 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  The right pentagon (Posted on 2004-06-12) 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?

 See The Solution Submitted by Gamer Rating: 3.3333 (6 votes) Comments: ( Back to comment list | You must be logged in to post comments.) And the answer is .... | Comment 4 of 11 | 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 2004-06-12 12:30:42 Please log in:

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