Minimum area of Inscribed SquareMaximum area of inscribed squareArea of a square inscribed in a circle of radius r, if area of the square inscribed in the semicircle is given.Find Area of Similar Right TriangleGRE Algebra answer is wrongMaximum area of inscribed squareShow that among all quadrilaterals of a given perimeter the square has the largest areaProve there is no maximum perimeter trapezoid inscribed in a circleInscribed Square in Inscribed Circle in SquareTriangle perimeter and areaInscribed irregular octagon areaCan area of rectangle be greater than the square of its diagonal?
How to make money from a browser who sees 5 seconds into the future of any web page?
What does "Scientists rise up against statistical significance" mean? (Comment in Nature)
How do I prevent inappropriate ads from appearing in my game?
What happens if I try to grapple an illusory duplicate from the Mirror Image spell?
How to leave product feedback on macOS?
Is there a RAID 0 Equivalent for RAM?
Pre-Employment Background Check With Consent For Future Checks
Personal or impersonal in a technical resume
Why is the Sun approximated as a black body at ~ 5800 K?
Echo with obfuscation
Unable to disable Microsoft Store in domain environment
Should I warn new/prospective PhD Student that supervisor is terrible?
What (the heck) is a Super Worm Equinox Moon?
Is there anyway, I can have two passwords for my wi-fi
How would you translate "more" for use as an interface button?
Sound waves in different octaves
Does Doodling or Improvising on the Piano Have Any Benefits?
How to test the sharpness of a knife?
ContourPlot — How do I color by contour curvature?
Why does the Persian emissary display a string of crowned skulls?
How do I tell my boss that I'm quitting in 15 days (a colleague left this week)
Why would five hundred and five be same as one?
Air travel with refrigerated insulin
Can I say "fingers" when referring to toes?
Minimum area of Inscribed Square
Maximum area of inscribed squareArea of a square inscribed in a circle of radius r, if area of the square inscribed in the semicircle is given.Find Area of Similar Right TriangleGRE Algebra answer is wrongMaximum area of inscribed squareShow that among all quadrilaterals of a given perimeter the square has the largest areaProve there is no maximum perimeter trapezoid inscribed in a circleInscribed Square in Inscribed Circle in SquareTriangle perimeter and areaInscribed irregular octagon areaCan area of rectangle be greater than the square of its diagonal?
$begingroup$
GRE study guide asks
The perimeter of square S is 40. Square T is inscribed in square S.
What is the least possible area of square T?
Choices are
- 45
- 48
- 49
- 50
- 52
They say answer is 50. How do they even get this? T has lengths less than 10, so it can be 1 x 1 square, or 9 x 9 square. Please guide.
geometry area gre-exam
asked Jan 23 '16 at 18:54
RhondaRhonda
2532512
$endgroup$
add a comment |
$begingroup$
GRE study guide asks
The perimeter of square S is 40. Square T is inscribed in square S.
What is the least possible area of square T?
Choices are
- 45
- 48
- 49
- 50
- 52
They say answer is 50. How do they even get this? T has lengths less than 10, so it can be 1 x 1 square, or 9 x 9 square. Please guide.
geometry area gre-exam
asked Jan 23 '16 at 18:54
RhondaRhonda
2532512
$endgroup$
$begingroup$
"Inscribed" means that all four vertices of $T$ lie on the boundary of $S$.
$endgroup$
– rogerl
Jan 23 '16 at 18:59
$begingroup$
@rogerl Wouldn't that be one square on top of another. Would it look like this cdn-1.analyzemath.com/middle_school_math/grade_9/graphs/… Then how it least area = 50?
$endgroup$
– Rhonda
Jan 23 '16 at 19:00
$begingroup$
Yes, that's a possibility. The four vertices don't have to be evenly placed, in the middle of the sides, like in that diagram, but that is indeed the kind of picture that the question asks about.
$endgroup$
– rogerl
Jan 23 '16 at 19:02
add a comment |
$begingroup$
GRE study guide asks
The perimeter of square S is 40. Square T is inscribed in square S.
