Lines and a Rectangle

Geometry Level pending

Lines perpendicular to the diagonals of a rectangle with side lengths 2 and 6 are drawn through the rectangle's vertices. What is the area of the quadrilateral formed by the intersection of the lines truncated to the tenth place? ex. If it was 600.8986... make it 600.8

Note: Figure is not to scale.


The answer is 66.6.

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1 solution

Chew-Seong Cheong
Mar 22, 2015

We can see that A G O E A O A O F \triangle AGO \equiv \triangle EAO \equiv \triangle AOF are similar right-angle triangles.

We note that: A O = O G 2 + G A 2 = 1 2 + 3 2 = 10 AO = \sqrt{OG^2+GA^2} = \sqrt{1^2+3^2} = \sqrt{10} .

O E = 10 A O = ( 10 ) 2 = 10 \Rightarrow OE = \sqrt{10}AO = (\sqrt{10})^2 = 10

E F = 10 3 O E = 10 10 3 \Rightarrow EF = \dfrac {\sqrt{10}}{3} OE = 10 \dfrac {\sqrt{10}}{3}

Area of the quadrilateral = E F × A C = E F × 2 × A O = 10 10 3 × 2 × 10 = 200 3 66.6 = EF \times AC = EF \times 2 \times AO = 10 \dfrac {\sqrt{10}}{3} \times 2 \times \sqrt{10} = \dfrac {200}{3} \approx \boxed{66.6}

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