A question from the Math B regents.. with the circle and the chords?

Question

inside the circle where there were 2 chords drawn. one chord the length was 7 (3+4) and the other with lenth 2x-4 (x-4+x) these 2 chords made 2 triangles (which i believe were similiar) was i to put the correspoding sides in proportion to find the value of x?

anyways can anyone fill me in on wat they might have gotten or how they procceded with the problem?

Answer 1

Hi,

When 2 chords intersect, the product of the lengths of 2 parts of one chord equals the product of the lengths of the 2 parts of the other chord. Since your first chord had parts with lengths of 3 and 4, and then your second chord had parts of x - 4 and x, then:

3(4) = x(x - 4)
12 = x² - 4x
x² - 4x - 12 = 0
(x - 6)(x + 2) = 0
x = 6 or x = -2

Clearly, a segment can not have a length of -2, so discard that answer. That means one part of the second segment had a length of 6 (x = 6) and the other section had a length of 2 ,
(x - 4 = 6 - 4 = 2).


I hope that helps!! :-)

Answer 2

Two chords cannot make one triangle much less than two triangles. I'm guessing that one chord was 7(3x+4) and the other was 2x-4. The triangles were made up by the chord and two radii each. If these triangles were similar, then they were also congruent.
Hence 7(3x+4)=2x-4
21x +28 =2x -4
19x = 24
x=24/19 = 1 5/24