ABCD is a trapezoid with AB < CD and AB parallel to CD. \Gamma is a circle inscribed in ABCD, such that \Gamma is tangent to all four sides. If AD = BC = 25 and the area of ABCD is 600, what is the radius of \Gamma?
in a trapezoid like this the sum of two opposite sides is equal to the sum of the other two
thus the sum of the bases is 50
area = (1/2)(sum of the bases) x height
height = (2 area)/(sum of the bases)
height = 1200/50 = 24
height is also the diameter of the circle
therefore its radius is 12
To find the radius of the circle inscribed in the trapezoid, we need to use the formula for the area of a trapezoid:
Area = (sum of the lengths of two parallel sides) * (distance between them) / 2.
In this case, the sum of the lengths of the two parallel sides is AB + CD = AD + BC = 25 + 25 = 50.
Now, we need to find the distance between these two parallel sides. Let's call it h.
Since the circle is inscribed in the trapezoid, it is tangent to all four sides. This means that the distance between AB and CD is equal to twice the radius of the circle (because the radius is perpendicular to the tangent line at the point of tangency). So, h = 2r.
Now we can substitute the values into the area formula:
600 = (50) * (2r) / 2
We can simplify this equation to:
600 = 25r
Now we can solve for r by dividing both sides by 25:
r = 24
Therefore, the radius of the circle inscribed in the trapezoid is 24.