Napkin

700 /www/nrich/html/content/00/03/six5/ 1 2011-02-01T00:00:01 <mdoxml version="1.0"> <div style="text-align: center;"><mdo:image height="217" width="460" src="napkin.png" alt="Folding diagram"></mdo:image></div> A square paper napkin is folded so that the corner P coincides with the midpoint of an opposite edge as shown in the diagram. Investigate the three triangles that are formed by folding in this way (where there is a single thickness of the paper). </mdoxml> <mdoxml version="1.0"> <comment> set var="TeX" value="l2h" </comment> This solution is a combination of the work of David from Trinity School and several others. A square is folded so that the corner $E$ coincides with the midpoint of an opposite edge as shown in the diagram. The length of the edge of the square is 1 unit and the problem is to find the lengths and areas of the three triangles. <mdo:image height="179" width="194" src="napkin2.gif" alt="Napkin"></mdo:image> The table shows the solutions: <table cellspacing="4" cellpadding="4" border="1" align="center"> <tbody> <tr> <td>Triangle</td> <td align="center">Lengths of sides 3 : 4 : 5</td> <td>Ratio of lengths in 3 triangles</td> <td>Area</td> <td>Ratio of areas of 3 triangles</td> </tr> <tr> <td align="center">EGF</td> <td>3/8 : 1/2 : 5/8</td> <td align="center">3</td> <td>3/32</td> <td align="center">9</td> </tr> <tr> <td align="center">CED</td> <td>1/2 : 2/3 : 5/6</td> <td align="center">4</td> <td>1/6</td> <td align="center">16</td> </tr> <tr> <td align="center">CAB</td> <td>1/8 : 1/6 : 5/24</td> <td align="center">1</td> <td>1/96</td> <td align="center">1</td> </tr> </tbody> </table> This is David's method: The edge, $EF = 1/2$. Taking $FG = a$ then $GE = 1-a$ As it has a corner of the paper as part of it, we know $EGF$ is a right angled triangle, and so $GE^2 = EF^2 + FG^2$, which gives \[ (1-a)^2 = 1/4 + a^2 \] so \[a = 3/8.\] Now that we have the length of $a$, we have the lengths of all the sides of this triangle: $EF = 1/2$ , $FG = 3/8$ and $GE = 5/8.$ At this point, David uses Trigonometry to complete the problem, but other solutions just used similar triangles. The area of triangle $EFG = 1/2$ base $\times$ height $ = 3/32$. Using this information and trigonometry, we can work out the angle $GEF$: $\angle GEF = \tan^{-1}(0.375 / 0.5) = \tan^{-1}0.75 = 36.9^{\circ}.$ taking the angle to the nearest tenth of a degree. Knowing this, we can work out the angle $DEC=53.1^{\circ}$ and use it with the length $DE$ to work out the length $DC=2/3.$ The last part needed of this triangle is $CE$ and, using Pythagoras' theorem, we get $CE =5/6.$ So $DE = {1 \over 2}$, $DC = {2 \over 3}$, $CE = {5 \over 6}$ and the area of triangle $DEC = 1/2 \times 2/3 \times 1/2 = {1 \over 6}.$ The last triangle needed is $ABC$. We know the length $BC = 1 - CE = 1/6.$ We can also work out the angle $BCA $ which is equal to angle $DCE$ and to angle $FEG.$ We can now use trigonometry to work out the length $AB = 1/6 \tan BCA = 1/8.$ And so once again, using Pythagoras, we can work out the length of line $AC$ which is $5/24.$ And so we have the final set of lengths: $AB = 1/8$, $BC = 1/ 6$ and $AC = 5/24.$ The area of triangle $ABC = 1/2 \times 1/8 \times 1/6 = 1/96$. This means the total area of all the paper with a single thickness = $26/96 = .270833..$ This can be backed up by working out the area of the trapezium $ABEG$, subtracting the area of triangle $ABC$ and then multiplying the result by 2 to give the area of the original square that is now double thickness. Area of trapezium = $1/2 [1/8 +5/8] \times 1 = 3/8.$ Double thickness area = $3/8 - 1/96 = 35/96.