Kite in a Square

8301 /www/nrich/html/content/id/8301/ 1 2012-07-01T00:00:00 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">ABCD is a square. M is the midpoint of the side AB. By constructing the lines AC, MC, BD and MD, the blue shaded quadrilateral is formed: <mdo:image src="diagram1.png"></mdo:image> What fraction of the total area is shaded? Below are three different methods for finding the shaded area. Unfortunately, the statements have been muddled up. Can you put them in the correct order? Coordinates <div class="toggle"> <mdo:image src="diagram4.png" style="float: right;"></mdo:image> (A) The shaded area is made up of two congruent triangles, one of which has vertices $(\frac{1}{3},\frac{2}{3}), (\frac{1}{2},\frac{1}{2}), (\frac{1}{2},1)$. (B) The line joining $(0,0)$ to $(\frac{1}{2},1)$ has equation $y=2x$ (C) Area of the triangle $= \frac{1}{2}(\frac{1}{2} \times \frac{1}{6}) = \frac{1}{24}$ (D) The line joining $(0,1)$ to $(1,0)$ has equation $y=1-x$. (E) Therefore the shaded area is $2 \times \frac{1}{24} = \frac{1}{12}$ (F) The point $(a,b)$ is at the intersection of the lines $y=2x$ and $y=1-x$. (G) Consider a unit square drawn on a coordinate grid. (H) The perpendicular height of the triangle is $\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$. (I) So $a = \frac{1}{3}, b=\frac{2}{3}$. The line joining $(0,0)$ to $(1,1)$ has equation $y=x$. To help you reorder the statements above, here is a set of <a class="pdflink" href="/content/id/8301/KiteInASquare1.pdf">printable cards</a> for you to cut out. </div> Similar Figures <div class="toggle"> (A) As line $AC$ intersects line $MD$ at point $E$, <mdo:image src="diagram2.png" style="float: right;"></mdo:image> the two opposite angles $\angle MEF$ and $\angle AED$ are equal. (B) The line $MF$ is half the length of $AD$. (C) Line $AD$ is parallel to line $MF$, so $\angle EDA$ and $\angle EMF$ are equal, and $\angle EAD$ and $\angle EFM$ are equal (alternate angles). (D) Therefore, $\triangle AED$ and $\triangle FEM$ are similar. (E) Therefore, the line $EH$ is half the length of $PE$. (F) Let ABCD be a unit square. (G) Therefore, the shaded area $MEFG = \frac{1}{24} \times 2 = \frac{1}{12}$ sq units. (H) $PH$ has length $\frac{1}{2}$ units, so $PE$ has length $\frac{1}{3}$ units and $EH$ has length $\frac{1}{6}$ units. (I) $\triangle MEF$ has area $\frac{1}{2}(\frac{1}{2}\times\frac{1}{6}) = \frac{1}{24}$ sq units. To help you reorder the statements above, here is a set of <a class="pdflink" href="/content/id/8301/KiteInASquare2.pdf">Printable cards</a> for you to cut out. </div> Pythagoras <div class="toggle"><mdo:image src="diagram3.png" style="float: right;"></mdo:image> (A) The area of $\triangle DMC = 2$ sq units. The area of $\triangle DFC = 1$ sq unit. Thus the combined area of $\triangle DFE$, $\triangle CFG$ and shaded area $MEFG$ is $1$ sq unit. (B) $(EH)^2+(HF)^2=(EF)^2 $ $EH = HF $ $(EH)^2 = \frac{1}{2}(EF)^2$ $EH = \frac{EF}{\sqrt 2}$ (C) Areas of $\triangle DFE$, $\triangle CFG$ and shaded area $MEFG$ are equal so each must have an area of $\frac{1}{3}$ sq units. (D) Area of $\triangle MEF = \frac{1}{2}(1 \times EH) = \frac{1}{2}(\frac{EF}{\sqrt 2})$ (E) By Pythagoras, $DF$ has length $\sqrt 2$. (F) The total area of the square is $4$ sq units, so the shaded area is $\frac{1}{12}$ the area of the whole square. (G) Area of $\triangle DFE = \frac{DF \times EF}{2}$ $= \frac{\sqrt 2 \times EF}{2} = \frac{EF}{\sqrt 2}$ (H) So the shaded area $MEFG$ is equal to the area of $\triangle DFE$. (I) Assume that the sides of the square are each $2$ units long. Thus, $DJ$ and $FJ$ are each $1$ unit long. To help you reorder the statements above, here is a set of <a class="pdflink" href="/content/id/8301/KiteInASquare3.pdf">Printable cards</a> for you to cut out. </div> Thanks to Jerome Foley for drawing our attention to this problem. