Points \(A\) and \(B\) are on the circumference of a circle with radius \(10\) cm and centre \(C\) such that \(AC \perp BC\). Point \(D\) lies inside the circle on \(BC\) such that \(BD=2\) cm. Point \(E\) lies on the circumference of the circle, on the minor arc \(AB\), such that \(DE \perp BC\). Point \(F\) lies inside the circle on \(AC\) such that \(CDEF\) forms a rectangle.
Determine the distance from \(A\) to \(F\).
Note: You may find the following useful:
The Pythagorean Theorem states, “In a right-angled triangle, the square of the length of hypotenuse (the side opposite the right angle) equals the sum of the squares of the lengths of the other two sides.”
In the right-angled triangle shown, \(c\) is the hypotenuse, \(a\) and \(b\) are the lengths of the other two sides, and \(c^2=a^2+b^2\).
First we draw the radius \(EC\). Since the radius of the circle is \(10\) cm, it follows that \(AC=BC=EC=10\). Then \(DC = BC - BD = 10-2=8\).
Since \(\angle CDE = 90\degree\), it follows that \(\triangle CDE\) is a right-angled triangle. Using the Pythagorean Theorem in \(\triangle CDE\), \[\begin{aligned} DE^2 + CD^2&= EC^2\\ DE^2 &= EC^2 - CD^2\\ &=10^2 - 8^2\\ &= 100- 64\\ &=36 \end{aligned}\] Then \(DE = \sqrt{36}=6\), since \(DE>0\).
Since \(CDEF\) is a rectangle, \(CF=DE=6\). Then \(AF = AC - CF = 10-6=4\). Thus, the distance from \(A\) to \(F\) is \(4\) cm.