Solved Name: Unit 1: Geometry Basics Date: Per: Homework 6 ... - Free Printable

Let's solve each problem step by step.

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Problem 11: Find the value of $ x $.

We have two angles formed by intersecting lines:
- One angle is $ (6x + 7)^\circ $
- The other is $ (8x - 42)^\circ $

These are vertical angles (opposite angles formed by intersecting lines), so they are equal:

$$
6x + 7 = 8x - 42
$$

Solve for $ x $:

$$
6x + 7 = 8x - 42 \\
7 + 42 = 8x - 6x \\
49 = 2x \\
x = \frac{49}{2} = 24.5
$$

✔ Answer: $ x = 24.5 $

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Problem 12: Find the value of $ x $.

Two angles form a straight line (linear pair), so their sum is $ 180^\circ $:

- $ (11x - 15)^\circ $
- $ (5x - 13)^\circ $

$$
(11x - 15) + (5x - 13) = 180 \\
16x - 28 = 180 \\
16x = 208 \\
x = \frac{208}{16} = 13
$$

✔ Answer: $ x = 13 $

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Problem 13: If $ \overline{BD} \perp \overline{AC} $, $ m\angle DBE = (2x - 1)^\circ $, and $ m\angle CBE = (3x - 42)^\circ $, find $ x $.

Since $ BD \perp AC $, the angle between them is $ 90^\circ $. So $ \angle DBC = 90^\circ $.

But $ \angle DBE $ and $ \angle CBE $ are adjacent angles that together make $ \angle DBC $, so:

$$
\angle DBE + \angle CBE = 90^\circ \\
(2x - 1) + (3x - 42) = 90 \\
5x - 43 = 90 \\
5x = 133 \\
x = \frac{133}{5} = 26.6
$$

Wait — let’s double-check the diagram interpretation.

Looking at the figure:
- Point $ B $ is the vertex.
- $ BD \perp AC $, so $ \angle ABC = 90^\circ $
- $ E $ is a point such that $ \angle DBE $ and $ \angle CBE $ are parts of the right angle.

So yes, $ \angle DBE + \angle CBE = 90^\circ $

So:

$$
(2x - 1) + (3x - 42) = 90 \\
5x - 43 = 90 \\
5x = 133 \\
x = 26.6
$$

But let’s check if this makes sense:

- $ \angle DBE = 2(26.6) - 1 = 53.2 - 1 = 52.2^\circ $
- $ \angle CBE = 3(26.6) - 42 = 79.8 - 42 = 37.8^\circ $
- Sum: $ 52.2 + 37.8 = 90^\circ $ ✔

✔ Answer: $ x = 26.6 $ or $ \frac{133}{5} $

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Problem 14: Find the value of $ x $ if $ \overrightarrow{QS} $ bisects $ \angle PQR $ and $ m\angle PQR = 82^\circ $.

If $ QS $ bisects $ \angle PQR $, then it divides the angle into two equal parts.

Each part is:

$$
\frac{82^\circ}{2} = 41^\circ
$$

The angle given is $ (10x + 1)^\circ $, which is one of the two equal parts.

So:

$$
10x + 1 = 41 \\
10x = 40 \\
x = 4
$$

✔ Answer: $ x = 4 $

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Problem 15: Find the values of $ x $ and $ y $.

Three angles meet at a point, forming a straight line? Or around a point?

From the diagram:
- Angles: $ (18y + 5)^\circ $, $ (10x - 61)^\circ $, $ (x + 10)^\circ $
- They appear to be adjacent angles forming a straight line (sum to $ 180^\circ $)

Assuming they lie on a straight line:

$$
(18y + 5) + (10x - 61) + (x + 10) = 180 \\
18y + 5 + 10x - 61 + x + 10 = 180 \\
18y + 11x - 46 = 180 \\
18y + 11x = 226 \quad \text{(Equation 1)}
$$

But we need another equation. Wait — maybe there’s more.

Alternatively, perhaps the middle angle is vertical to something?

Wait — looking again: the three angles are likely adjacent on a straight line, but only two expressions are given: $ (10x - 61) $ and $ (x + 10) $, and $ (18y + 5) $.

Wait — maybe $ (18y + 5) $ and $ (x + 10) $ are vertical angles? Or perhaps $ (10x - 61) $ is the middle angle?

