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
    2 hours ago










  • $begingroup$
    Hmmn! It's equilateral!
    $endgroup$
    – Abdulhameed
    2 hours ago






  • 1




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










  • $begingroup$
    Hmmn! It's equilateral!
    $endgroup$
    – Abdulhameed
    2 hours ago






  • 1




    $begingroup$
    Then just use one of the previous solutions, since you know the angle.
    $endgroup$
    – Andrei
    2 hours 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
    2 hours 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
    2 hours 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 2 hours ago







Abdulhameed

















asked 2 hours ago









AbdulhameedAbdulhameed

122113




122113








  • 1




    $begingroup$
    Notice anything special about $triangle ABC$?
    $endgroup$
    – Blue
    2 hours ago










  • $begingroup$
    Hmmn! It's equilateral!
    $endgroup$
    – Abdulhameed
    2 hours ago






  • 1




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














  • 1




    $begingroup$
    Notice anything special about $triangle ABC$?
    $endgroup$
    – Blue
    2 hours ago










  • $begingroup$
    Hmmn! It's equilateral!
    $endgroup$
    – Abdulhameed
    2 hours ago






  • 1




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








1




1




$begingroup$
Notice anything special about $triangle ABC$?
$endgroup$
– Blue
2 hours ago




$begingroup$
Notice anything special about $triangle ABC$?
$endgroup$
– Blue
2 hours ago












$begingroup$
Hmmn! It's equilateral!
$endgroup$
– Abdulhameed
2 hours ago




$begingroup$
Hmmn! It's equilateral!
$endgroup$
– Abdulhameed
2 hours ago




1




1




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




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










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










  • $begingroup$
    This is so beautiful! Its well appreciated
    $endgroup$
    – Abdulhameed
    1 hour 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
    1 hour 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
    1 hour ago










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










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










  • $begingroup$
    This is so beautiful! Its well appreciated
    $endgroup$
    – Abdulhameed
    1 hour 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
    1 hour 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
    1 hour ago










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










  • $begingroup$
    This is so beautiful! Its well appreciated
    $endgroup$
    – Abdulhameed
    1 hour 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
    1 hour 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 1 hour ago

























answered 2 hours 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
    1 hour ago










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










  • $begingroup$
    This is so beautiful! Its well appreciated
    $endgroup$
    – Abdulhameed
    1 hour 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
    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
    1 hour ago










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










  • $begingroup$
    This is so beautiful! Its well appreciated
    $endgroup$
    – Abdulhameed
    1 hour 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
    1 hour 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
1 hour ago




$begingroup$
@Abdulhameed Added a picture
$endgroup$
– Bram28
1 hour ago












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




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












$begingroup$
This is so beautiful! Its well appreciated
$endgroup$
– Abdulhameed
1 hour ago




$begingroup$
This is so beautiful! Its well appreciated
$endgroup$
– Abdulhameed
1 hour 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
1 hour 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
1 hour 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
    1 hour ago










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










  • $begingroup$
    @ZubinMukerjee oh, yeah... mental slip
    $endgroup$
    – Daniel Mathias
    1 hour 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 58 mins ago

























answered 1 hour ago









Daniel MathiasDaniel Mathias

81917




81917












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










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


















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










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
















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




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












$begingroup$
@ZubinMukerjee oh, yeah... mental slip
$endgroup$
– Daniel Mathias
1 hour ago




$begingroup$
@ZubinMukerjee oh, yeah... mental slip
$endgroup$
– Daniel Mathias
1 hour 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

الفوسفات في المغرب

Image
قطار ينقل فوسفات من منطقة خريبكة متجه نحو الدار البيضاء. ديار عمال بمصانع الفوسفات بمدينة آسفي. المحتوى هنا ينقصه الاستشهاد بمصادر. يرجى إيراد مصادر موثوق بها. أي معلومات غير موثقة يمكن التشكيك بها وإزالتها. (أغسطس 2015) الفوسفات في المغرب هو ثالث مصدر عالمي للفوسفات، ويتوفر المغرب على %75 من الاحتياطي العالمي  [بحاجة لمصدر] ويتوزع بنسب متفاوتة بين عدة مناطق أساسية وهي: وادي زم بن جرير بوكراع خريبكة اليوسفية أولاد عبدون كنتور واد الذهب مستقلة. هذه الأخيرة لم تفتح بعد للاستغلال. الصادرات الحامض الفوسفوري P205 تمتص الهند %50 من ا لصادرات الإجمالية آسيا (دون روسيا)%59 أوروبا %30 أمريكا اللاتينية %11 أسمدة صلبة أوروبا 38 % أمريكا اللاتينية %30 آسيا (دون روسيا) 22 % أستراليا 9 % بوابة المغرب هذه بذرة مقالة عن علم الاقتصاد أو موضوع متعلق به بحاجة للتوسيع. شارك في تحريرها. This page is only for reference, If you need detailed information, please check here
Read more

بطل الاتحاد السوفيتي

Image
بطل الاتحاد السوفيتي   البلد الاتحاد السوفيتي  أول جائزة 16 أبريل 1934  العدد الممنوح 12776 [1]   تعديل   وسام بطل الاتحاد السوفيتي بطل الاتحاد السوفيتي (بالروسية:Герой Советского Союза؛ جيروي سويتسكوفو سيوزا) كان أعلى وسام في الاتحاد السوفيتي، منح لمجموعات وأشخاصا منفردين قدموا خدمات جليلة للمجتمع والدولة السوفيتية. مجموع الأوسمة التي مُنحت هو 12,775 وسامًا، مُنح أولها في 16 أبريل 1939، وآخرها في 20 ديسمبر 1991. بعض من نال اللقب مرة واحدة Rouslan Aouchev يوري جاجارين Valentina Grizodoubova Anna Iegorova ميليتون كنتاريا Vladimir Konovalov Zoïa Kosmodemianskaïa نيقولاي غيراسيموفيتش كوزنيتسوف Alexandre Marinesko Nikolai Melnik أوتو شميدت ريخارد زورغه فالنتينا تريشكوفا Evgueni Tolstikov فاسيلي زايتسيف Ievgueni Pepeliaïev ناتاليا ميكلين Наталья Мекли مرتان إيفان باغراميان Sydir Kovpak Amet-Khan Soultan Ivan Papanine سيميون تيموشينكو 3 مرات Semion Boudienny Ivan Kojedoub ألكسندر بوكريشكين 4 مرات ليون
Read more