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# Folded Square

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Age 14 to 16

ShortChallenge Level

- 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.

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.

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.