Actually, from the diagram (based on standard problems), it looks like:

- $ (18y + 5)^\circ $ and $ (x + 10)^\circ $ are vertical angles, so equal.
- $ (10x - 61)^\circ $ is adjacent to both.

Wait — no, vertical angles would be opposite.

But in the diagram, likely:
- $ (18y + 5)^\circ $ and $ (x + 10)^\circ $ are vertical angles, so:

$$
18y + 5 = x + 10 \quad \text{(Equation 1)}
$$

And the adjacent angles form a linear pair with $ (10x - 61)^\circ $, so:

Suppose $ (10x - 61)^\circ $ and $ (x + 10)^\circ $ are supplementary:

$$
(10x - 61) + (x + 10) = 180 \\
11x - 51 = 180 \\
11x = 231 \\
x = 21
$$

Now plug into Equation 1:

$$
18y + 5 = x + 10 = 21 + 10 = 31 \\
18y = 26 \\
y = \frac{26}{18} = \frac{13}{9} \approx 1.44
$$

But let's verify:

- $ x = 21 $
- $ x + 10 = 31^\circ $
- $ 18y + 5 = 31 \Rightarrow y = \frac{26}{18} = \frac{13}{9} $
- $ 10x - 61 = 210 - 61 = 149^\circ $
- Now check: $ 149^\circ + 31^\circ = 180^\circ $? Yes! ✔

Also, vertical angles: $ 18y + 5 = 31^\circ $, $ x + 10 = 31^\circ $ → equal ✔

✔ Answer: $ x = 21 $, $ y = \frac{13}{9} $

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Problem 16: Find the values of $ x $ and $ y $.

We have a right triangle or perpendicular lines.

From the diagram:
- A right angle is shown at the intersection.
- Angles: $ (2y + 5)^\circ $, $ (5x - 17)^\circ $, $ (3x - 11)^\circ $

Likely, $ (2y + 5)^\circ $ is one angle, and $ (5x - 17)^\circ $ and $ (3x - 11)^\circ $ are adjacent angles forming a right angle.

Wait — since there’s a right angle symbol, the two adjacent angles must add to $ 90^\circ $.

Possibility:
- $ (5x - 17) $ and $ (3x - 11) $ are complementary (add to $ 90^\circ $)

So:

$$
(5x - 17) + (3x - 11) = 90 \\
8x - 28 = 90 \\
8x = 118 \\
x = \frac{118}{8} = 14.75
$$

Then:
- $ 5x - 17 = 5(14.75) - 17 = 73.75 - 17 = 56.75^\circ $
- $ 3x - 11 = 44.25 - 11 = 33.25^\circ $
- Sum: $ 56.75 + 33.25 = 90^\circ $ ✔

Now, $ (2y + 5)^\circ $ is the other angle, possibly vertical to one of them?

But wait — the right angle is formed by the two lines, and $ (2y + 5)^\circ $ is likely adjacent to one of the angles.

Wait — actually, in many such diagrams, $ (2y + 5)^\circ $ is equal to $ (3x - 11)^\circ $ or $ (5x - 17)^\circ $ as vertical angles?

But without exact diagram, assume $ (2y + 5)^\circ $ is vertical to $ (5x - 17)^\circ $, so equal:

$$
2y + 5 = 5x - 17
$$

Plug $ x = 14.75 $:

$$
2y + 5 = 5(14.75) - 17 = 73.75 - 17 = 56.75 \\
2y = 51.75 \\
y = 25.875
$$

But this seems messy. Let’s see if there’s a better interpretation.

Wait — perhaps $ (2y + 5)^\circ $ is adjacent to $ (5x - 17)^\circ $, and together they form $ 90^\circ $? But we already used the two angles.

Alternative idea: The angle $ (2y + 5)^\circ $ might be the same as $ (5x - 17)^\circ $, or $ (3x - 11)^\circ $, depending on labeling.

But more likely: $ (2y + 5)^\circ $ and $ (3x - 11)^\circ $ are vertical angles, so:

$$
2y + 5 = 3x - 11
$$

We already found $ x = 14.75 $, so:

$$
2y + 5 = 3(14.75) - 11 = 44.25 - 11 = 33.25 \\
2y = 28.25 \\
y = 14.125
$$

Still fractional.

