+269 A rectangle contains a strip of width $1,$ as shown below. Find the area of the strip.
+9664 
Label the diagram as shown. Let \(\angle FCE = \theta\) and \(EC = x\).
Let H be a point on AC such that FH is perpendicular to AC. Then \(FH = 8\).
Note that \(CE = FD = HA = x\). Then \(CH = 10 - x\).
\(CF = \sqrt{8^2 + (10 - x)^2} = \sqrt{x^2 - 20x + 164}\) by using Pythagoras theorem in \(\triangle CFH\).
In \(\triangle ECG\), \(\sin \theta = \dfrac{EG}{EC} = \dfrac1x\).
In \(\triangle FHC\), \(\sin \theta = \dfrac{FH}{CF} = \dfrac8{\sqrt{x^2 -20x+164}}\).
So we have \(\dfrac1x = \dfrac8{\sqrt{x^2 - 20x + 164}}\). Squaring gives \(\dfrac1{x^2} = \dfrac{64}{x^2 - 20x + 164}\).
Rearranging,
\(64x^2 = x^2 - 20x + 164\\ 63x^2 + 20x - 164 = 0\\ x = \dfrac{-20 \pm \sqrt{20^2 - 4(63)(-164)}}{2(63)}\\ x = \dfrac{8 \sqrt{163} - 10}{63}\text{ (reject negative root)} \)
Hence, the area is \(\dfrac{8x}2 = \dfrac{32 \sqrt{163} - 40}{63}\).
+9664
Label the diagram as shown. Let \(\angle FCE = \theta\) and \(EC = x\).
Let H be a point on AC such that FH is perpendicular to AC. Then \(FH = 8\).
Note that \(CE = FD = HA = x\). Then \(CH = 10 - x\).
\(CF = \sqrt{8^2 + (10 - x)^2} = \sqrt{x^2 - 20x + 164}\) by using Pythagoras theorem in \(\triangle CFH\).
In \(\triangle ECG\), \(\sin \theta = \dfrac{EG}{EC} = \dfrac1x\).
In \(\triangle FHC\), \(\sin \theta = \dfrac{FH}{CF} = \dfrac8{\sqrt{x^2 -20x+164}}\).
So we have \(\dfrac1x = \dfrac8{\sqrt{x^2 - 20x + 164}}\). Squaring gives \(\dfrac1{x^2} = \dfrac{64}{x^2 - 20x + 164}\).
Rearranging,
\(64x^2 = x^2 - 20x + 164\\ 63x^2 + 20x - 164 = 0\\ x = \dfrac{-20 \pm \sqrt{20^2 - 4(63)(-164)}}{2(63)}\\ x = \dfrac{8 \sqrt{163} - 10}{63}\text{ (reject negative root)} \)
Hence, the area is \(\dfrac{8x}2 = \dfrac{32 \sqrt{163} - 40}{63}\).
