ABCD is a trapezoid with parallel sides AB and CD. Γ is an inscribed circle of ABCD, and tangential to sides AB,BC,CD and AD at the points E,F,G and H respectively. If AE=2,BE=3, and the radius of Γ is 12, what is the length of CD?
72
To find the length of CD, we can use the fact that the sum of the lengths of the two parallel sides of a trapezoid is equal to the product of the height (distance between the parallel sides) and the average of the lengths of the parallel sides.
Let x be the length of CD. So, the average of the lengths of AB and CD is (3 + x)/2.
Since Γ is an inscribed circle, we know that the distance between the center of the circle and any tangent is equal to the radius of the circle. Therefore, the radius of Γ is 12, and FH = 12.
We can now use the property of tangent lines to find the lengths of FC and GD.
In the right triangle FHC, using the Pythagorean theorem, we have FH² = FC² + HC².
Substituting the known values, we get:
12² = FC² + (BE + AE)²
144 = FC² + (3 + 2)²
144 = FC² + 5²
144 = FC² + 25
FC² = 144 - 25
FC² = 119
FC ≈ √119
In the right triangle GDH, using the Pythagorean theorem, we have GH² = GD² + HD².
Substituting the known values, we get:
12² = GD² + (BE + AE)²
144 = GD² + (3 + 2)²
144 = GD² + 5²
144 = GD² + 25
GD² = 144 - 25
GD² = 119
GD ≈ √119
Now that we have the lengths of FC and GD, we can find the length of CD:
CD = FC + GD
CD ≈ √119 + √119
CD ≈ 2√119
Therefore, the length of CD is approximately 2√119.
Note: The lengths in the solution are given as approximations due to the use of square roots.