What is the least possible area of square T?
Choices are
- 45
- 48
- 49
- 50
- 52
They say answer is 50. How do they even get this? T has lengths less than 10, so it can be 1 x 1 square, or 9 x 9 square. Please guide.
geometry area gre-exam
asked Jan 23 '16 at 18:54
RhondaRhonda
2532512
$endgroup$
GRE study guide asks
The perimeter of square S is 40. Square T is inscribed in square S.
What is the least possible area of square T?
Choices are
- 45
- 48
- 49
- 50
- 52
They say answer is 50. How do they even get this? T has lengths less than 10, so it can be 1 x 1 square, or 9 x 9 square. Please guide.
geometry area gre-exam
geometry area gre-exam
asked Jan 23 '16 at 18:54
RhondaRhonda
2532512
asked Jan 23 '16 at 18:54
RhondaRhonda
2532512
asked Jan 23 '16 at 18:54
RhondaRhonda
2532512
asked Jan 23 '16 at 18:54
RhondaRhonda
2532512
asked Jan 23 '16 at 18:54
RhondaRhonda
2532512
2532512
$begingroup$
"Inscribed" means that all four vertices of $T$ lie on the boundary of $S$.
$endgroup$
– rogerl
Jan 23 '16 at 18:59
$begingroup$
@rogerl Wouldn't that be one square on top of another. Would it look like this cdn-1.analyzemath.com/middle_school_math/grade_9/graphs/… Then how it least area = 50?
$endgroup$
– Rhonda
Jan 23 '16 at 19:00
$begingroup$
Yes, that's a possibility. The four vertices don't have to be evenly placed, in the middle of the sides, like in that diagram, but that is indeed the kind of picture that the question asks about.
$endgroup$
– rogerl
Jan 23 '16 at 19:02
add a comment |
$begingroup$
"Inscribed" means that all four vertices of $T$ lie on the boundary of $S$.
$endgroup$
– rogerl
Jan 23 '16 at 18:59
$begingroup$
@rogerl Wouldn't that be one square on top of another. Would it look like this cdn-1.analyzemath.com/middle_school_math/grade_9/graphs/… Then how it least area = 50?
$endgroup$
– Rhonda
Jan 23 '16 at 19:00
$begingroup$
Yes, that's a possibility. The four vertices don't have to be evenly placed, in the middle of the sides, like in that diagram, but that is indeed the kind of picture that the question asks about.
$endgroup$
– rogerl
Jan 23 '16 at 19:02
$begingroup$
"Inscribed" means that all four vertices of $T$ lie on the boundary of $S$.
$endgroup$
– rogerl
Jan 23 '16 at 18:59
$begingroup$
"Inscribed" means that all four vertices of $T$ lie on the boundary of $S$.
$endgroup$
– rogerl
Jan 23 '16 at 18:59
$begingroup$
@rogerl Wouldn't that be one square on top of another. Would it look like this cdn-1.analyzemath.com/middle_school_math/grade_9/graphs/… Then how it least area = 50?
$endgroup$
– Rhonda
Jan 23 '16 at 19:00
$begingroup$
@rogerl Wouldn't that be one square on top of another. Would it look like this cdn-1.analyzemath.com/middle_school_math/grade_9/graphs/… Then how it least area = 50?
$endgroup$
– Rhonda
Jan 23 '16 at 19:00
$begingroup$
Yes, that's a possibility. The four vertices don't have to be evenly placed, in the middle of the sides, like in that diagram, but that is indeed the kind of picture that the question asks about.
$endgroup$
– rogerl
Jan 23 '16 at 19:02
$begingroup$
Yes, that's a possibility. The four vertices don't have to be evenly placed, in the middle of the sides, like in that diagram, but that is indeed the kind of picture that the question asks about.
$endgroup$
– rogerl
Jan 23 '16 at 19:02
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
If it is inscribed it means that all vertices of $T$ lie on $S$. Like this
This divides the square into 4 triangle pieces and one square piece. We need to find the hypotenuse of the triangle piece to find the area of the inner square. The triangle has 2 sides of length 5. Using the pythagorean theorem yields $h = sqrt(5^2 + 5^2) = sqrt50$. The area of a square with side length $h$ is $h^2$ so the area of the inner square is $sqrt50^2 = 50$. Now you may wonder why the inscription has to be like this to yield a minimum.
