Question 231878
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{{{drawing(
500, 500, -5, 5, -5, 5,
triangle(-3,4,-3,-4,3,-4),
red(circle(-3,1.6,.05),
circle(-1.2,1.6,.05),
line(-3,1.6,-1.2,1.6),
locate(-3.2,1.6,O),
locate(-1,1.6,A)),
blue(locate(-3.2,4.2,E),
locate(-3.2,-4.2,R),
locate(3.2,-4.2,D))
)}}}


Notice a few things right off:


1.  EA = 6 and AD = 14 means that ED = 20.


2.  OA perpendicular to RE means OA parallel to RD, hence:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ]*  Triangle OEA similar to triangle RED


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ ]*  Quadrilateral ROAD is a trapeziod 


3.  20 is 4 times 5 and 16 is 4 times 4, hence RD has to be 12: 4 times 3 -- 3-4-5 right triangle.
(verify by using Pythagoras and calculating *[tex \Large RD\ =\ \sqrt{20^2\ -\ 16^2}] if you want)



The hypotenuse of OEA is 6, so *[tex \Large \ \ \frac{6}{5}\ =\ \frac{OE}{4}\ =\ \frac{OA}{3}\ \ \ ] because of similar triangles.  Hence OE = 4.8 and OA = 3.6.  Verification left as an exercise for the student.


Since OE = 4.8, RO = 16 - 4.8 = 11.2.  And because of right angle R, RO is an altitude of the trapezoid


The formula for the area of a trapezoid is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A_t\ = a\left(\frac{b_1\ +\ b_2}{2}\right)]


Plug in the values:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ A_{ROAD}\ = 11.2\left(\frac{3.6\ +\ 12}{2}\right)]


You can do your own arithmetic.


John
*[tex \LARGE e^{i\pi} + 1 = 0]
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