station 13: solve for x.
station 14: which of the following angles has the smallest measure?
station 15: what is the area of the smallest square?
station 16: determine the height of the triangle. round to the nearest hundredth.

Question

Accepted Answer

### Station 13: Solve for $ x $
# Explanation:
## Step 1: Recall Triangle Angle Sum
The sum of angles in a triangle is $ 180^\circ $. So, $ 45^\circ + (8x - 7)^\circ + (3x + 8)^\circ = 180^\circ $.
## Step 2: Combine Like Terms
Simplify the left - hand side: $ 45 + 8x - 7+3x + 8 = 180 $. Combine the $ x $ terms: $ (8x+3x)=11x $, and the constant terms: $ 45 - 7 + 8=46 $. So the equation becomes $ 11x + 46 = 180 $.
## Step 3: Solve for $ x $
Subtract 46 from both sides: $ 11x=180 - 46=134 $. Then divide both sides by 11: $ x=\frac{134}{11}\approx12.18 $ (if we consider the angle expressions correctly, maybe there was a typo, but following the steps: if the angles are $ 45^\circ $, $ (8x - 7)^\circ $ and $ (3x + 8)^\circ $)

### Station 15: Area of the smallest square
# Explanation:
## Step 1: Recall Pythagorean Theorem (for squares on sides of right triangle)
In a right - triangle, if we have three squares drawn on the sides (two legs and the hypotenuse), the sum of the areas of the squares on the legs is equal to the area of the square on the hypotenuse. Let the areas of the squares be $ A_1 = 16\space cm^2 $, $ A_2=\text{area of smallest square} $, and $ A_3 = 25\space cm^2 $.
## Step 2: Apply the Relationship
If the square with area $ 25\space cm^2 $ is on the hypotenuse, and the square with area $ 16\space cm^2 $ is on one leg, then $ A_1+A_2 = A_3 $ (wait, no: actually, if the right - triangle has legs $ a,b $ and hypotenuse $ c $, then $ a^{2}+b^{2}=c^{2} $, where $ a^{2} $ is the area of the square on $ a $, $ b^{2} $ on $ b $, and $ c^{2} $ on $ c $). So if $ c^{2}=25 $ and $ a^{2}=16 $, then $ b^{2}=c^{2}-a^{2} $ (assuming the square with area 25 is on the hypotenuse and 16 on one leg). So $ b^{2}=25 - 16 = 9\space cm^2 $. So the area of the smallest square is $ 9\space cm^2 $.

### Station 16: Height of the triangle
# Explanation:
## Step 1: Identify the Triangle Type
The triangle is isosceles (two sides are equal, as indicated by the marks). The base is $ 13\space m $, and the equal sides are $ 15\space m $. When we draw the height $ h $ from the top vertex to the base, it bisects the base. So the base of each right - triangle formed is $ \frac{13}{2}=6.5\space m $, and the hypotenuse is $ 15\space m $.
## Step 2: Apply Pythagorean Theorem
Using the Pythagorean theorem $ h^{2}+(6.5)^{2}=15^{2} $. So $ h^{2}=15^{2}-6.5^{2}=225 - 42.25 = 182.75 $. Then $ h=\sqrt{182.75}\approx13.52\space m $

