Finding Lengths of Segments in Chords

When two chords intersect in the interior of a circle, each chord is divided into two segments which are called segments of a chord.

Theorem 10.15If two chords intersect in the interior of a circle, then the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord.

THEOREM

The following theorem gives a relationship between the lengths of the four segments that are formed.

EA • EB = EC • ED

Finding Lengths of Segments in Chords

You can use similar triangles to prove Theorem 10.15.

PROVE EA • EB = EC • ED

GIVEN AB, CD are chords that intersect at E.

Paragraph Proof Draw DB and AC.

The lengths of the sides are proportional.EAED

=ECEB

Cross Product PropertyEA • EB = EC • ED

Because C and B intercept the same arc, C B.

Likewise, A D.

So, the lengths of corresponding sides are proportional.

By the AA Similarity Postulate, AEC ~ DEB.

Finding Segment Lengths

Use Theorem 10.15.RQ • RP = RS • RT

Substitute.RQ • RP = RS • RT

Simplify.9x = 18

Divide each side by 9. x = 2

63x9

Chords ST and PQ intersect inside the circle. Find the value of x.

Using Segments of Tangents and Secants

In the figure shown below, PS is called a tangent segment because it is tangent to the circle at the endpoint. Similarly, PR is a secant segment and PQ is the external segment of PR.

EA • EB = EC • ED

THEOREMS

Using Segments of Tangents and Secants

Theorem 10.16If two secant segments share the same endpoint outside a circle, then the product of the length of one secant segment and the length of its external segment equals the product of the length of the other secant segment and the length of its external segment.

(EA) 2 = EC • ED

Theorem 10.17If a secant segment and a tangent segment share an endpoint outside a circle, then the product of the length of the secant segment and the length of its external segment equals the square of the length of the tangent segment.

Finding Segment Lengths

Find the value of x.

Use Theorem 10.16.RP • RQ = RS • RT

Substitute.RP • RQ = RS • RT

Simplify.180 = 10x + 100

Divide each side by 10. 8 = x

(x + 10)10(11 + 9)9

Subtract 100 from each side. 80 = 10x

(CB) 2 CE • CD

Estimating the Radius of a Circle

AQUARIUM TANK You are standing at point C, about 8 feet from a circular aquarium tank. The distance from you to a point of tangency on the tank is about 20 feet. Estimate the radius of the tank.

Use Theorem 10.17.(CB) 2 = CE • CD

Substitute.

Simplify.400 16r + 64

21 r

(2r + 8)820 2

Subtract 64 from each side. 336 16r

SOLUTION You can use Theorem 10.17 to find the radius.

So, the radius of the tank is about 21 feet.

Divide each side by 16.

(BA) 2 = BC • BD

Finding Segment Lengths

Use the figure to find the value of x.

Use Theorem 10.17.(BA) 2 = BC • BD

Substitute.

Simplify.25 = x 2 + 4x

Simplify.

(x + 4)x5 2

Write in standard form.0 = x 2 + 4x – 25

SOLUTION

Use Quadratic Formula.

Use the positive solution, because lengths cannot be negative.

3.39.So, x = –2 + 29

x =–4 ± 4

2 – 4(1)(–25)2

x = –2 ± 29