To classify the triangle using the side lengths \( a = 10 \), \( b = 16 \), and \( c = \sqrt{239} \) (assuming \( c \) is the longest side), we will use the properties of triangles and the Pythagorean theorem.
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Check if it's a triangle: We need to ensure that it satisfies the triangle inequality:
- \( a + b > c \)
- \( a + c > b \)
- \( b + c > a \)
Calculating using \( c = \sqrt{239} \):
- Is \( 10 + 16 > \sqrt{239} \)?
- \( 26 > \sqrt{239} \) → \( 26^2 > 239 \) → \( 676 > 239 \) → True.
- Is \( 10 + \sqrt{239} > 16 \)?
- We need \( \sqrt{239} > 6 \).
- \( 239 > 36 \) → True (since \( 15.5^2 = 240.25 \)).
- Is \( 16 + \sqrt{239} > 10 \)?
- Always true since both numbers are positive.
So, it can be formed as a triangle.
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Classify the triangle: To classify whether it's acute, right, or obtuse, we will compare the squares of the sides.
We need to check the following:
- If \( a^2 + b^2 > c^2 \), the triangle is acute.
- If \( a^2 + b^2 = c^2 \), the triangle is right.
- If \( a^2 + b^2 < c^2 \), the triangle is obtuse.
First, calculate \( a^2 \), \( b^2 \), and \( c^2 \):
- \( a^2 = 10^2 = 100 \)
- \( b^2 = 16^2 = 256 \)
- \( c^2 = (\sqrt{239})^2 = 239 \)
Now compute \( a^2 + b^2 \): \[ a^2 + b^2 = 100 + 256 = 356 \]
Now compare \( a^2 + b^2 \) with \( c^2 \): \[ 356 > 239 \]
Since \( a^2 + b^2 > c^2 \), the triangle is classified as acute.
So the answer is: These side lengths form an acute triangle.
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