determine what possible measures the third side of the triangle will be given the two sides.
24. 9 and 17.
$9 + 17 = 26$
$9 - 17 = -8$ (but length cant be negative, so $17 - 9 = 8$)
$8

Question

determine what possible measures the third side of the triangle will be given the two sides.
24. 9 and 17.
$9 + 17 = 26$
$9 - 17 = -8$ (but length cant be negative, so $17 - 9 = 8$)
$8 < x < 26$
25. 30 and 11
$30 + 11 = 41$
$30 - 11 = 19$
$19 < x < 41$
26. 4 and 7
$4 + 7 = 11$
$4 - 7 = -3$ (length cant be negative, so $7 - 4 = 3$)
$3 < x < 11$

each problems shows a triangle with one or more of its special segments. identify the special segment & find the indicated measure.
5. $\angle 1 \cong \angle 2$ & $\angle sut = 25^\circ$
special segment:
$\angle sup = $

6. $\overline{fe} = 15$, t is the midpoint of $\overline{fe}$
special segment:
$\overline{tf} = $

7. $\overline{cd} = 5$ & $\angle ade \cong \angle cde$
special segment:
$\overline{ca} = $

$\overline{xz} \perp \overline{yb}$ & $\angle ybz = (6x - 6)^\circ$
special segment:
$x = $

(image of a triangle with a segment labeled $8x - 2$ and a smaller segment labeled $2x + 1$, and two segments labeled $3z$ and $z + 22$)
special segment:
$x = $
$z = $

(image of a triangle with two marked congruent segments and angles $60^\circ$ and $x + 15^\circ$)
special segment:
$x = $

Accepted Answer

Let's solve each problem one by one:

### Problem 24: Sides 9 and 17
# Explanation:
## Step 1: Recall Triangle Inequality Theorem
The triangle inequality theorem states that the length of the third side $ x $ of a triangle with two sides $ a $ and $ b $ must satisfy $ |a - b| < x < a + b $.

## Step 2: Apply the theorem
For sides $ a = 9 $ and $ b = 17 $:
- Calculate the difference: $ |17 - 9| = 8 $
- Calculate the sum: $ 17 + 9 = 26 $
So, $ 8 < x < 26 $

# Answer:
The third side must satisfy $ 8 < x < 26 $

### Problem 25: Sides 30 and 11
# Explanation:
## Step 1: Recall Triangle Inequality Theorem
The triangle inequality theorem states that the length of the third side $ x $ of a triangle with two sides $ a $ and $ b $ must satisfy $ |a - b| < x < a + b $.

## Step 2: Apply the theorem
For sides $ a = 30 $ and $ b = 11 $:
- Calculate the difference: $ |30 - 11| = 19 $
- Calculate the sum: $ 30 + 11 = 41 $
So, $ 19 < x < 41 $

# Answer:
The third side must satisfy $ 19 < x < 41 $

### Problem 26: Sides 4 and 7
# Explanation:
## Step 1: Recall Triangle Inequality Theorem
The triangle inequality theorem states that the length of the third side $ x $ of a triangle with two sides $ a $ and $ b $ must satisfy $ |a - b| < x < a + b $.

## Step 2: Apply the theorem
For sides $ a = 4 $ and $ b = 7 $:
- Calculate the difference: $ |7 - 4| = 3 $
- Calculate the sum: $ 7 + 4 = 11 $
So, $ 3 < x < 11 $

# Answer:
The third side must satisfy $ 3 < x < 11 $

### Problem 5: $ \angle 1 \cong \angle 2 $ and $ \angle SUT = 25^\circ $
# Explanation:
Since $ \angle 1 \cong \angle 2 $, $ UP $ is the angle bisector of $ \angle SUT $.
So, $ \angle SUP = \angle PUT $
And $ \angle SUT = \angle SUP + \angle PUT = 2 \angle SUP $
Given $ \angle SUT = 25^\circ $, then $ \angle SUP = \frac{25^\circ}{2} = 12.5^\circ $

# Answer:
Special Segment: Angle Bisector
$ \angle SUP = 12.5^\circ $

### Problem 6: $ FE = 15 $, $ T $ is the midpoint of $ FE $
# Explanation:
Since $ T $ is the midpoint of $ FE $, $ FT = TE $ and $ FT = \frac{FE}{2} $
Given $ FE = 15 $, then $ FT = \frac{15}{2} = 7.5 $

# Answer:
Special Segment: Median
$ \overline{TF} = 7.5 $

### Problem 7: $ CD = 5 $ and $ \angle ADE \cong \angle CDE $
# Explanation:
Since $ \angle ADE \cong \angle CDE $, $ DE $ is the angle bisector and also the perpendicular bisector (since it's a triangle with $ DE $ as the angle bisector and likely isosceles). So, $ AD = CD $
Given $ CD = 5 $, then $ AD = 5 $, so $ CA = AD + CD = 5 + 5 = 10 $

# Answer:
Special Segment: Angle Bisector (and Perpendicular Bisector)
$ \overline{CA} = 10 $

### Problem: $ \overline{XZ} \perp \overline{YB} $ and $ \angle YBZ = (6x - 6)^\circ $
# Explanation:
Since $ \overline{XZ} \perp \overline{YB} $ and $ B $ is the midpoint (as $ XZ $ is perpendicular and $ YB $ is the altitude and median), triangle $ XYZ $ is isosceles with $ XY = YZ $. So, $ \angle YBZ = 90^\circ $ (since $ XZ \perp YB $)
Set $ 6x - 6 = 90 $
$ 6x = 96 $
$ x = 16 $

# Answer:
Special Segment: Altitude (and Median)
$ x = 16 $

### Problem: $ 8x - 2 $ and $ 2x + 1 $ (Midsegment Theorem)
# Explanation:
The midsegment of a triangle is parallel to the third side and half its length. So, $ 2x + 1 = \frac{8x - 2}{2} $
Solve:
$ 2x + 1 = 4x - 1 $
$ 1 + 1 = 4x - 2x $
$ 2 = 2x $
$ x = 1 $
For $ z $: The segments $ 3z $ and $ z + 22 $ are equal (since it's a midsegment), so $ 3z = z + 22 $
$ 3z - z = 22 $
$ 2z = 22 $
$ z = 11 $

# Answer:
Special Segment: Midsegment
$ x = 1 $, $ z = 11 $

### Problem: Angles $ 60^\circ $ and $ x + 15^\circ $
# Explanation:
Since the lines are parallel (marked with congruent segments), the triangles are similar. So, the angles are equal.
Set $ x + 15 = 60 $
$ x = 60 - 15 = 45 $

# Answer:
Special Segment: Midsegment (or Parallel Segment)
$ x = 45 $

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Problem 24: Sides 9 and 17

Step 1: Recall Triangle Inequality Theorem

Step 2: Apply the theorem

Step 1: Recall Triangle Inequality Theorem

Step 2: Apply the theorem

Step 1: Recall Triangle Inequality Theorem

Step 2: Apply the theorem

Answer:

Problem 25: Sides 30 and 11