## Rectangle Construction

#### Problem

Given rectangle $ABCD$, $AB$ is extended to $E$ such that $BE = BC$.
$AE$ forms the diameter of a circle and $CB$ is extended to $F$ which lies on the circle.

How does the length $BF$ relate to the original rectangle?

#### Solution

Consider the following diagram.

As triangle $AEF$ is in a semi-circle, $\angle AFE = 90 \Rightarrow \angle FAB + \angle FEB = 90$

But $\angle FAB + \angle AFB = 90 \Rightarrow \angle AFB = \angle FEB$. In the same way, $\angle BFE = \angle FAB$.

Thus the right angled triangles $FAB$ and $FEB$ are similar.

$\therefore \dfrac{AB}{FB} = \dfrac{FB}{BE} \Rightarrow AB \centerdot BE = (FB)^2$

But $BE = BC$, so $AB \centerdot BC = (FB)^2$

Hence the square on $FB$ is equal to the area of the rectangle $ABCD$ and it can be seen that this construction process has "squared" the rectangle.

Show how this method can be adapted to construct the square root of a given length $AB = x$.

Problem ID: 376 (17 Oct 2010)     Difficulty: 2 Star

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