Lets consider the general case.
The area of the inner square is minimized when the area of the outer triangles is maximized. You can consider the vertices of the inner square dividing the sides of the outer square into two pieces. In the optimal case the length of these pieces are the same. This intuitively seems in the first diagram we can prove this to be true in general.
Let $L$ be the side length of the square. Say the vertices divides the side into pieces of length $a$ and $b$. We know that $a + b = L$ so $b = L - a$. Notice that all the triangles have the same area. So we only need to minimize one. The area of the triangle, $A$, will be $A = frac12ab = frac12a(L - a)$. We can find the minimum without calculus because this is a parabola. We just need to find the vertex of the parabola. Instead of completing the square, we can use the roots of the parabola as we have it in factored form. Recall that if the parabola has roots $r_1$ and $r_2$ then its vertex has an $x$ coordinate of $fracr_1 + r_22$(Note that we dont care about the actual value of the maximum area for this proof, but rather the value of $a$ that maximizes the area of the triangle). Since our quadratic has roots $0$ and $L$, then the x coordinate of the vertex is $frac0 + L2 = fracL2$. Therefore the values of $a$ and $b$ that maximize the triangle area is $a = fracL2$, $b = L - fracL2 = fracL2$. This yields the result that the minimum case has $a = b$.
answered Jan 23 '16 at 19:21
Jeevan DevaranjanJeevan Devaranjan
2,237616
$endgroup$
$begingroup$
This is quite thorough and I'm still soaking in your explanation. What happens if you want to find maximum area of inscribed square? This means you'd have to minimize outside triangles. Will post follow-up question and link here.
$endgroup$
– Rhonda
Jan 23 '16 at 19:43
1
$begingroup$
@Rhonda Say we have a shape $S$, and another shape $D$ that is completely inscribed in it(no part of $D$ is outside of $S$). We know that $textArea(D) leq textArea(S)$. From this we can see that the maximum possible area for $D$ is the area of $S$. So the maximum inscribed square would be the square itself. You can also think of this in terms of the minimization. We know the area of a triangle cannot be negative, which means its minimum can be 0. If $a = 0$ or $b = 0$ then the area of the triangle is 0. Think of what the diagram looks like if $a = 0$.
$endgroup$
– Jeevan Devaranjan
Jan 23 '16 at 21:33
add a comment |
$begingroup$
The definition of an "inscribed" square in a square is that all of the smaller square's vertices lies on the boundaries of the larger square.
Notice that for this to happen, the smaller square's vertices will divide each side of the larger square into two segments, the same on each side. Let's call their lengths $a$ and $b.$ We have that $a + b = 10.$ The side length of the smaller square is, by Pythagoras, $s = sqrta^2 + b^2.$ Then the area of the smaller square must be $s^2 = a^2 + b^2.$
You can use calculus here, but common sense also tells us that the minimum occurs when $a = b = frac102 = 5.$ So the minimum area of the smaller square is $5^2 + 5^2 = boxed50.$ This is answer choice $4.$
answered Jan 23 '16 at 19:15
K. JiangK. Jiang
3,0311513
$endgroup$
add a comment |
$begingroup$
Another explanation [Continuation of Jeevan Devaranjan ] A= (1/2) *a(L-a)=-(1/2)(a^2-aL)= -(1/2)(a^2-2*aL/2+L^2/4)+L^2/8=-(1/2)(a-L/2)^2+L^2/8; In order to maximize A; a-L/2=0 Hence a=(1/2)*L
answered Aug 19 '17 at 17:41
user473244user473244
1
$endgroup$
$begingroup$
Please use MathJax.
