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  <resource>
  <id>383</id>
  <path>/www/nrich/html/content/01/11/15plus3/</path>
  <resourceTypeID>1</resourceTypeID>
  <last_published>2011-02-01T00:00:01</last_published>
  <indexXML>&lt;mdoxml version=&quot;1.0&quot;&gt;&lt;br&gt;&lt;/br&gt;
&lt;p&gt;A tetrahedron has two identical equilateral triangles faces, of side length 1 unit. The other two faces are right angled isosceles triangles. Find the exact volume of the tetrahedron.&lt;/p&gt;
&lt;p&gt;&lt;br&gt;&lt;/br&gt;
 &lt;/p&gt;&lt;/mdoxml&gt;</indexXML>
  <solutionXML>&lt;mdoxml version=&quot;1.0&quot;&gt;&lt;br&gt;&lt;/br&gt;
&lt;span class=&quot;editorial&quot;&gt;This excellent solution came from Ruth from
Manchester High School for Girls.&lt;/span&gt; &lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
In the tetrahedron $ABCD$, let $ABC$ and $ACD$ be right angled. If
you position the tetrahedron so that $ABC$ is the base, then the
vertex $D$ is directly above the edge $AC$. This means that the
height of the tetrahedron is the height of the triangle $ACD$ which
is $\frac{1}{\sqrt{2}}$ and the area of the base of the tetrahedron
is the area of the triangle $ABC$ which is $\frac{1}{2}$. The
volume of a pyramid is one third base times height. Therefore
&lt;br&gt;&lt;/br&gt;
&lt;br&gt;&lt;/br&gt;
$$\begin{eqnarray} V &amp;amp;=&amp;amp; \frac{1}{3}\frac{1}{2}
\frac{1}{\sqrt{2}} \\ &amp;amp;=&amp;amp;\frac{1}{6 \sqrt{2}} \\
&amp;amp;=&amp;amp; \frac{\sqrt{2}}{12} \end{eqnarray}$$&lt;br&gt;&lt;/br&gt;&lt;/mdoxml&gt;</solutionXML>
  <noteXML/>
  <clueXML>&lt;mdoxml version=&quot;1.0&quot;&gt;&lt;br&gt;&lt;/br&gt;
To discover the easy way to find this volume you can very quickly
make for yourself a model of the tetrahedron. Take a square $ABCD$
of stiff paper and fold it along the diagonal $AC$ . Open it until
the distance between $B$ and $D$ is equal to the length of the
sides of the square and then the four vertices are the vertices of
the tetrahedron with two equilateral faces and two isosceles faces.
You can either stand this 'tetrahedron' on an equilateral face (
$ABD$ or $CBD$ ) as its base or on an isosceles face ( $ABC$ or
$ADC$ ) as its base. One choice makes it easy to find the area of
the base and the height and so to find the volume.&lt;br&gt;&lt;/br&gt;&lt;/mdoxml&gt;</clueXML>
  <canonXML/>
  <end_user_role>2</end_user_role>
  <difficulty>3</difficulty>
  <keystage1>0</keystage1>
  <keystage2>0</keystage2>
  <keystage3>0</keystage3>
  <keystage4>0</keystage4>
  <keystage4plus>1</keystage4plus>
  <title>Reach for Polydron</title>
  <description>In a tetrahedron two of the faces are identical equilateral
triangles - side length 1 unit and two are right angled isosceles
triangles. Find the exact volume of the tetrahedron.</description>
  <spec_group>2D Geometry, Shape and Space
    <specifier>Right angled triangles</specifier>
  </spec_group>
  <spec_group>2D Geometry, Shape and Space
    <specifier>Pythagoras' theorem</specifier>
  </spec_group>
  <spec_group>3D Geometry, Shape and Space
    <specifier>Tetrahedra</specifier>
  </spec_group>
  <spec_group>2D Geometry, Shape and Space
    <specifier>Equilateral triangles</specifier>
  </spec_group>
  <spec_group>Measures and Mensuration
    <specifier>Volume and capacity</specifier>
  </spec_group>
</resource>