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Folded Square

Age 14 to 16
ShortChallenge Level Yellow starYellow starYellow star
Secondary curriculum
  • Problem
  • Solutions
In the diagram below, the line EB has been added. The dotted line must be the perpendicular bisector of EB, since for B to fold onto E, B and E must be the same distance on either side of the fold.



Using gradient
The distance CE is half of the distance BC, so the gradient of EB must be 2.
This means that the gradient of the dotted line must be $-\frac{1}{2}$.

”‹The dotted line and the folded line cross halfway between E and B (since EB is bisected), so 4 cm from the top (and from the bottom) and 2 cm from the right (and from E). So the distance marked on the diagram below is 6 cm.
 
So the vertical distance between the point where the triangle meets AD and the point where the dotted line and the folded line cross is $\frac{1}{2}\times$6 = 3 cm.

So the dotted line meets AD 4$-$3 = 1 cm below A.


Using similar triangles
The two triangles coloured blue are similar, as they both contain a right angle and their other two angles are 90$^\text{o}$ rotations of each other.

The dotted line and the folded line cross halfway between E and B (since EB is bisected), so 4 cm from the top (and from the bottom) and 2 cm from the right (and from E). So the distance marked on the diagram below is 6 cm.
 
The sides of the larger triangle are in the ratio 1:2 (from 4:8), so the vertical side of the smaller triangle must be 3 cm. So the dotted line meets AD 4$-$3 = 1 cm below A.

You can find more short problems, arranged by curriculum topic, in our short problems collection.

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The NRICH Project aims to enrich the mathematical experiences of all learners. To support this aim, members of the NRICH team work in a wide range of capacities, including providing professional development for teachers wishing to embed rich mathematical tasks into everyday classroom practice.

NRICH is part of the family of activities in the Millennium Mathematics Project.

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