area-of-the-small-shaded-portion-and-i%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







-

Popular posts from this blog

A CLEAN and SIMPLE way to add appendices to Table of Contents and bookmarks

Image
0 I found a lot of questions abount appendices and ToC. Many users want appendices to be grouped in an Appendix part, however some problems arise with ToC, hyperref, PDF viewer bookmarks, and so on. There are different solutions which require extra packages, command patching and other extra code, however none of them satisfies me. I almost found an easy way to accomplish a good result, where appendices are added to bookmarks in the right way and hyperref links point to the right page. However, the number of the "Appendix" part page is wrong (it's the number of appendix A). Is there any EASY way to fix that? This is a MWE: documentclass{book} usepackage[nottoc,notlot,notlof]{tocbibind} usepackage{hyperref} begin{document} frontmatter tableofcontents mainmatter part{First} chapter{...
Read more

Calculate evaluation metrics using cross_val_predict sklearn

Image
1 In the sklearn.model_selection.cross_val_predict page it is stated: Generate cross-validated estimates for each input data point. It is not appropriate to pass these predictions into an evaluation metric. Can someone explain what does it mean? If this gives estimate of Y (y prediction) for every Y (true Y), why can't I calculate metrics such as RMSE or coefficient of determination using these results? python scikit-learn cross-validation share | improve this question edited Nov 28 '18 at 17:52 desertnaut 20.3k 7 43 79 ...
Read more

Insert data from modal to MySQL (multiple modal on website)

Image
0 I have problem with inserting data to MySQL from modal. My modal: <a href="#" class="badge badge-pill badge-success">6 komentarzy</a> <a href="#" class="badge badge-pill badge-danger">brak komentarzy</a> <a data-toggle="modal" href="#add_desk_comm_{$desk_['desk_id']}" data-target="#add_desk_comm_{$desk_['desk_id']}" class="ediiit">(dodaj)</a> <div class="modal fade" id="add_desk_comm_{$desk_['desk_id']}" tabindex="-1" role="dialog" aria-labelledby="edit_printer" aria-hidden="true"> <div class="modal-dialog" role="document"> ...
Read more