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# Question

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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!

Nov 13, 2020

#1
0  To be honest im not sure so I hope these help in anyway.

Nov 13, 2020
#2
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i think since it is an isosceles trapezoid and that is a right trapezoid it might not be correct

Guest Nov 13, 2020
#3
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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  :    Nov 13, 2020
#4
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I understand. Thank you so much!

Guest Nov 13, 2020
#5
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∠(x) = arctan(10 / 8) = 51.34019175º

∠(y) = 90 - ∠x = 38.65980825º

AB = 2 [tan(y) * 8] = 12.8  Nov 14, 2020
#6
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thanks for giving another approach!

Guest Nov 16, 2020