$endgroup$
– José Carlos Santos
Aug 19 '17 at 18:00
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%2f1623865%2fminimum-area-of-inscribed-square%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If it is inscribed it means that all vertices of $T$ lie on $S$. Like this
This divides the square into 4 triangle pieces and one square piece. We need to find the hypotenuse of the triangle piece to find the area of the inner square. The triangle has 2 sides of length 5. Using the pythagorean theorem yields $h = sqrt(5^2 + 5^2) = sqrt50$. The area of a square with side length $h$ is $h^2$ so the area of the inner square is $sqrt50^2 = 50$. Now you may wonder why the inscription has to be like this to yield a minimum.
Lets consider the general case.
The area of the inner square is minimized when the area of the outer triangles is maximized. You can consider the vertices of the inner square dividing the sides of the outer square into two pieces. In the optimal case the length of these pieces are the same. This intuitively seems in the first diagram we can prove this to be true in general.
Let $L$ be the side length of the square. Say the vertices divides the side into pieces of length $a$ and $b$. We know that $a + b = L$ so $b = L - a$. Notice that all the triangles have the same area. So we only need to minimize one. The area of the triangle, $A$, will be $A = frac12ab = frac12a(L - a)$. We can find the minimum without calculus because this is a parabola. We just need to find the vertex of the parabola. Instead of completing the square, we can use the roots of the parabola as we have it in factored form. Recall that if the parabola has roots $r_1$ and $r_2$ then its vertex has an $x$ coordinate of $fracr_1 + r_22$(Note that we dont care about the actual value of the maximum area for this proof, but rather the value of $a$ that maximizes the area of the triangle). Since our quadratic has roots $0$ and $L$, then the x coordinate of the vertex is $frac0 + L2 = fracL2$. Therefore the values of $a$ and $b$ that maximize the triangle area is $a = fracL2$, $b = L - fracL2 = fracL2$. This yields the result that the minimum case has $a = b$.
answered Jan 23 '16 at 19:21
Jeevan DevaranjanJeevan Devaranjan
2,237616
$endgroup$
$begingroup$
This is quite thorough and I'm still soaking in your explanation. What happens if you want to find maximum area of inscribed square? This means you'd have to minimize outside triangles. Will post follow-up question and link here.
$endgroup$
– Rhonda
Jan 23 '16 at 19:43
1
$begingroup$
@Rhonda Say we have a shape $S$, and another shape $D$ that is completely inscribed in it(no part of $D$ is outside of $S$). We know that $textArea(D) leq textArea(S)$. From this we can see that the maximum possible area for $D$ is the area of $S$. So the maximum inscribed square would be the square itself. You can also think of this in terms of the minimization. We know the area of a triangle cannot be negative, which means its minimum can be 0. If $a = 0$ or $b = 0$ then the area of the triangle is 0. Think of what the diagram looks like if $a = 0$.
$endgroup$
– Jeevan Devaranjan
Jan 23 '16 at 21:33
add a comment |
$begingroup$
If it is inscribed it means that all vertices of $T$ lie on $S$. Like this
This divides the square into 4 triangle pieces and one square piece. We need to find the hypotenuse of the triangle piece to find the area of the inner square. The triangle has 2 sides of length 5. Using the pythagorean theorem yields $h = sqrt(5^2 + 5^2) = sqrt50$. The area of a square with side length $h$ is $h^2$ so the area of the inner square is $sqrt50^2 = 50$. Now you may wonder why the inscription has to be like this to yield a minimum.
Lets consider the general case.
The area of the inner square is minimized when the area of the outer triangles is maximized. You can consider the vertices of the inner square dividing the sides of the outer square into two pieces. In the optimal case the length of these pieces are the same. This intuitively seems in the first diagram we can prove this to be true in general.
