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?

(In reply to Drawing by Tristan)

Click on the "In reply to..." link above to see the picture I used.

Draw a perpendicular to AB and ED through point D.  Call its intersection with AB "F."  FD=AE=5.  Notice that BDF and DBC are congruent by hypotenuse-leg theorem.

AREA=
AEDB-DBC
AEDB-BDF
AEDF

Since the total area is equal to the area of AEDF, triangle AED is exactly half the area.  ACB must be the other half.  Since both of those right triangles' areas are equal, and they both have a 5 inch leg, they are congruent.  Therefore, AD=AC+1=ED+1.

AE² + ED² = AD²
25 + ED² = (ED+1)²
24 = 2ED
ED = 12.
AREA = AEDF = 2(AED) = 2 * 12 * 5 / 2 = 60

So the area is 60 inē

Posted by Tristan on 2004-06-13 13:20:26