490 /www/nrich/html/content/97/02/six5/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> <p><em>ABC</em> and <em>DEF</em> are equilateral triangles of side 3 and 4 respectively. Construct an equilateral triangle whose area is the sum of the area of <em>ABC</em> and <em>DEF</em> .</p> <mdo:image border="0" align="texttop" alt="" src="feb6.gif"></mdo:image> <br></br> <p>Does this work for any whole number side lengths?</p> <p>If not, under what circumstances does it work?</p> <p>What if the lengths of the sides of the triangles had been <em>a</em> and <em>b</em> instead of 3 and 4, can you construct an equilateral triangle whose area is the sum of the areas of <em>ABC</em> and <em>DEF</em> ? What is the new area?</p> <mdo:image border="0" align="texttop" alt="" src="feb7.gif"></mdo:image> <br></br></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><p class="editorial">John wrote:</p> <br></br> Areas of triangles using triangluar measure generate the square numbers<br></br> $1, 4, 9, 16, 25$.<br></br> So the two triangles $3$ and $4$ were a fairly special case as $3^2 + 4^2 = 5^2$<br></br> <br></br> But there are others that work such as $5, 12,13$ - that is Pythagorean Triples.<br></br> <br></br> In the original problem $a = 3$ and $b = 4$, so $3^2 + 4^2 = c^2$ giving $c = 5$.<br></br> <p>This was essentially just another way of looking at Pythagoras&#39;s theorem.</p> <p>In general:</p> <p>The formula for the area of an equilateral triangle with side $x$ is</p> $\text{Area} = \frac{x^2\sqrt3}{4}$ <p>So with the two triangles with sides <em>a</em> and <em>b</em> respectively, we are looking for a third triangle with area:<br></br> $$\frac{c^2\sqrt3}{4} = \frac{a^2\sqrt3}{4} + \frac{b^2\sqrt3}{4} $$</p> <div>This simplifies to give $c^2 = a^2 + b^2$, which is Pythagoras&#39;s theorem. This also means that it is possible to find a triangle whose area is the sum of any two triangles, although the sides will not necessarily be ineger lengths.</div> <br></br></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><h3>Why do this problem</h3> <div><a href="http://nrich.maths.org/public/viewer.php?obj_id=490&amp;part=490">This problem</a> encourages learners to considerfamiliar mathematical ideas in less less familar contexts. The problem will make connections with triangluar numbers and Pythagoras&#39; theorem. If non-standard units of area (equilateral triangles) are used calculation can be made easier.</div> <br></br> <h3>Possible approach</h3> <div>Draw the two trianglesand ask the group to consider how they would calculate their areas. Share ideas. If the suggestion does not arise from the group - draw one of the triangles on an isometric grid and ask the question again, seeking to draw attention to areas using areas of triangles as the unit of measurement.</div> <div>Now pose the problem and leave the group to experiment, raise and test conjectures.</div> <br></br> <h3>Key questions</h3> <div>What about triangles whose sides are not of integer length?</div> <h3>Possible support</h3> <div>Find a rule for the areas of all equilateral triangles with a side of integer length. Can they use this to generate some examples?</div> <br></br> <h3>Possible extension</h3> <div>Do the findings work when using areas measured in square units?</div> <br></br> <br></br> <br></br></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> Drawing the triangles on isometric paper and using areas bsed on triangles rather than squares might help.<br></br> <br></br></mdoxml> 2 4 0 0 0 1 0 ABC and DEF are equilateral triangles of side 3 and 4 respectively. Construct an equilateral triangle whose area is the sum of the area of ABC and DEF. 2D Geometry, Shape and Space Pythagoras' theorem Measures and Mensuration Area Using, Applying and Reasoning about Mathematics Generalising