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# perpendicular chords

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137
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Perpendicular cords AC and DE intersect at B so that the lengths of AB,BC and BD are 3, 4 and 2 respectively.  Find the radius of the circle.

Dec 10, 2020

#1
+1164
+2

2 * BE = 3 * 4   ==>  BE = 6

Radius   r = sqrt(82 +1) / 2         r = √65 / 2

Dec 10, 2020
edited by jugoslav  Dec 10, 2020
#3
+1

Can you explain where th expresion for the radius comes from ?

Guest Dec 10, 2020
#5
+1164
+1

Line segments ED and E'D' are congruent. The distance between them is 1 unit.

So, we have (ED')2 = (E'D')2 + (EE')2

Radius        r = (ED') / 2

( I hope it's clear enough.)

jugoslav  Dec 12, 2020
#2
+117546
+1

By the intersecting  chords theorem

EB * DB   =  AB * CB

EB * 2  =    3 * 4

EB    = 6

Angle EBC  =(1/2)  (minor arc EC + minor arc AD)

90  = (1/2)  ( minor arc  EC  + minor arc  AD)

180 =  minor arc EC + minor arc AD

EC^2  = 4^2 + 6^2   = 52

AD^2  =  3^2  + 2^2  =  13

Using the Law of  Cosines

EC^2 =  2 r^2 - 2r^2 cos A

AD^2  = 2r^2  - 2r^2  cos (180 - A)           where A is a central angle....simplify

Note.....cos (180 - A)  =  -cos A

So....the  second equation  becomes

AD^2  = 2r^2 - 2r^2 (-cos A)

13 = 2r^2  + 2r^2 cos A

So we have

52  =2r^2  -2r^2 cos A

13  = 2r^2 +2r^2 cos A         add these equations

65 = 4r^2

65/4  = r^2

r  = sqrt (65) / 2

Dec 10, 2020
#4
+1

Here's a different method.

Take a line parallel to DE, a half a unit to the right.

It's the perependicular bisector of AC so will pass through the centre of the circle.

Drop a perpendicular from the centre of the circle onto DE, bisecting it.

Then, by Pythagoras, r^2 = 4^2 + (1/2)^2 = 16 + (1/4) = 65/4, so r = sqrt(65)/2.

Dec 10, 2020