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# 9.6: Segments of Chords, Secants, and Tangents

Difficulty Level: At Grade Created by: CK-12

## Learning Objectives

• Find the lengths of segments within circles.

## Review Queue

1. What do you know about mDAC\begin{align*}m\angle DAC\end{align*} and mDBC\begin{align*}m\angle DBC\end{align*}? Why?
2. What do you know about mAED\begin{align*}m\angle AED\end{align*} and mBEC\begin{align*}m\angle BEC\end{align*}? Why?
3. Is AEDBEC\begin{align*}\triangle AED \sim \triangle BEC\end{align*}? How do you know?
4. If AE=8, ED=7,\begin{align*}AE = 8, \ ED = 7,\end{align*} and BE=6\begin{align*}BE = 6\end{align*}, find EC\begin{align*}EC\end{align*}.

Know What? At a particular time during its orbit, 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 we see that Jakarta, Indonesia is collinear. If the distance from the moon to Jakarta is 240,128 miles, what is the distance from Cape Horn 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 in each triangle proportional.

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

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

ab=cd\begin{align*}ab = cd\end{align*}

Example 1: Find x\begin{align*}x\end{align*} in each diagram below.

a)

b)

Solution: Use the ratio from Theorem 9-14.

a) 128=10x96=10x9.6=x\begin{align*}12 \cdot 8=10 \cdot x\!\\ 96 = 10x\!\\ 9.6 = x\end{align*}

b) x15=5915x=45x=3\begin{align*}x \cdot 15=5 \cdot 9\!\\ 15x=45\!\\ x=3\end{align*}

Example 2: Algebra Connection Solve for x\begin{align*}x\end{align*}.

a)

b)

Solution: Use Theorem 9-13.

a) 824=(3x+1)12192=36x+12180=36x5=x\begin{align*}8 \cdot 24=(3x+1)\cdot 12\!\\ 192=36x+12\!\\ 180=36x\!\\ 5=x\end{align*}

b) (x5)21=(x9)2421x105=24x216111=3x37=x\begin{align*}(x-5)21=(x-9)24\!\\ 21x-105=24x-216\!\\ 111=3x\!\\ 37=x \end{align*}

## 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.

Theorem 9-15: If two secants are drawn from a common point outside a circle and the segments are labeled as below, then a(a+b)=c(c+d)\begin{align*}a(a+b)=c(c+d)\end{align*}.

The product of the outer segment and the whole of one secant equals the product of the outer segment and the whole of the other secant.

a(a+b)=c(c+d)\begin{align*}a(a+b)=c(c+d)\end{align*}

Example 3: Find the value of the missing variable.

a)

b)

Solution: Use Theorem 9-15 to set up an equation.

a) 18(18+x)=16(16+24)324+18x=256+38418x=316x=1759\begin{align*}18 \cdot (18+x)=16 \cdot (16+24)\!\\ 324+18x=256+384\!\\ 18x=316\!\\ x=17\frac{5}{9}\end{align*}

b) x(x+x)=9322x2=288x2=144x=12, x12 (length is not negative)\begin{align*}x \cdot (x+x)=9 \cdot 32\!\\ 2x^2=288\!\\ x^2=144\!\\ x =12, \ x\ne -12 \ (\text{length is not negative})\end{align*}

## 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.

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 below), then a2=b(b+c)\begin{align*}a^2=b(b+c)\end{align*}.

The product of the outside segment of the secant and the whole is equal to the square of the tangent.

a2=b(b+c)\begin{align*}a^2=b(b+c)\end{align*}

Example 4: Find the value of the missing segment.

a)

b)

Solution: Use Theorem 9-16.

a) x2=4(4+12)x2=416=64x=8\begin{align*}x^2=4(4+12)\!\\ x^2=4 \cdot 16 = 64\!\\ x=8\end{align*}

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

When you have to factor a quadratic equation to find an answer, always eliminate the negative answer because length is never negative.

Example 5: Ishmael found a broken piece of a CD in his car. He places a ruler across two points on the rim, and the length of the chord is 9.5 cm. The distance from the midpoint of this chord to the nearest point on the rim is 1.75 cm. Find the diameter of the CD.