$ The total area is $2 \times 35/96 + 26/96= 1$ and so all areas are worked out and recorded, and the total area of single thickness paper and double thickness areas are recorded. <div>Here is the Similar Triangles method:</div> <div><mdo:image height="161" width="188" alt="" src="napkin2.gif"></mdo:image></div> <div>Triangles EFG, CDE and ABC are similar triangles. This is how I know: firstly all three are right-angled triangles. Angle GEF and EGF add up to 90, as do angles CED and GEF. Therefore angles EGF and CED must be the same.This tells me that angle DCE must be the same as GEF, because triangle CDE is right-angled. Angle BCA and DCE are opposite angles so they must be equal, meaning that angle BAC is equal to CED. All three triangles have the same three angles, therefore they are similar.</div> <mdo:image height="297" width="417" alt="" src="Napkin-solution.png"></mdo:image> Triangle EFG has sides in the ratio $3:4:5$, so triangle CDE must also have sides in the ratio $3:4:5$. $CD = {1 \over 2} \div 3 \times 4$ $CD = {2 \over 3}$ $CE = {1 \over 2} \div 4 \times 5$ $CE = {5 \over 8}$ $EB$ is a side of the square, so $EB$ is 1 unit. $BC = 1- {5 \over 6}$ $BC = {1 \over 6}$ $BC$ is the $4$ side of the 3:4:5 triangle. $BA = {1 \over 6} \div 4 \times 3$ $BA = {3 \over {24}}$ $AC = {1 \over 6} \div 4 \times 5$ $AC = {5 \over {24}}$ </mdoxml> <mdoxml version="1.0"> <h3>Why do this problem </h3> <div><a href="http://nrich.maths.org/public/viewer.php?obj_id=700&part=">This problem</a> offers a simple context which can generate lots of questions. Inviting learners to make conjectures and form convincing arguments. Demonstrating the similarity of triangles is relatively straight forward and calculating lengths offers opportunities for links with Pythagoras' theorem and ratios, bringing together important geometrical concepts.</div> <h3>Possible approach </h3> <div>Time to engage in, and become familiar with, the context is important. Early on, encourage learners to list and share what they notice, using large squares of paper on a display board can encourage discussion of key features and ideas and conjectures which they might explore.</div> <div>Identify questions about the triangles that learners will work on.</div> <div>As learners work <a href="/content/00/03/six5/Napkinnewlogo.doc">this document</a> may help them discuss possibilities and focus on some possible approaches.</div> <div>It is likely that learners will arrive at results in different ways. These journeys and findings form opportunities to share and discuss good and elegant solutions and different ways of "seeing". </div> <h3>Key questions</h3> <ul> <li>What do you think might be true?</li> <li>What do you know?</li> <li>What do you need to know?</li> <li>What mathematical ideas and techniques might be of use in order to answer that question? </li> </ul> <h3>Possible extension</h3> If the square paper napkin is folded so that the corner P does not coincide with the midpoint of an opposite edge, where would you place the fold for a 5, 12, 13 or an 8, 15, 17 or a 7, 24, 25 triangle? Are any of these findings extendable to other quadrilaterals? See also the problem <a href="http://nrich.maths.org/public/viewer.php?obj_id=1956&part=">Origami</a> <h3>Possible support </h3> <div><a href="http://nrich.maths.org/public/viewer.php?obj_id=6355&part=">Making Sixty</a> is a similar context involving a similar fold but with more accessible results. It also has the potential to lead to some practical mathematics.