+0

# Geometry

0
2
1
+48

Chords  UV,WX andYZ  of a circle are parallel. The distance between chords UV and WX is  2 and the distance between chords WX and  YZ is also 2. If UV=64  and YZ=48 then find WX.

Jun 26, 2024

### 1+0 Answers

#1
+129725
+1

Let  d  be the distance between  the center of the circle and UV

We can  form a right triangle  with legs of d , (UV/2)   and a hypotenuse of r

So  we have

d^2  + (UV/2)^2  = r^3

d^2  + 32^2  = r^2      (1)

Likewise  let  the distance between the center of the circle and  YZ = d + 2 + 2 = d + 4

And similar to  the first case we have

(d + 4)^2 + (YZ / 2)^2  = r^2

(d + 4)^2 + 24^2 = r^2    (2)

Set (1)  = (2)

d^2 + 32^2  = (d + 4)^2 + 24^2       simplify

d^2 + 1024 = d^2 + 8d + 16 + 576

1024 - 16 - 576  =  8d

432  = 8d

d =432 / 8   =  54

And

d^2 + 32^2  = r^2

54^2 + 32^2 = r^2

3940  = r^2

So  we have another right triangle  such that the distance from the center of the circle to WX = d + 2

So

(d + 2)^2  + ( WX / 2)^2  = r^2

(54 + 2)^2  +  WX^2 / 4  = 3940

56^2  + WX^2 / 4  =3940

3136 + WX^2 /  4  = 3940

WX^2 =  4 ( 3940 - 3136)

WX^2  = 4 * 804

WX^2  = 3216

WX = sqrt (3216)   ≈  56.7

This makes sense.....WX should  be > 48  but < 64

Jun 26, 2024