Let $L$ be the side length of the square. Say the vertices divides the side into pieces of length $a$ and $b$. We know that $a + b = L$ so $b = L - a$. Notice that all the triangles have the same area. So we only need to minimize one. The area of the triangle, $A$, will be $A = frac12ab = frac12a(L - a)$. We can find the minimum without calculus because this is a parabola. We just need to find the vertex of the parabola. Instead of completing the square, we can use the roots of the parabola as we have it in factored form. Recall that if the parabola has roots $r_1$ and $r_2$ then its vertex has an $x$ coordinate of $fracr_1 + r_22$(Note that we dont care about the actual value of the maximum area for this proof, but rather the value of $a$ that maximizes the area of the triangle). Since our quadratic has roots $0$ and $L$, then the x coordinate of the vertex is $frac0 + L2 = fracL2$. Therefore the values of $a$ and $b$ that maximize the triangle area is $a = fracL2$, $b = L - fracL2 = fracL2$. This yields the result that the minimum case has $a = b$.
answered Jan 23 '16 at 19:21
Jeevan DevaranjanJeevan Devaranjan
2,237616
$endgroup$
$begingroup$
This is quite thorough and I'm still soaking in your explanation. What happens if you want to find maximum area of inscribed square? This means you'd have to minimize outside triangles. Will post follow-up question and link here.
$endgroup$
– Rhonda
Jan 23 '16 at 19:43
1
$begingroup$
@Rhonda Say we have a shape $S$, and another shape $D$ that is completely inscribed in it(no part of $D$ is outside of $S$). We know that $textArea(D) leq textArea(S)$. From this we can see that the maximum possible area for $D$ is the area of $S$. So the maximum inscribed square would be the square itself. You can also think of this in terms of the minimization. We know the area of a triangle cannot be negative, which means its minimum can be 0. If $a = 0$ or $b = 0$ then the area of the triangle is 0. Think of what the diagram looks like if $a = 0$.
$endgroup$
– Jeevan Devaranjan
Jan 23 '16 at 21:33
add a comment |
$begingroup$
If it is inscribed it means that all vertices of $T$ lie on $S$. Like this
This divides the square into 4 triangle pieces and one square piece. We need to find the hypotenuse of the triangle piece to find the area of the inner square. The triangle has 2 sides of length 5. Using the pythagorean theorem yields $h = sqrt(5^2 + 5^2) = sqrt50$. The area of a square with side length $h$ is $h^2$ so the area of the inner square is $sqrt50^2 = 50$. Now you may wonder why the inscription has to be like this to yield a minimum.
Lets consider the general case.
The area of the inner square is minimized when the area of the outer triangles is maximized. You can consider the vertices of the inner square dividing the sides of the outer square into two pieces. In the optimal case the length of these pieces are the same. This intuitively seems in the first diagram we can prove this to be true in general.
Let $L$ be the side length of the square. Say the vertices divides the side into pieces of length $a$ and $b$. We know that $a + b = L$ so $b = L - a$. Notice that all the triangles have the same area. So we only need to minimize one. The area of the triangle, $A$, will be $A = frac12ab = frac12a(L - a)$. We can find the minimum without calculus because this is a parabola. We just need to find the vertex of the parabola. Instead of completing the square, we can use the roots of the parabola as we have it in factored form. Recall that if the parabola has roots $r_1$ and $r_2$ then its vertex has an $x$ coordinate of $fracr_1 + r_22$(Note that we dont care about the actual value of the maximum area for this proof, but rather the value of $a$ that maximizes the area of the triangle). Since our quadratic has roots $0$ and $L$, then the x coordinate of the vertex is $frac0 + L2 = fracL2$. Therefore the values of $a$ and $b$ that maximize the triangle area is $a = fracL2$, $b = L - fracL2 = fracL2$. This yields the result that the minimum case has $a = b$.
answered Jan 23 '16 at 19:21
Jeevan DevaranjanJeevan Devaranjan
2,237616
$endgroup$
If it is inscribed it means that all vertices of $T$ lie on $S$. Like this
This divides the square into 4 triangle pieces and one square piece. We need to find the hypotenuse of the triangle piece to find the area of the inner square. The triangle has 2 sides of length 5. Using the pythagorean theorem yields $h = sqrt(5^2 + 5^2) = sqrt50$. The area of a square with side length $h$ is $h^2$ so the area of the inner square is $sqrt50^2 = 50$. Now you may wonder why the inscription has to be like this to yield a minimum.
Lets consider the general case.
