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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
avatar+383 
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
avatar+114098 
+2

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  :

 

 

 

cool cool cool

 Nov 13, 2020
 #4
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+1

I understand. Thank you so much!

Guest Nov 13, 2020
 #5
avatar+824 
+3

∠(x) = arctan(10 / 8) = 51.34019175º

 

∠(y) = 90 - ∠x = 38.65980825º

 

AB = 2 [tan(y) * 8] = 12.8  smiley

 

 Nov 14, 2020
 #6
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0

thanks for giving another approach! 

Guest Nov 16, 2020

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