Mixing
Simple geometry problem regarding fitting square inside a circle
$begingroup$
Consider a square of side $a$. Fit the largest possible circle inside it and a largest possible square inside the circle. What is the side length of the square.
Above is how I picture the situation 
And so the side length of the innermost square equals $sqrt{frac{a^2}{4}+frac{a^2}{4}}=frac{a}{sqrt{2}}$. Is this true. Need help!
geometry
edited Jun 19 '17 at 7:11
callculus
18.5k31428
asked Jun 19 '17 at 6:52
Castor GodinhoCastor Godinho
854
$endgroup$
add a comment |
$begingroup$
Consider a square of side $a$. Fit the largest possible circle inside it and a largest possible square inside the circle. What is the side length of the square.
Above is how I picture the situation
And so the side length of the innermost square equals $sqrt{frac{a^2}{4}+frac{a^2}{4}}=frac{a}{sqrt{2}}$. Is this true. Need help!
geometry
edited Jun 19 '17 at 7:11
callculus
18.5k31428
asked Jun 19 '17 at 6:52
Castor GodinhoCastor Godinho
854
$endgroup$
$begingroup$
The side length of the square is $a$, as it says in the first sentence! :)
$endgroup$
– Rahul
Jun 19 '17 at 8:09
add a comment |
$begingroup$
Consider a square of side $a$. Fit the largest possible circle inside it and a largest possible square inside the circle. What is the side length of the square.
Above is how I picture the situation
And so the side length of the innermost square equals $sqrt{frac{a^2}{4}+frac{a^2}{4}}=frac{a}{sqrt{2}}$. Is this true. Need help!
geometry
edited Jun 19 '17 at 7:11
callculus
18.5k31428
asked Jun 19 '17 at 6:52
Castor GodinhoCastor Godinho
854
$endgroup$
Consider a square of side $a$. Fit the largest possible circle inside it and a largest possible square inside the circle. What is the side length of the square.
Above is how I picture the situation
And so the side length of the innermost square equals $sqrt{frac{a^2}{4}+frac{a^2}{4}}=frac{a}{sqrt{2}}$. Is this true. Need help!
geometry
geometry
edited Jun 19 '17 at 7:11
callculus
18.5k31428
asked Jun 19 '17 at 6:52
Castor GodinhoCastor Godinho
854
edited Jun 19 '17 at 7:11
callculus
18.5k31428
asked Jun 19 '17 at 6:52
Castor GodinhoCastor Godinho
854
edited Jun 19 '17 at 7:11
callculus
18.5k31428
edited Jun 19 '17 at 7:11
callculus
18.5k31428
edited Jun 19 '17 at 7:11
callculus
18.5k31428
18.5k31428
asked Jun 19 '17 at 6:52
Castor GodinhoCastor Godinho
854
asked Jun 19 '17 at 6:52
Castor GodinhoCastor Godinho
854
asked Jun 19 '17 at 6:52
Castor GodinhoCastor Godinho
854
854
$begingroup$
The side length of the square is $a$, as it says in the first sentence! :)
$endgroup$
– Rahul
Jun 19 '17 at 8:09
add a comment |
$begingroup$
The side length of the square is $a$, as it says in the first sentence! :)
$endgroup$
– Rahul
Jun 19 '17 at 8:09
$begingroup$
The side length of the square is $a$, as it says in the first sentence! :)
$endgroup$
– Rahul
Jun 19 '17 at 8:09
$begingroup$
The side length of the square is $a$, as it says in the first sentence! :)
$endgroup$
– Rahul
Jun 19 '17 at 8:09
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes, it is.
We have that the diameter of the circle is equal to the side length of the bigger square.
We also have that the diagonal of the square is equal to the diameter of the circle, which is the side length of the square.
We have that $ssqrt{2}=a$, so $s=displaystyle frac{a}{sqrt{2}}$, which is exactly what you said.
Good job!
answered Jun 19 '17 at 7:07
Saketh MalyalaSaketh Malyala
7,4231534
$endgroup$
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2328162%2fsimple-geometry-problem-regarding-fitting-square-inside-a-circle%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes, it is.
We have that the diameter of the circle is equal to the side length of the bigger square.
We also have that the diagonal of the square is equal to the diameter of the circle, which is the side length of the square.
We have that $ssqrt{2}=a$, so $s=displaystyle frac{a}{sqrt{2}}$, which is exactly what you said.
Good job!
answered Jun 19 '17 at 7:07
Saketh MalyalaSaketh Malyala
7,4231534
$endgroup$
add a comment |
$begingroup$
Yes, it is.
We have that the diameter of the circle is equal to the side length of the bigger square.
We also have that the diagonal of the square is equal to the diameter of the circle, which is the side length of the square.
We have that $ssqrt{2}=a$, so $s=displaystyle frac{a}{sqrt{2}}$, which is exactly what you said.
Good job!
answered Jun 19 '17 at 7:07
Saketh MalyalaSaketh Malyala
7,4231534
$endgroup$
add a comment |
$begingroup$
Yes, it is.
We have that the diameter of the circle is equal to the side length of the bigger square.
We also have that the diagonal of the square is equal to the diameter of the circle, which is the side length of the square.
We have that $ssqrt{2}=a$, so $s=displaystyle frac{a}{sqrt{2}}$, which is exactly what you said.
Good job!
answered Jun 19 '17 at 7:07
Saketh MalyalaSaketh Malyala
7,4231534
$endgroup$
Yes, it is.
We have that the diameter of the circle is equal to the side length of the bigger square.
We also have that the diagonal of the square is equal to the diameter of the circle, which is the side length of the square.
We have that $ssqrt{2}=a$, so $s=displaystyle frac{a}{sqrt{2}}$, which is exactly what you said.
Good job!
answered Jun 19 '17 at 7:07
Saketh MalyalaSaketh Malyala
7,4231534
answered Jun 19 '17 at 7:07
Saketh MalyalaSaketh Malyala
7,4231534
answered Jun 19 '17 at 7:07
Saketh MalyalaSaketh Malyala
7,4231534
answered Jun 19 '17 at 7:07
Saketh MalyalaSaketh Malyala
7,4231534
7,4231534
add a comment |
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2328162%2fsimple-geometry-problem-regarding-fitting-square-inside-a-circle%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
$begingroup$
The side length of the square is $a$, as it says in the first sentence! :)
$endgroup$
– Rahul
Jun 19 '17 at 8:09