The area of the inner square is minimized when the area of the outer triangles is maximized. You can consider the vertices of the inner square dividing the sides of the outer square into two pieces. In the optimal case the length of these pieces are the same. This intuitively seems in the first diagram we can prove this to be true in general.
Let $L$ be the side length of the square. Say the vertices divides the side into pieces of length $a$ and $b$. We know that $a + b = L$ so $b = L - a$. Notice that all the triangles have the same area. So we only need to minimize one. The area of the triangle, $A$, will be $A = frac12ab = frac12a(L - a)$. We can find the minimum without calculus because this is a parabola. We just need to find the vertex of the parabola. Instead of completing the square, we can use the roots of the parabola as we have it in factored form. Recall that if the parabola has roots $r_1$ and $r_2$ then its vertex has an $x$ coordinate of $fracr_1 + r_22$(Note that we dont care about the actual value of the maximum area for this proof, but rather the value of $a$ that maximizes the area of the triangle). Since our quadratic has roots $0$ and $L$, then the x coordinate of the vertex is $frac0 + L2 = fracL2$. Therefore the values of $a$ and $b$ that maximize the triangle area is $a = fracL2$, $b = L - fracL2 = fracL2$. This yields the result that the minimum case has $a = b$.
answered Jan 23 '16 at 19:21
Jeevan DevaranjanJeevan Devaranjan
2,237616
answered Jan 23 '16 at 19:21
Jeevan DevaranjanJeevan Devaranjan
2,237616
answered Jan 23 '16 at 19:21
Jeevan DevaranjanJeevan Devaranjan
2,237616
answered Jan 23 '16 at 19:21
Jeevan DevaranjanJeevan Devaranjan
2,237616
2,237616
$begingroup$
This is quite thorough and I'm still soaking in your explanation. What happens if you want to find maximum area of inscribed square? This means you'd have to minimize outside triangles. Will post follow-up question and link here.
$endgroup$
– Rhonda
Jan 23 '16 at 19:43
1
$begingroup$
@Rhonda Say we have a shape $S$, and another shape $D$ that is completely inscribed in it(no part of $D$ is outside of $S$). We know that $textArea(D) leq textArea(S)$. From this we can see that the maximum possible area for $D$ is the area of $S$. So the maximum inscribed square would be the square itself. You can also think of this in terms of the minimization. We know the area of a triangle cannot be negative, which means its minimum can be 0. If $a = 0$ or $b = 0$ then the area of the triangle is 0. Think of what the diagram looks like if $a = 0$.
$endgroup$
– Jeevan Devaranjan
Jan 23 '16 at 21:33
add a comment |
$begingroup$
This is quite thorough and I'm still soaking in your explanation. What happens if you want to find maximum area of inscribed square? This means you'd have to minimize outside triangles. Will post follow-up question and link here.
$endgroup$
– Rhonda
Jan 23 '16 at 19:43
1
$begingroup$
@Rhonda Say we have a shape $S$, and another shape $D$ that is completely inscribed in it(no part of $D$ is outside of $S$). We know that $textArea(D) leq textArea(S)$. From this we can see that the maximum possible area for $D$ is the area of $S$. So the maximum inscribed square would be the square itself. You can also think of this in terms of the minimization. We know the area of a triangle cannot be negative, which means its minimum can be 0. If $a = 0$ or $b = 0$ then the area of the triangle is 0. Think of what the diagram looks like if $a = 0$.
$endgroup$
– Jeevan Devaranjan
Jan 23 '16 at 21:33
$begingroup$
This is quite thorough and I'm still soaking in your explanation. What happens if you want to find maximum area of inscribed square? This means you'd have to minimize outside triangles. Will post follow-up question and link here.
$endgroup$
– Rhonda
Jan 23 '16 at 19:43
$begingroup$
This is quite thorough and I'm still soaking in your explanation. What happens if you want to find maximum area of inscribed square? This means you'd have to minimize outside triangles. Will post follow-up question and link here.
