If line FL is a radius of circle F and point T is in the exterior of circle F, then Ft is greater than FL
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ΔABC is a 45 - 45 - 90 triangle. The pattern of its sides is as follows:
Each leg = 1 unit (and both legs are that way, since the triangle is isosceles - so two sides are the same)
Hypotenuse = √2 units.
So if we know either leg, we multiply by √2 to get the hypotenuse. In reverse, we divide by √2 if we know the hypotenuse to get the measurement of a leg.
Our problem tells us that the hypotenuse AC is 10 units. We divide 10 by √2 to get the measurement of leg AB. Since it's a 45 -45 - 90 triangle, AB = BC.
to rationalize the radical
Thus, each leg is 5\sqrt{2} [/tex].