But maybe I made a mistake.

Wait — perhaps the right angle is between $ (2y + 5)^\circ $ and $ (5x - 17)^\circ $? That is, they are adjacent and form $ 90^\circ $?

Then:

$$
(2y + 5) + (5x - 17) = 90 \quad \text{(Equation 1)}
$$

And the other angle $ (3x - 11)^\circ $ is vertical to $ (2y + 5)^\circ $, so:

$$
3x - 11 = 2y + 5 \quad \text{(Equation 2)}
$$

Now solve:

From Equation 2:
$$
3x - 11 = 2y + 5 \\
2y = 3x - 16 \quad \text{(A)}
$$

Substitute into Equation 1:

$$
(2y + 5) + (5x - 17) = 90 \\
(3x - 16 + 5) + 5x - 17 = 90 \\
(3x - 11) + 5x - 17 = 90 \\
8x - 28 = 90 \\
8x = 118 \\
x = 14.75
$$

Then from (A): $ 2y = 3(14.75) - 16 = 44.25 - 16 = 28.25 \Rightarrow y = 14.125 $

Same as before.

But maybe the problem expects integers.

Wait — could $ (2y + 5) $ and $ (3x - 11) $ be the two angles forming the right angle?

Then:

$$
(2y + 5) + (3x - 11) = 90 \\
2y + 3x - 6 = 90 \\
2y + 3x = 96 \quad \text{(1)}
$$

And $ (5x - 17) $ is vertical to $ (2y + 5) $, so:

$$
5x - 17 = 2y + 5 \\
5x - 2y = 22 \quad \text{(2)}
$$

Now solve system:

From (1): $ 2y = 96 - 3x $

Plug into (2):

$$
5x - (96 - 3x) = 22 \\
5x - 96 + 3x = 22 \\
8x = 118 \\
x = 14.75
$$

Again same result.

So likely correct.

But let’s write as fractions:

$ x = \frac{118}{8} = \frac{59}{4} = 14.75 $

$ y = \frac{28.25}{2} = \frac{113}{8} = 14.125 $

So:

✔ Answer: $ x = \frac{59}{4} $, $ y = \frac{113}{8} $

Or decimals: $ x = 14.75 $, $ y = 14.125 $

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Problem 17: If $ \overrightarrow{NP} $ bisects $ \angle MNQ $, $ m\angle MNQ = (8x + 12)^\circ $, $ m\angle PNQ = 78^\circ $, and $ m\angle RNM = (3y - 9)^\circ $, find $ x $ and $ y $.

Given:
- $ NP $ bisects $ \angle MNQ $
- So $ \angle MNP = \angle PNQ $
- $ \angle PNQ = 78^\circ $, so $ \angle MNP = 78^\circ $
- Therefore, total $ \angle MNQ = 78 + 78 = 156^\circ $

But $ \angle MNQ = (8x + 12)^\circ $, so:

$$
8x + 12 = 156 \\
8x = 144 \\
x = 18
$$

Now, $ \angle RNM = (3y - 9)^\circ $

Note: $ \angle RNM $ and $ \angle MNQ $ are adjacent angles forming a straight line? Or vertical?

Looking at the diagram: Points $ R, N, M $ and $ O, N, Q $

Likely, $ \angle RNM $ and $ \angle MNQ $ are adjacent angles forming a straight line at point $ N $, so they are supplementary:

$$
\angle RNM + \angle MNQ = 180^\circ \\
(3y - 9) + 156 = 180 \\
3y - 9 = 24 \\
3y = 33 \\
y = 11
$$

✔ Answer: $ x = 18 $, $ y = 11 $

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✔ Final Answers:

| Problem | Answer |
|--------|--------|
| 11 | $ x = 24.5 $ |
| 12 | $ x = 13 $ |
| 13 | $ x = 26.6 $ or $ \frac{133}{5} $ |
| 14 | $ x = 4 $ |
| 15 | $ x = 21 $, $ y = \frac{13}{9} $ |
| 16 | $ x = \frac{59}{4} = 14.75 $, $ y = \frac{113}{8} = 14.125 $ |
| 17 | $ x = 18 $, $ y = 11 $ |

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