### Station 14: Smallest angle measure
# Explanation:
## Step 1: Recall Triangle Angle Sum and Exterior Angles
First, find the interior angles of the triangle. The exterior angle is equal to the sum of the two non - adjacent interior angles. The exterior angles given are $ 105^\circ $ and $ 75^\circ $. The interior angle adjacent to $ 105^\circ $ is $ 180 - 105=75^\circ $, the interior angle adjacent to $ 75^\circ $ is $ 180 - 75 = 105^\circ $. Then the third interior angle $ d $ is $ 180-(75 + 105)=0 $? No, wait, the triangle's interior angles: let's assume the triangle has angles $ a,b,c $ (interior) and exterior angles. The sum of interior angles is $ 180^\circ $. If we have two exterior angles as $ 105^\circ $ and $ 75^\circ $, the adjacent interior angles are $ 75^\circ $ and $ 105^\circ $, which is impossible because $ 75 + 105=180 $, so the third angle would be $ 0 $, which is wrong. Maybe the exterior angles are $ 105^\circ $ and $ 75^\circ $ and we need to find the interior angles. Wait, the angles at the vertices: angle $ a $ is supplementary to $ 105^\circ $ (if $ 105^\circ $ is an exterior angle), so $ a = 180 - 105 = 75^\circ $, angle $ c $ is supplementary to $ 75^\circ $, so $ c=180 - 75 = 105^\circ $, and angle $ d $ (interior) is $ 180-(75 + 105)=0 $ (which is wrong). Maybe the triangle has angles: let's look at the diagram again. If the triangle has an exterior angle of $ 105^\circ $ (so interior angle $ 75^\circ $) and another exterior angle of $ 75^\circ $ (interior angle $ 105^\circ $), then the third interior angle is $ 180-(75 + 105)=0 $, which is incorrect. Maybe the angles are: the triangle has angles, and we need to find which of $ a,b,c $ (the angles at the vertices) is the smallest. Let's assume the triangle has interior angles: if one exterior angle is $ 105^\circ $, the interior angle is $ 75^\circ $, another exterior angle is $ 75^\circ $, interior angle is $ 105^\circ $, then the third interior angle is $ 0 $, which is wrong. Maybe the angles are: angle $ a $ is supplementary to $ 105^\circ $, so $ a = 75^\circ $, angle $ c $ is supplementary to $ 75^\circ $, so $ c = 105^\circ $, and angle $ d $ (the third interior angle) is $ 180-(75 + 105)=0 $ (no). Maybe the diagram is a triangle with two exterior angles marked as $ 105^\circ $ and $ 75^\circ $ on the base, so the interior angles at the base are $ 75^\circ $ and $ 105^\circ $, and the vertex angle $ d $ is $ 180-(75 + 105)=0 $ (wrong). Maybe there is a typo, but if we consider the angles: the smallest angle among the options (assuming $ a,b,c $ are the angles) - if $ a $ is supplementary to $ 105^\circ $ (so $ a = 75^\circ $), $ c $ is supplementary to $ 75^\circ $ (so $ c = 105^\circ $), and the interior angle at the top $ d=180-(75 + 105)=0 $ (no). Alternatively, if the triangle has angles: the exterior angle $ a $ is equal to the sum of the two non - adjacent interior angles. If the two non - adjacent interior angles are $ 75^\circ $ and the vertex angle, then $ a=75^\circ+\text{vertex angle} $, $ b = 180 - \text{vertex angle}-75^\circ $, $ c = 180 - 75^\circ=105^\circ $. This is getting too confusing without a clear diagram, but assuming the standard problem: in a triangle, the smallest angle is opposite the shortest side. If we can find the angles, the smallest angle will be the one with the smallest measure.

# Answers:
- Station 13: $ x=\frac{134}{11}\approx12.18 $ (or if there was a typo in angle expressions, re - check. If the angles were $ 45^\circ $, $ (8x + 7)^\circ $ and $ (3x - 8)^\circ $, then $ 45+8x + 7+3x - 8 = 180\Rightarrow11x + 44 = 180\Rightarrow11x = 136\Rightarrow x=\frac{136}{11}\approx12.36 $)
- Station 15: $ 9\space cm^2 $
- Station 16: $ \approx13.52\space m $
- Station 14: (Assuming the angles: if the interior angles are $ 75^\circ $, $ 105^\circ $ and $ 0^\circ $ (wrong), maybe the correct approach is: the sum of exterior angles of a triangle is $ 360^\circ $, but for a triangle, each exterior angle is supplementary to an interior angle. If two exterior angles are $ 105^\circ $ and $ 75^\circ $, the third exterior angle is $ 360-(105 + 75)=180^\circ $ (wrong). So there must be a mis - interpretation. If the triangle has angles: the two given angles on the base are $ 105^\circ $ and $ 75^\circ $ as interior angles, then the third angle is $ 0^\circ $ (impossible). So likely a typo, but if we assume the triangle has angles: the exterior angle $ a = 180 - 105 = 75^\circ $, $ c=180 - 75 = 105^\circ $, and the interior angle at the top $ d = 180-(75 + 105)=0 $ (no). So maybe the correct problem is that the triangle has angles: the two non - adjacent interior angles to exterior angle $ a $ are $ 75^\circ $ and $ 30^\circ $, then $ a = 105^\circ $, $ b = 180 - 105 = 75^\circ $, $ c = 105^\circ $, and the smallest angle is $ 30^\circ $ (but this is guesswork).

(Note: The answers are provided based on the best - effort analysis of the problems. The clarity of the diagram would help in more accurate solutions.)

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QUESTION IMAGE

Question

Explanation:

Station 13: Solve for \( x \)

Step 1: Recall Triangle Angle Sum

Step 2: Combine Like Terms

Step 3: Solve for \( x \)

Station 15: Area of the smallest square

Step 1: Recall Pythagorean Theorem (for squares on sides of right triangle)

Step 2: Apply the Relationship

Station 16: Height of the triangle

Step 1: Identify the Triangle Type

Step 2: Apply Pythagorean Theorem

Station 14: Smallest angle measure

Answer: