There is a pentagon ABCDE such that angles A, C, and E are all right angles and it is concave at angle C.
If a ray starting at D was aimed at C, it would go exactly 1 inch before going through C, and would eventually go through A.
If BC and AE are both 5 inches, what is the area of this pentagon?
I'm getting 60.
AB is 13 in length
ED is 12 in length
So it's like a rectangle 13 by 5 minus 2 little triangles that are each 1 by 5 by sqrt(26) right triangles. or, 65 minus 5.
Consider A to be at the origin, B is along the positive x axis, E is on the negative y axis. To find the length of ED, one must find the slope (m) of segment DC, and then it's equation is y=mx since the line goes through the origin. The equation for AB is y=0. The equation for ED is y= 5. To figure where C is, first imagine a long rectangle A,B,D',E where D' is directly below B. Point D is 1 inch to the left of D'. Consider the triangle B,D',D which is a 1 by 5 by sqrt(26) right triangle with hypotenuse BD. Now reflect that triangle around BD, and you get triangle BDC. The angle of D,D' is zero; angle DB is arctan(5); angle DC is 2*arctan(5). The slope m is TAN(2*arctan(5)) = .41667
So y=mx ... 5 = .41667 x ... x=12 so
D is at (12,5) and
B is at (13,0)
E is at (0,5)
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My figure is oriented differently than Tristan's
Edited on June 12, 2004, 12:49 pm
Edited on June 12, 2004, 12:53 pm

Posted by Larry
on 20040612 12:30:42 