Question

ch 1 & 2 practice test
name: _
geometry
distance formula
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
midpoint formula
$(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2})$

1. give an example of each from the diagram
2. where do plane p and line bm intersect?
3. name plane p in 3 different ways.
4. give an example of each from the diagram at right
5. find the measure of $
6. find the measure of $m
7. · a is the midpoint of segment bc. if $ab = 3x + 6$ and $ac = 2x + 14$ find x and bc.

Explanation:

Question 5

Step1: Use complementary angles property

Complementary angles sum to \(90^{\circ}\). Let \(\angle x\) and \(50^{\circ}\) be complementary. So \(x + 50=90\).

Step2: Solve for \(x\)

Subtract \(50\) from both sides: \(x=90 - 50\).

Step1: Use supplementary angles property

\(\angle AFE\) and \(50^{\circ}\) are supplementary (they form a linear - pair). Supplementary angles sum to \(180^{\circ}\). So \(m\angle AFE+50 = 180\).

Step2: Solve for \(m\angle AFE\)

Subtract \(50\) from both sides: \(m\angle AFE=180 - 50\).

Step1: Use mid - point property

Since \(A\) is the mid - point of \(BC\), \(AB = AC\). So \(3x + 6=2x + 14\).

Step2: Solve for \(x\)

Subtract \(2x\) from both sides: \(3x-2x + 6=2x-2x + 14\), which gives \(x+6 = 14\). Then subtract \(6\) from both sides: \(x=14 - 6=8\).

Step3: Find \(AB\) and \(AC\)

Substitute \(x = 8\) into \(AB\): \(AB=3\times8 + 6=24 + 6=30\). Since \(AB = AC\), \(AC = 30\).

Step4: Find \(BC\)

\(BC=AB + AC\). So \(BC=30+30\).

Answer:

\(x = 40^{\circ}\)

Question 6