$endgroup$
– Rhonda
Jan 23 '16 at 19:43
1
1
$begingroup$
@Rhonda Say we have a shape $S$, and another shape $D$ that is completely inscribed in it(no part of $D$ is outside of $S$). We know that $textArea(D) leq textArea(S)$. From this we can see that the maximum possible area for $D$ is the area of $S$. So the maximum inscribed square would be the square itself. You can also think of this in terms of the minimization. We know the area of a triangle cannot be negative, which means its minimum can be 0. If $a = 0$ or $b = 0$ then the area of the triangle is 0. Think of what the diagram looks like if $a = 0$.
$endgroup$
– Jeevan Devaranjan
Jan 23 '16 at 21:33
$begingroup$
@Rhonda Say we have a shape $S$, and another shape $D$ that is completely inscribed in it(no part of $D$ is outside of $S$). We know that $textArea(D) leq textArea(S)$. From this we can see that the maximum possible area for $D$ is the area of $S$. So the maximum inscribed square would be the square itself. You can also think of this in terms of the minimization. We know the area of a triangle cannot be negative, which means its minimum can be 0. If $a = 0$ or $b = 0$ then the area of the triangle is 0. Think of what the diagram looks like if $a = 0$.
$endgroup$
– Jeevan Devaranjan
Jan 23 '16 at 21:33
add a comment |
$begingroup$
The definition of an "inscribed" square in a square is that all of the smaller square's vertices lies on the boundaries of the larger square.
Notice that for this to happen, the smaller square's vertices will divide each side of the larger square into two segments, the same on each side. Let's call their lengths $a$ and $b.$ We have that $a + b = 10.$ The side length of the smaller square is, by Pythagoras, $s = sqrta^2 + b^2.$ Then the area of the smaller square must be $s^2 = a^2 + b^2.$
You can use calculus here, but common sense also tells us that the minimum occurs when $a = b = frac102 = 5.$ So the minimum area of the smaller square is $5^2 + 5^2 = boxed50.$ This is answer choice $4.$
answered Jan 23 '16 at 19:15
K. JiangK. Jiang
3,0311513
$endgroup$
add a comment |
$begingroup$
The definition of an "inscribed" square in a square is that all of the smaller square's vertices lies on the boundaries of the larger square.
Notice that for this to happen, the smaller square's vertices will divide each side of the larger square into two segments, the same on each side. Let's call their lengths $a$ and $b.$ We have that $a + b = 10.$ The side length of the smaller square is, by Pythagoras, $s = sqrta^2 + b^2.$ Then the area of the smaller square must be $s^2 = a^2 + b^2.$
You can use calculus here, but common sense also tells us that the minimum occurs when $a = b = frac102 = 5.$ So the minimum area of the smaller square is $5^2 + 5^2 = boxed50.$ This is answer choice $4.$
answered Jan 23 '16 at 19:15
K. JiangK. Jiang
3,0311513
$endgroup$
add a comment |
$begingroup$
The definition of an "inscribed" square in a square is that all of the smaller square's vertices lies on the boundaries of the larger square.
Notice that for this to happen, the smaller square's vertices will divide each side of the larger square into two segments, the same on each side. Let's call their lengths $a$ and $b.$ We have that $a + b = 10.$ The side length of the smaller square is, by Pythagoras, $s = sqrta^2 + b^2.$ Then the area of the smaller square must be $s^2 = a^2 + b^2.$
You can use calculus here, but common sense also tells us that the minimum occurs when $a = b = frac102 = 5.$ So the minimum area of the smaller square is $5^2 + 5^2 = boxed50.$ This is answer choice $4.$
answered Jan 23 '16 at 19:15
K. JiangK. Jiang
3,0311513
$endgroup$
The definition of an "inscribed" square in a square is that all of the smaller square's vertices lies on the boundaries of the larger square.