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> Olivia from St Paul's Girls' School sent us <a class="pdflink" href="/content/id/8301/Olivia%20Kite.pdf">this solution</a> and Niharika from Leicester sent us <a class="pdflink" href="/content/id/8301/Niharika%20Kite.pdf">this solution</a>. Well done to both of you. Below are the three "muddled up methods" with all the steps in the correct order. Similar triangles <mdo:image src="diagram2.png" style="float: right;"></mdo:image> Let ABCD be a unit square. Line $AD$ is parallel to line $MF$, so $\angle EDA$ and $\angle EMF$ are equal, and $\angle EAD$ and $\angle EFM$ are equal (alternate angles). As line $AC$ intersects line $MD$ at point $E$, the two opposite angles $\angle MEF$ and $\angle AED$ are equal. Therefore, $\triangle AED$ and $\triangle FEM$ are similar. The line $MF$ is half the length of $AD$. Therefore, the line $EH$ is half the length of $PE$. $PH$ has length $\frac{1}{2}$ units, so $PE$ has length $\frac{1}{3}$ units and $EH$ has length $\frac{1}{6}$ units. $\triangle MEF$ has area $\frac{1}{2}(\frac{1}{2}\times\frac{1}{6}) = \frac{1}{24}$ sq units. Therefore, the shaded area $MEFG = \frac{1}{24} \times 2 = \frac{1}{12}$ sq units. Pythagoras <mdo:image src="diagram3.png" style="float: right;"></mdo:image> Assume that the sides of the square are each $2$ units long. Thus, $DJ$ and $FJ$ are each $1$ unit long. By Pythagoras, $DF$ has length $\sqrt 2$. Area of $\triangle DFE = \frac{DF \times EF}{2}$ $= \frac{\sqrt 2 \times EF}{2} = \frac{EF}{\sqrt 2}$ $(EH)^2+(HF)^2=(EF)^2 $ $EH = HF $ $(EH)^2 = \frac{1}{2}(EF)^2$ $EH = \frac{EF}{\sqrt 2}$ Area of $\triangle MEF = \frac{1}{2}(1 \times EH) = \frac{1}{2}(\frac{EF}{\sqrt 2})$ So the shaded area $MEFG$ is equal to the area of $\triangle DFE$. The area of $\triangle DMC = 2$ sq units. The area of $\triangle DFC = 1$ sq unit. Thus the combined area of $\triangle DFE$, $\triangle CFG$ and shaded area $MEFG$ is $1$ sq unit. Areas of $\triangle DFE$, $\triangle CFG$ and shaded area $MEFG$ are equal so each must have an area of $\frac{1}{3}$ sq units. The total area of the square is $4$ sq units, so the shaded area is $\frac{1}{12}$ the area of the whole square. Coordinates <mdo:image src="diagram4.png" style="float: right;"></mdo:image> Consider a unit square drawn on a coordinate grid. The line joining $(0,0)$ to $(1,1)$ has equation $y=x$. The line joining $(0,0)$ to $(\frac{1}{2},1)$ has equation $y=2x$ The line joining $(0,1)$ to $(1,0)$ has equation $y=1-x$. The point $(a,b)$ is at the intersection of the lines $y=2x$ and $y=1-x$. So $a = \frac{1}{3}, b=\frac{2}{3}$. The shaded area is made up of two congruent triangles, one of which has vertices $(\frac{1}{3},\frac{2}{3}), (\frac{1}{2},\frac{1}{2}), (\frac{1}{2},1)$. The perpendicular height of the triangle is $\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$. Area of the triangle $= \frac{1}{2}(\frac{1}{2} \times \frac{1}{6}) = \frac{1}{24}$ Therefore the shaded area is $2 \times \frac{1}{24} = \frac{1}{12}$ </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><h3>Why do this problem?</h3> Students often find geometric proofs quite intractable. In this problem, three different ways of proving the same result are presented, jumbled up, so that students can engage with the proofs without having to start from scratch. <h3>Possible approach</h3> Show the <a href="/content/id/8301/diagram1.png">image</a> from the problem. "ABCD is a square. M is the midpoint of AB. What fraction of the total area is shaded?" Give students some time to have a go at the problem. While they are working, circulate and see the methods they are trying. After a while, bring the class together again and acknowledge that the answer may not be immediately obvious. "I've been given the methods used by three different people. Unfortunately each method has got jumbled up. Can you put the statements in the right order to build a logical argument?" Hand out envelopes with each method (<a class="pdflink" href="/content/id/8301/KiteInASquare1.pdf">Coordinates</a>, <a class="pdflink" href="/content/id/8301/KiteInASquare2.pdf">Similar Figures</a>, and <a class="pdflink" href="/content/id/8301/KiteInASquare3.pdf">Pythagoras</a>) to pairs or threes. Coordinates is the most accessible method, and Pythagoras the most challenging. It is a good idea to print each method on different coloured card to avoid them getting muddled up. "With your partner, make sense of each step and put the cards in the right order. Can you recreate each method for your partner without looking at the cards?" Once students have spent a good long time engaging with the three methods, making sense of them and recreating them for themselves, bring the class together. Invite students to present each method to the class, and finally discuss the merits and disadvantages of each. <h3>Key questions</h3> For Coordinates method: <ul> <li>what are the equations of the lines?</li> <li>where do they intersect?</li> </ul> For Similar Figures method: <ul> <li>which angles are the same?</li> <li>what lengths do we know?