Feed on
Posts
Comments

Geometry Challenge

March 3rd, 2006 by SplineGuy

I came across this at think again! I had fun solving it so I’ll pass it along.

Consider the following quadrilateral

The midpoint of opposite sides have been connected, dividing it into 4 parts. We know the areas of three of the regions. The question is what is the area of the fourth?

SOLUTION: This spoils the ending so give it a shot yourself before you take my word for it. Then you can scroll down.

.

.

.

.

.

.

.

.

.

.

Consider the following. Let us label the points A,\ B,\ C,\ D,\ E,\ F,\ and O.

Notice that C is the point that we get if we drop a line straight done from O and form a line \overline{OC} which is perpendicular to \overline{AD}. This line \overline{OC} is the altitude of the two triangles \Delta OAB and \Delta OBD. Let h = | \overline{OC} |. Thus, we can say

\mbox{Area}(\Delta OAB) = \displaystyle \frac{1}{2} h | \overline{AB} |

and

\mbox{Area}(\Delta ODB) = \displaystyle \frac{1}{2} h | \overline{BD} |

But | \overline{BD} | = | \overline{AB} |, since B is the midpoint of \overline{AD}. Therefore

\mbox{Area}(\Delta ODB) = \mbox{Area}(\Delta OAB) = 6

where the last equality is because we have split the lower right quadrilateral into two equal pieces and half of 12 is 6.

Similarly we can argue that \mbox{Area}(\Delta OED) = 5

So the final answer is: The area of the remaining piece of the quadrilateral is 6+5 = 11. Tada!

UPDATED: (Thanks to Jan Nordgreen) While it is true that we do not have the lower right quadrilateral cut into two equal pieces we do know that as we cut each quadrilateral in two, adjacent triangles of different quadrilaterals will always have the same area. This can be used to determine the missing area. In fact, it can be shown that sum of areas of catty-corner pieces is the same, i.e. the area of the upper left plus lower right is equal to the lower left plus upper right.

Share and Enjoy:
  • Digg
  • del.icio.us
  • Slashdot
  • StumbleUpon
  • Technorati
  • Facebook
  • Google
  • Pownce

Posted in Math is Everywhere, Mathematics |

4 Responses to “Geometry Challenge”

  1. on 04 Mar 2006 at 5:13 pm   Jan Nordgreen

    ‘we have split the lower right quadrilateral into two equal pieces’

    Sure?

    If the areas had been 10, 11, 10, ? instead of 10, 11, 12, ? one would find, with the reasoning above, that the first ? is 5 + 5, while it is 9.

  2. on 04 Mar 2006 at 7:09 pm   SplineGuy

    Ahh!! Yes. Stupid mistake. When time allows, the correction will be posted.

  3. on 04 Mar 2006 at 10:04 pm   Jan Nordgreen

    It is a nice example of reaching the correct result with wrong methods. I wouldn’t change it if I were you.

    http://simpler-solutions.net/pmachinefree/thinkagain/comments.php?id=1867_0_3_0_C

  4. on 05 Mar 2006 at 1:01 pm   SplineGuy

    Thanks again, Jan. Both for commenting and directing a bit of traffic this way. Note that I used the strikeout feature to correct the original posting so people can see where the mistake was made.

Trackback URI | Comments RSS

Leave a Reply