A circle of radius 8 is inscribed in an isosceles trapezoid. The length of the larger base of the trapezoid is 20. Find the length of the shorter base.
I got 12 4/5 but I am still confused so if anyone can explain, that would be great!
To be honest im not sure so I hope these help in anyway.
Here's one way of solving this.....
Note that from a point ourside a circle, two tangents drawn to a circle have the same length
We have two circles
One has the equation x^2 + (y-8)^2 = 64 [ this is the inscribed circle ]
The other has the equation (x - 10)^2 + y^2 = 100
Solving these two equations for their intersections we have that (x,y) = (320/41, 400/41) = point C in the figure below
We have two points (10,0) and (320/41, 400/41)
The slope of this tangent line to the inscribed circle = (400/41 - 0) / ( 320/41 -10) = (400/41) / ( -90/41) =
-400/90 = -40/9
So.....the equation of this tangent line through these rwo points is
y = (-40/9) ( x - 10)
y =( -40/9)x + 400/9
Since the inscribed circle touches the top of the trapezoid, the line will pass through y= 16
So....to find the x value at this point we can solve this
16 = (-40/9)x + 400/9
16 - 400/9 = (-40/9)
-256/9 = *(-40/9)
-256/-40 = 6.4
Double this to find the upper base length = 12.8 [ your answer is correct !!! ]
See the following figure :