Question

use the figure below for problem 7 below and 8 – 10 on the back. $overline{ec}$ bisects $overline{ad}$ at c. find the value of “x” and the measure of the indicated segment.

1. $ac = 3x + 6$ and $cd = 2x + 14$

$x = \underline{\quad\quad}$
$ac = \underline{\quad\quad}$

1. $ac = 5x - 8$ and $cd = 16 - 3x$

$x = \underline{\quad\quad}$
$ad = \underline{\quad\quad}$

1. $ad = 6x - 4$ and $ac = 4x - 3$

$x = \underline{\quad\quad}$
$cd = \underline{\quad\quad}$

1. $ac = 3x - 1$ and $ad = 12 - x$

$x = \underline{\quad\quad}$
$cd = \underline{\quad\quad}$

Explanation:

Problem 7

Step1: Since EC bisects AD, AC = CD.

Set \( 3x + 6 = 2x + 14 \)

Step2: Solve for x. Subtract 2x and 6 from both sides.

\( 3x - 2x = 14 - 6 \)
\( x = 8 \)

Step3: Find AC by substituting x = 8 into AC = 3x + 6.

\( AC = 3(8) + 6 = 24 + 6 = 30 \)

Step1: EC bisects AD, so AC = CD. Set \( 5x - 8 = 16 - 3x \)

Step2: Solve for x. Add 3x and 8 to both sides.

\( 5x + 3x = 16 + 8 \)
\( 8x = 24 \)
\( x = 3 \)

Step3: Find AC: \( 5(3) - 8 = 15 - 8 = 7 \). AD = AC + CD = 7 + 7 = 14 (or AD = 2*AC = 14)

Step1: EC bisects AD, so AC = CD and AD = 2*AC. Thus, \( 6x - 4 = 2(4x - 3) \)

Step2: Expand and solve for x.

\( 6x - 4 = 8x - 6 \)
\( -4 + 6 = 8x - 6x \)
\( 2 = 2x \)
\( x = 1 \)

Step3: Find CD. AC = 4(1) - 3 = 1, so CD = AC = 1 (or CD = AD - AC = (6(1)-4) - 1 = 2 - 1 = 1)

Answer:

x = 8, AC = 30

Problem 8