Question

state if the triangles are acute, obtuse, or right.

Explanation:

⚡ Using what you learned: the pythagorean theorem and its converse

Step 1: Analyze the first triangle

Identify the side lengths:

$$ a = 3, \quad b = 3\sqrt{15}, \quad c = 12 $$

Verify the longest side \( c \):

$$ (3\sqrt{15})^2 = 9 \times 15 = 135 $$

$$ 12^2 = 144 $$

Since \( 135 < 144 \), the longest side is indeed \( c = 12 \).

Compare \( a^2 + b^2 \) with \( c^2 \):

$$ a^2 + b^2 = 3^2 + (3\sqrt{15})^2 = 9 + 135 = 144 $$

$$ c^2 = 12^2 = 144 $$

Since \( a^2 + b^2 = c^2 \), the triangle is right.

Step 2: Analyze the second triangle

Identify the side lengths:

$$ a = 5, \quad b = \sqrt{26}, \quad c = 7 $$

Verify the longest side \( c \):

$$ 5^2 = 25, \quad (\sqrt{26})^2 = 26, \quad 7^2 = 49 $$

The longest side is \( c = 7 \).

Compare \( a^2 + b^2 \) with \( c^2 \):

$$ a^2 + b^2 = 5^2 + (\sqrt{26})^2 = 25 + 26 = 51 $$

$$ c^2 = 7^2 = 49 $$

Since \( a^2 + b^2 > c^2 \) (\( 51 > 49 \)), the triangle is acute.

Step 3: Analyze the third triangle

Identify the side lengths:

$$ a = 3, \quad b = \sqrt{51}, \quad c = 10 $$

Verify the longest side \( c \):

$$ 3^2 = 9, \quad (\sqrt{51})^2 = 51, \quad 10^2 = 100 $$

The longest side is \( c = 10 \).

Compare \( a^2 + b^2 \) with \( c^2 \):

$$ a^2 + b^2 = 3^2 + (\sqrt{51})^2 = 9 + 51 = 60 $$

$$ c^2 = 10^2 = 100 $$

Since \( a^2 + b^2 < c^2 \) (\( 60 < 100 \)), the triangle is obtuse.

Answer:

• First triangle: Right
• Second triangle: Acute
• Third triangle: Obtuse