SEGMENT SEGMENT PRODUCT THEOREM

Find x.

Problem 1:

Solution :

Using theorem,

JH ∙ JG = JK ∙ JL

8 ∙ (8 + x) = 6 ∙ (6 + 10)

64 + 8x = 96

8x = 32

x = 32/8

x = 4

Problem 2 :

Solution :

Using theorem,

RT ∙ RU = RS ∙ RV

x ∙ (x + 9) = 4 ∙ (4 + 5)

x² + 9x = 36

x² + 9x - 36 = 0

(x + 12) (x - 3) = 0

x = -12 or x = 3

We can use only positive value for x. because lengths cannot be negative.

So, we have

x = 3

Problem 3 :

Solution :

Using theorem,

TW ∙ TX = TY ∙ TZ

7 ∙ (7 + 12) = 6 ∙ (6 + x)

133 = 36 + 6x

6x = 97

x = 97/6

x ≈ 16

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

Problem 4 :

Solution :

DC ∙ DB = DE ∙ DF

5 ∙ (5 + 4.5) = 5 ∙ (5 + x)

47.5 = 25 + 5x

5x = 47.5 - 25

5x = 22.5

x = 4.5

Problem 5 :

Solution :

Using theorem,

GI ∙ GK = GH ∙ GJ

6 ∙ (6 + y) = 5 ∙ (5 + 16)

36 + 6y = 105

6y = 105 - 36

6y = 69

y = 11.5

Problem 6 :

Solution :

Using theorem,

SR ∙ SQ = ST ∙ SU

14 ∙ (14 + 4) = 9 ∙ (9 + z)

252 = 81 + 9z

9z = 252 - 81

9z = 171

z = 19

Problem 7:

Solution :

Using theorem,

CD ∙ CE = CG ∙ CF

12 ∙ (12 + n) = 9 ∙ (9 + 18)

144 + 12n = 243

12n = 243 - 144

12n = 99

n = 8.25