Solution: Think of this as two chords intersecting each other. If we were to extend the 1.75 cm segment, it would be a diameter. So, if we find \begin{align*}x\end{align*}, in the diagram to the left, and add it to 1.75 cm, we would find the diameter.

\begin{align*}4.25 \cdot 4.25&=1.75\cdot x\\ 18.0625&=1.75x\\ x & \approx 10.3 \ cm,\ \text{making the diameter} \ 12 \ cm, \ \text{which is the}\\ & \qquad \qquad \qquad \text{actual diameter of a CD.}\end{align*}

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

\begin{align*}238857\cdot 251308&=240128(240128+x)\\ 60026674956&=57661456380+240128x\\ 2365218572&=240128x\\ x & \approx 9849.8 \ miles\end{align*}

## Review Questions

• Questions 1-25 are similar to Examples 1, 3, and 4.
• Questions 26-28 are similar to Example 2.
• Questions 29 is similar to Example 5.
• Questions 30 and 31 are proofs of Theorem 9-14 and 9-15.

Fill in the blanks for each problem below. Then, solve for the missing segment.

\begin{align*}\underline{\;\;\;\;\;\;} \cdot 4=\underline{\;\;\;\;\;\;\;} \cdot x\end{align*}

\begin{align*}3(\underline{\;\;\;\;\;\;\;}+\underline{\;\;\;\;\;\;\;})=2(2+7)\end{align*}

\begin{align*}20x=\underline{\;\;\;\;\;\;\;}\end{align*}

\begin{align*}x \cdot \underline{\;\;\;\;\;\;\;}=8(\underline{\;\;\;\;\;\;\;}+\underline{\;\;\;\;\;\;\;})\end{align*}

\begin{align*}\underline{\;\;\;\;\;\;\;}=\underline{\;\;\;\;\;\;\;}(4+5)\end{align*}

\begin{align*}10^2=x(\underline{\;\;\;\;\;\;\;}+\underline{\;\;\;\;\;\;\;})\end{align*}

Find \begin{align*}x\end{align*} in each diagram below. Simplify any radicals.

1. Error Analysis Describe and correct the error in finding \begin{align*}y\end{align*}. \begin{align*}10 \cdot 10&=y\cdot 15y\\ 100&=15y^2\\ \frac{20}{3}&=y^2\\ \frac{2\sqrt{15}}{3}&=y \quad {\color{red}\longleftarrow \ y} \ {\color{red}\text{in \underline{not} correct}}\end{align*}

Algebra Connection Find the value of \begin{align*}x\end{align*}.

1. 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 6 inches. The distance from the midpoint of this chord to the nearest point on the rim is 1 inch. Find the diameter of the plate.
2. Fill in the blanks of the proof of Theorem 9-14. Given: Intersecting chords \begin{align*}\overline{AC}\end{align*} and \begin{align*}\overline{BE}\end{align*}. Prove: \begin{align*}ab=cd\end{align*}
Statement Reason
1. Intersecting chords \begin{align*}\overline{AC}\end{align*} and \begin{align*}\overline{BE}\end{align*} with segments \begin{align*}a, \ b, \ c,\end{align*} and \begin{align*}d\end{align*}.
2. Theorem 9-8
3. \begin{align*}\triangle ADE \sim \triangle BDC\end{align*}
4. Corresponding parts of similar triangles are proportional
5. \begin{align*}ab=cd\end{align*}
1. Fill in the blanks of the proof of Theorem 9-15. Given: Secants \begin{align*}\overline{PR}\end{align*} and \begin{align*}\overline{RT}\end{align*} Prove: \begin{align*}a(a+b)=c(c+d)\end{align*}
Statement Reason
1. Secants \begin{align*}\overline{PR}\end{align*} and \begin{align*}\overline{RT}\end{align*} with segments \begin{align*}a, \ b, \ c,\end{align*} and \begin{align*}d\end{align*}. given
2. \begin{align*}\angle R \cong \angle R\end{align*} Reflexive PoC
3. \begin{align*}\angle QPS \cong \angle STQ\end{align*} Theorem 9-8
4. \begin{align*}\triangle RPS \sim \triangle RTQ\end{align*} AA Similarity Postulate
5. \begin{align*}\frac{a}{c+d}=\frac{c}{a+b}\end{align*} Corresponding parts of similar triangles are proportional
6. \begin{align*}a(a+b)=c(c+d)\end{align*} Cross multiplication

1. \begin{align*}m \angle DAC = m \angle DBC\end{align*} by Theorem 9-8, they are inscribed angles and intercept the same arc.
2. \begin{align*}m \angle AED = m \angle BEC\end{align*} by the Vertical Angles Theorem.
3. Yes, by AA Similarity Postulate.
4. \begin{align*}{\;} \qquad \ \ \frac{8}{6} = \frac{7}{EC}\!\\ {\;} \ \ 8 \cdot EC = 42\!\\ {\;} \quad \ \ EC = \frac{21}{4} = 5.25\end{align*}

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