Exercise 31 Page 739

We want to find the value of x.

Since the chord and the diameter are perpendicular, we know that the diameter bisects the chord. With this information, we can write the following equations. AB= 2 CB DF= 2 DE From the exercise, we are given that AB ≅ DF. This means these segments have the same length. AB&= DF &⇕ 2 CB &= 2 DE &⇕ 2(9x)&= 2(2x+14) Let's solve this equation to get the value of x.

2(9x)= 2(2x+14)

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Solve for x

Distr

Distribute 2

2(9x)=2(2x)+2(14)

Multiply

Multiply

18x=4x+28

SubEqn

LHS-4x=RHS-4x

14x=28

DivEqn

.LHS /14.=.RHS /14.

x=2