</div> </mdoxml> <mdoxml version="1.0">Can you find equal lengths? Can you spot similar triangles? <a href="/content/00/03/six5/Napkin.doc">These diagrams</a> might help you to move forward. Cut the images out and sort them in a way that helps you see any connections. </mdoxml> <mdoxml version="1.0"> <comment> set var="TeX" value="l2h" </comment> This solution is from David from Trinity School. A square is folded so that the corner $E$ coincides with the midpoint of an opposite edge as shown in the diagram. The length of the edge of the square is 1 unit and the problem is to find the lengths and areas of the three triangles. David uses trigonometry but alternatively, once you have found the dimensions of triangle $EGF$ you can find all the other lengths and areas using the fact that the triangles $EGF$, $CED$ and $CAB$ are similar 3:4:5 triangles. <mdo:image height="179" width="194" src="napkin2.gif" alt="Napkin"></mdo:image> The table shows the solutions: <table cellspacing="4" cellpadding="4" border="1" align="center"> <tbody> <tr> <td>Triangle</td> <td align="center">Lengths of sides 3 : 4 : 5</td> <td>Ratio of lengths in 3 triangles</td> <td>Area</td> <td>Ratio of areas of 3 triangles</td> </tr> <tr> <td align="center">EGF</td> <td>3/8 : 1/2 : 5/8</td> <td align="center">3</td> <td>3/32</td> <td align="center">9</td> </tr> <tr> <td align="center">CED</td> <td>1/2 : 2/3 : 5/6</td> <td align="center">4</td> <td>1/6</td> <td align="center">16</td> </tr> <tr> <td align="center">CAB</td> <td>1/8 : 1/6 : 5/24</td> <td align="center">1</td> <td>1/96</td> <td align="center">1</td> </tr> </tbody> </table> This is David's method: The edge, $EF = 1/2$. Taking $FG = a$ then $GE = 1-a$ As it has a corner of the paper as part of it, we know $EGF$ is a right angled triangle, and so $GE^2 = EF^2 + FG^2$, which gives \[ (1-a)^2 = 1/4 + a^2 \] so \[a = 3/8.\] Now that we have the length of $a$, we have the lengths of all the sides of this triangle: $EF = 1/2$ , $FG = 3/8$ and $GE = 5/8.$ The area of triangle $EFG = 1/2$ base $\times$ height $ = 3/32$. Using this information and trigonometry, we can work out the angle $GEF$: $\angle GEF = \tan^{-1}(0.375 / 0.5) = \tan^{-1}0.75 = 36.9^{\circ}.$ taking the angle to the nearest tenth of a degree. Knowing this, we can work out the angle $DEC=53.1^{\circ}$ and use it with the length $DE$ to work out the length $DC=2/3.$ The last part needed of this triangle is $CE$ and, using Pythagoras' theorem, we get $CE =5/6.$ So $DE = {1 \over 2}$, $DC = {2 \over 3}$, $CE = {5 \over 6}$ and the area of triangle $DEC = 1/2 \times 2/3 \times 1/2 = {1 \over 6}.$ The last triangle needed is $ABC$. We know the length $BC = 1 - CE = 1/6.$ We can also work out the angle $BCA $ which is equal to angle $DCE$ and to angle $FEG.$ We can now use trigonometry to work out the length $AB = 1/6 \tan BCA = 1/8.$ And so once again, using Pythagoras, we can work out the length of line $AC$ which is $5/24.$ And so we have the final set of lengths: $AB = 1/8$, $BC = 1/ 6$ and $AC = 5/24.$ The area of triangle $ABC = 1/2 \times 1/8 \times 1/6 = 1/96$. This means the total area of all the paper with a single thickness = $26/96 = .270833..$ This can be backed up by working out the area of the trapezium $ABEG$, subtracting the area of triangle $ABC$ and then multiplying the result by 2 to give the area of the original square that is now double thickness. Area of trapezium = $1/2 [1/8 +5/8] \times 1 = 3/8.$ Double thickness area = $3/8 - 1/96 = 35/96.$ The total area is $2 \times 35/96 + 26/96= 1$ and so all areas are worked out and recorded, and the total area of single thickness paper and double thickness areas are recorded. </mdoxml> 2 5 0 0 0 1 0 Napkin A napkin is folded so that a corner coincides with the midpoint of an opposite edge . Investigate the three triangles formed . Transformations and their Properties Similarity 2D Geometry, Shape and Space Pythagoras' theorem Information and Communications Technology smartphone