<meta http-equiv="refresh" content="1; url=/nojavascript/"> Segments of Chords, Secants, and Tangents | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: CK-12 Geometry - Second Edition Go to the latest version.

9.6: Segments of Chords, Secants, and Tangents

Created by: CK-12
0  0  0

Learning Objectives

• Find the lengths of segments associated with circles.

Review Queue

1. What can you say about $m \angle DAC$ and $m \angle DBC$? What theorem do you use?
2. What do you know about $m \angle AED$ and $m \angle BEC$? Why?
3. Is $\triangle AED \sim \triangle BEC$? How do you know?
4. If $AE = 8, ED = 7$, and $BE = 6$, find $EC$.
5. If $\overline{AD}$ and $\overline{BC}$ are not in the circle, would the ratios from #4 still be valid?

Know What? As you know, the moon orbits the earth. At a particular time, the moon is 238,857 miles from Beijing, China. On the same line, Yukon is 12,451 miles from Beijing. Drawing another line from the moon to Cape Horn (the southernmost point of South America), we see that Jakarta, Indonesia is collinear. If the distance from Cape Horn to Jakarta is 9849 miles, what is the distance from the moon to Jakarta?

Segments from Chords

In the Review Queue above, we have two chords that intersect inside a circle. The two triangles are similar, making the sides of each triangle in proportion with each other. If we remove $\overline{AD}$ and $\overline{BC}$ the ratios between $\overline{AE}, \overline{EC}, \overline{DE}$, and $\overline{EB}$ will still be the same. This leads us to our first theorem.

Theorem 9-14: If two chords intersect inside a circle so that one is divided into segments of length $a$ and $b$ and the other into segments of length $c$ and $d$ then $ab = cd$.

The product of the segments of one chord is equal to the product of segments of the second chord.

Example 1: Find $x$ in each diagram below.

a)

b)

Solution: Use the ratio from Theorem 9-13. The product of the segments of one chord is equal to the product of the segments of the other.

a) $12 \cdot 8=10 \cdot x\!\\96=10x\!\\ 9.6=x$

b) $x \cdot 15=5 \cdot 9\!\\15x=45\!\\x=3$

Example 2: Algebra Connection Solve for $x$.

a)

b)

Solution: Again, we can use Theorem 9-13. Set up an equation and solve for $x$.

a) $8 \cdot 24=(3x+1) \cdot 12\!\\192=36x+12\!\\180=36x\!\\5=x$

b) $32 \cdot 21=(x-9)(x-13)\!\\672=x^2-22x+117\!\\0=x^2-22x-555\!\\0=(x-37)(x+15)\!\\x=37, -15$

However, $x \neq -15$ because length cannot be negative, so $x=37$.

Segments from Secants

In addition to forming an angle outside of a circle, the circle can divide the secants into segments that are proportional with each other.

If we draw in the intersecting chords, we will have two similar triangles.

From the inscribed angles and the Reflexive Property $( \angle R \cong \angle R), \triangle PRS \sim \triangle TRQ$.

Because the two triangles are similar, we can set up a proportion between the corresponding sides. Then, cross-multiply. $\frac{a}{c+d}=\frac{c}{a+b} \Rightarrow a(a+b)=c(c+d)$

Theorem 9-15: If two secants are drawn from a common point outside a circle and the segments are labeled as above, then $a(a+b)=c(c+d)$.

In other words, the product of the outer segment and the whole of one secant is equal to the product of the outer segment and the whole of the other secant.

Example 3: Find the value of the missing variable.

a)

b)

Solution: Use Theorem 9-15 to set up an equation. For both secants, you multiply the outer portion of the secant by the whole.

a) $18 \cdot (18+x)=16 \cdot (16+24)\!\\324+18x=256+384\!\\18x=316\!\\ x=17 \frac{5}{9}$

b) $x \cdot (x+x)=9 \cdot 32\!\\2x^2=288\!\\x^2=144\!\\x=12$

$x \neq -12$ because length cannot be negative.

Segments from Secants and Tangents

If a tangent and secant meet at a common point outside a circle, the segments created have a similar relationship to that of two secant rays in Example 3. Recall that the product of the outer portion of a secant and the whole is equal to the same of the other secant. If one of these segments is a tangent, it will still be the product of the outer portion and the whole. However, for a tangent line, the outer portion and the whole are equal.

Theorem 9-16: If a tangent and a secant are drawn from a common point outside the circle (and the segments are labeled like the picture to the left), then $a^2=b(b+c)$.

This means that the product of the outside segment of the secant and the whole is equal to the square of the tangent segment.

Example 4: Find the value of the missing segment.

a)

b)

Solution: Use Theorem 9-16. Square the tangent and set it equal to the outer part times the whole secant.

a) $x^2=4(4+12)\!\\x^2=4 \cdot 16=64\!\\x=8$

b) $20^2=y(y+30)\!\\400=y^2+30y\!\\0=y^2+30y-400\!\\0=(y+40)(y-10)\!\\y=\xcancel{-40},10$

When you have to factor a quadratic equation to find an answer, always eliminate the negative answer(s). Length is never negative.

Know What? Revisited The given information is to the left. Let’s set up an equation using Theorem 9-15.

$238857 \cdot 251308 &= x \cdot (x+9849)\\60026674956 &= x^2+9849x\\0 &= x^2+9849x-60026674956\\Use \ the \ Quadratic \ Formula \ x & \approx \frac{-9849 \pm \sqrt{9849^2-4(-60026674956)}}{2}\\x & \approx 240128.4 \ miles$

Review Questions

Find $x$ in each diagram below. Simplify any radicals.

1. Error Analysis Describe and correct the error in finding $y$. $10 \cdot 10 &= y \cdot 15y\\100 &= 15y^2\\\frac{20}{3} &= y^2\\\frac{2\sqrt{15}}{3} &= y \quad {\color{red}\leftarrow y \ \text{is \underline{not} correct}}$
2. Suzie found a piece of a broken plate. She places a ruler across two points on the rim, and the length of the chord is found to be 6 inches. The distance from the midpoint of this chord to the nearest point on the rim is found to be 1 inch. Find the diameter of the plate.

Algebra Connection For problems 21-30, solve for $x$.

1. Find $x$ and $y$.

1. $m \angle DAC = m \angle DBC$ by Theorem 9-8, they are inscribed angles and intercept the same arc.
2. $m \angle AED = m \angle BEC$ by the Vertical Angles Theorem.
3. Yes, by AA Similarity Postulate.
4. $\frac{8}{6} = \frac{7}{EC}\!\\8 \cdot EC = 42\!\\{\;} \quad EC = \frac{21}{4} = 5.25$
5. Yes, the $EC$ would be the same and the ratio would still be valid.

Feb 22, 2012

Aug 22, 2014