Question 188200
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Can't do it.  Not enough information.  You have described an isoceles triangle, but the length of the third side could be anything.  Now if you happen to have failed to mention that the included angle between the two 8" sides was a right angle, which might be indicated by a little square in the angle, then we have enough.


I will proceed on that assumption.


If it is an isoceles right triangle, then either of the 8" sides can be considered the base of the triangle, and the other 8" side will be the altitude.


The area of a triangle is the measure of the base times the measure of the altitude (or height, if you will) and the product divided by 2.  8 times 8 is 64 divided by 2 is 32 square inches.  To the nearest tenth, that would be 32.0 square inches.


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