</li> </ul> For Pythagoras method: <ul> <li>where are the right angles?</li> <li>what lengths do we know?</li> </ul> <h3> Possible extension</h3> <a href="/763">Enclosing Squares</a> offers a follow-up activity linked to Coordinates. <a href="/743">Take a Square</a> offers a follow-up activity linked to Similar Figures. <a href="/6553">Pythagoras Proofs</a> offers a follow-up activity linked to Pythagoras. <h3>Possible support</h3> Start by drawing a square on dotty paper (2 by 2 to start with) and explain that vertices and mid-points can be joined with straight lines. Challenge students to find the different fractions of the square that can be shaded. </mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">For Coordinates method: <ul> <li>what are the equations of the lines?</li> <li>where do they intersect?</li> </ul> For Similar Figures method: <ul> <li>which angles are the same?</li> <li>what lengths do we know?</li> </ul> For Pythagoras method: <ul> <li>where are the right angles?</li> <li>what lengths do we know?</li> </ul> <h3> </h3></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">Problem suggested by Jerome Foley: jfoley1@williamallitt.derbyshire.sch.uk (22 March 2012 email) Two different methods for solving the problem: <a href="/content/id/8301/Area%20of%20kite.doc">Area of kite.doc</a> <mdo:image src="diagram2.png" style="float: right;"></mdo:image>Similar triangles Let ABCD be a unit square. Line $AD$ is parallel to line $MF$, so $\angle EDA$ and $\angle EMF$ are equal, and $\angle EAD$ and $\angle EFM$ are equal (alternate angles). As line $AC$ intersects line $MD$ at point $E$, the two opposite angles $\angle MEF$ and $\angle AED$ are equal. Therefore, $\triangle AED$ and $\triangle FEM$ are similar. The line $MF$ is half the length of $AD$. Therefore, the line $EH$ is half the length of $PE$. $PH$ has length $\frac{1}{2}$ units, so $PE$ has length $\frac{1}{3}$ units and $EH$ has length $\frac{1}{6}$ units. $\triangle MEF$ has area $\frac{1}{2}(\frac{1}{2}\times\frac{1}{6}) = \frac{1}{24}$ sq units. Therefore, the shaded area $MEFG = \frac{1}{24} \times 2 = \frac{1}{12}$ sq units. Pythagoras <mdo:image src="diagram3.png" style="float: right;"></mdo:image> Assume that the sides of the square are each $2$ units long. Thus, $DJ$ and $FJ$ are each $1$ unit long. By Pythagoras, $DF$ has length $\sqrt 2$. Area of $\triangle DFE = \frac{DF \times EF}{2}$ $= \frac{\sqrt 2 \times EF}{2} = \frac{EF}{\sqrt 2}$ $(EH)^2+(HF)^2=(EF)^2 $ $EH = HF $ $(EH)^2 = \frac{1}{2}(EF)^2$ $EH = \frac{EF}{\sqrt 2}$ Area of $\triangle MEF = \frac{1}{2}(1 \times EH) = \frac{1}{2}(\frac{EF}{\sqrt 2})$ So the shaded area $MEFG$ is equal to the area of $\triangle DFE$. The area of $\triangle DMC = 2$ sq units. The area of $\triangle DFC = 1$ sq unit. Thus the combined area of $\triangle DFE$, $\triangle CFG$ and shaded area $MEFG$ is $1$ sq unit. Areas of $\triangle DFE$, $\triangle CFG$ and shaded area $MEFG$ are equal so each must have an area of $\frac{1}{3}$ sq units. The total area of the square is $4$ sq units, so the shaded area is $\frac{1}{12}$ the area of the whole square. Coordinates <mdo:image src="diagram4.png" style="float: right;"></mdo:image> Consider a unit square drawn on a coordinate grid. The line joining $(0,0)$ to $(1,1)$ has equation $y=x$. The line joining $(0,0)$ to $(\frac{1}{2},1)$ has equation $y=2x$ The line joining $(0,1)$ to $(1,0)$ has equation $y=1-x$. The point $(a,b)$ is at the intersection of the lines $y=2x$ and $y=1-x$. So $a = \frac{1}{3}, b=\frac{2}{3}$. The shaded area is made up of two congruent triangles, one of which has vertices $(\frac{1}{3},\frac{2}{3}), (\frac{1}{2},\frac{1}{2}), (\frac{1}{2},1)$. The perpendicular height of the triangle is $\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$. Area of the triangle $= \frac{1}{2}(\frac{1}{2} \times \frac{1}{6}) = \frac{1}{24}$ Therefore the shaded area is $2 \times \frac{1}{24} = \frac{1}{12}$ </mdoxml> 2 3 0 0 0 1 0 Kite in a Square Can you make sense of the three methods to work out the area of the kite in the square? 2D Geometry, Shape and Space Pythagoras' theorem 2D Geometry, Shape and Space Similar triangles Coordinates and Coordinate Geometry Cartesian equations of lines Using, Applying and Reasoning about Mathematics Proof Sorting Using, Applying and Reasoning about Mathematics Mathematical reasoning & proof Secondary Mapping Document Geometrical reasoning US Secondary Mapping Document DisplayCabinet