Notice that for this to happen, the smaller square's vertices will divide each side of the larger square into two segments, the same on each side. Let's call their lengths $a$ and $b.$ We have that $a + b = 10.$ The side length of the smaller square is, by Pythagoras, $s = sqrta^2 + b^2.$ Then the area of the smaller square must be $s^2 = a^2 + b^2.$
You can use calculus here, but common sense also tells us that the minimum occurs when $a = b = frac102 = 5.$ So the minimum area of the smaller square is $5^2 + 5^2 = boxed50.$ This is answer choice $4.$
answered Jan 23 '16 at 19:15
K. JiangK. Jiang
3,0311513
answered Jan 23 '16 at 19:15
K. JiangK. Jiang
3,0311513
answered Jan 23 '16 at 19:15
K. JiangK. Jiang
3,0311513
answered Jan 23 '16 at 19:15
K. JiangK. Jiang
3,0311513
3,0311513
add a comment |
add a comment |
$begingroup$
Another explanation [Continuation of Jeevan Devaranjan ] A= (1/2) *a(L-a)=-(1/2)(a^2-aL)= -(1/2)(a^2-2*aL/2+L^2/4)+L^2/8=-(1/2)(a-L/2)^2+L^2/8; In order to maximize A; a-L/2=0 Hence a=(1/2)*L
answered Aug 19 '17 at 17:41
user473244user473244
1
$endgroup$
$begingroup$
Please use MathJax.
$endgroup$
– José Carlos Santos
Aug 19 '17 at 18:00
add a comment |
$begingroup$
Another explanation [Continuation of Jeevan Devaranjan ] A= (1/2) *a(L-a)=-(1/2)(a^2-aL)= -(1/2)(a^2-2*aL/2+L^2/4)+L^2/8=-(1/2)(a-L/2)^2+L^2/8; In order to maximize A; a-L/2=0 Hence a=(1/2)*L
answered Aug 19 '17 at 17:41
user473244user473244
1
$endgroup$
$begingroup$
Please use MathJax.
$endgroup$
– José Carlos Santos
Aug 19 '17 at 18:00
add a comment |
$begingroup$
Another explanation [Continuation of Jeevan Devaranjan ] A= (1/2) *a(L-a)=-(1/2)(a^2-aL)= -(1/2)(a^2-2*aL/2+L^2/4)+L^2/8=-(1/2)(a-L/2)^2+L^2/8; In order to maximize A; a-L/2=0 Hence a=(1/2)*L
answered Aug 19 '17 at 17:41
user473244user473244
1
$endgroup$
Another explanation [Continuation of Jeevan Devaranjan ] A= (1/2) *a(L-a)=-(1/2)(a^2-aL)= -(1/2)(a^2-2*aL/2+L^2/4)+L^2/8=-(1/2)(a-L/2)^2+L^2/8; In order to maximize A; a-L/2=0 Hence a=(1/2)*L
answered Aug 19 '17 at 17:41
user473244user473244
1
answered Aug 19 '17 at 17:41
user473244user473244
1
answered Aug 19 '17 at 17:41
user473244user473244
1
answered Aug 19 '17 at 17:41
user473244user473244
1
1
$begingroup$
Please use MathJax.
$endgroup$
– José Carlos Santos
Aug 19 '17 at 18:00
add a comment |
$begingroup$
Please use MathJax.
$endgroup$
– José Carlos Santos
Aug 19 '17 at 18:00
$begingroup$
Please use MathJax.
$endgroup$
– José Carlos Santos
Aug 19 '17 at 18:00
$begingroup$
Please use MathJax.
$endgroup$
– José Carlos Santos
Aug 19 '17 at 18:00
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%2f1623865%2fminimum-area-of-inscribed-square%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
$begingroup$
"Inscribed" means that all four vertices of $T$ lie on the boundary of $S$.
$endgroup$
– rogerl
Jan 23 '16 at 18:59
$begingroup$
@rogerl Wouldn't that be one square on top of another. Would it look like this cdn-1.analyzemath.com/middle_school_math/grade_9/graphs/… Then how it least area = 50?
$endgroup$
– Rhonda
Jan 23 '16 at 19:00
$begingroup$
Yes, that's a possibility. The four vertices don't have to be evenly placed, in the middle of the sides, like in that diagram, but that is indeed the kind of picture that the question asks about.
$endgroup$
– rogerl
Jan 23 '16 at 19:02