CHORD CHORD PRODUCT THEOREM

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The products of the lengths of the line segments on each chord are equal.

Proof :

Considering the chords BC and DE,

 βˆ DFC =  βˆ BFE (vertically opposite angles)

∠CDE = βˆ CBE (angle between same arc CE)

∠DCB = βˆ DEB (angle between same arc DB)

Triangles FDC and FBE are similar, then

CF/FE = DF/BF

Doing cross multiplication, we get

CF β‹… BF = DF β‹… EF

For each figure, determine the value of the variable and the indicated lengths by applying the chord-chord product theorem.

Problem 1:

Find the following.

i) x     ii)  AD and  iii) BE

Solution :

Intersecting chord theorem,

CA βˆ™ CD = CB βˆ™ CE

2 βˆ™ 4 = x βˆ™ 8

8 = 8x

x = 1

AD = CA + CD

AD = 2 + 4

AD = 6

BE = CB + CE

= x + 8

= 1 + 8

BE = 9

Problem 2 :

Find the following.

i) y     ii)  FH and  iii) GL

Solution :

Intersecting chord theorem,

JG βˆ™ JI = JF βˆ™ JH

2.4 βˆ™ y = 3.5 βˆ™ 4.8

2.4y = 16.8

y = 16.8/2.4

y = 7

FH = JF + JH

= 3.5 + 4.8

FH = 8.3

GI = JG + JI

= 2.4 + y

= 2.4 + 7

GI = 9.4

Problem 3 :

Find the following.

i) z     ii)  PS   and  iii) TR

Solution :


Intersecting chord theorem,

QT βˆ™ QR = QP βˆ™ QS

2.4 βˆ™ z = 7 βˆ™ 2.4

2.4z = 16.8

z = 16.8/2.4

z = 7

PS = QP + QS

= 7 + 2.4

PS = 9.4

RT = QR + QT

= z + 2.4

= 7 + 2.4

RT = 9.4

Problem 4 :

Find m.

Solution :

Intersecting chord theorem,

YU βˆ™ YW = YV βˆ™ YX

m βˆ™ 4 = 3 βˆ™ 6

4m = 18

m = 18/4

m = 4.5

UW = YU + YW

= m + 4

= 4.5 + 4

UW = 8.5

VX = YV + YX

= 3 + 6

VX = 9

Problem 5 :

Find x.

Solution :

Intersecting chord theorem,

BA βˆ™ BC = BD βˆ™ BE

5 βˆ™ 12 = 10 βˆ™ x

60 = 10x

x = 60/10

x = 6

Problem 6 :

Solve for x.

Solution :

Intersecting chord theorem,

KJ βˆ™ KL = KP βˆ™ KM

(x + 10) βˆ™ (x) = (x + 1) βˆ™ (x + 8)

xΒ² + 10x = xΒ² + 9x + 8

xΒ² - xΒ² + 10x - 9x = 8

x = 8

Problem 7 :

Solve for x.

Solution :

Intersecting chord theorem,

PT βˆ™ PR = PQ βˆ™ PS

4 βˆ™ 15 = 6 βˆ™ x

60 = 6x

x = 60/6

x = 10

Problem 8 :

Solve for x.

Solution :

Intersecting chord theorem,

SX βˆ™ SZ = SW βˆ™ SY

(x + 2) βˆ™ (x + 6) = x βˆ™ (x + 12)

xΒ² + 8x + 12 = xΒ² + 12x

4x = 12

x = 12/4

x = 3

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