Question

name: kawlee m. date: dilations and proportions quiz review 1. for each set of side lengths, determine if they would create a triangle. if so, say whether it would be acute, right, or obtuse 5, 12, 19 6, 10, 14 not a δ not a δ acute acute right right obtuse obtuse 2. solve for “x” if bf = 28 in the diagram below 2b. verify if \\(\overline{ab} \parallel \overline{cd}\\). show all work.

Explanation:

Step1: Check triangle inequality (5,12,19)

For a triangle, the sum of any two sides must be greater than the third side.
Check \( 5 + 12 \) vs \( 19 \): \( 5 + 12 = 17 \), and \( 17 < 19 \). So, \( 5, 12, 19 \) do not form a triangle (Not a \( \Delta \)).

Step2: Check triangle inequality (6,10,14)

Check \( 6 + 10 = 16 \), \( 16 > 14 \); \( 6 + 14 = 20 \), \( 20 > 10 \); \( 10 + 14 = 24 \), \( 24 > 6 \). So, it forms a triangle.

Step3: Classify triangle (6,10,14)

Use the Pythagorean theorem extension: For sides \( a \leq b \leq c \),

• If \( a^2 + b^2 = c^2 \): Right
• If \( a^2 + b^2 > c^2 \): Acute
• If \( a^2 + b^2 < c^2 \): Obtuse

Here, \( a = 6 \), \( b = 10 \), \( c = 14 \).
Calculate \( 6^2 + 10^2 = 36 + 100 = 136 \), and \( 14^2 = 196 \).
Since \( 136 < 196 \) (\( a^2 + b^2 < c^2 \)), the triangle is obtuse.

Answer:

For \( 5, 12, 19 \): Not a \( \Delta \)
For \( 6, 10, 14 \): Obtuse (and it is a triangle)