To determine which triangles are right triangles, we can apply the Pythagorean theorem. For a triangle with sides of lengths \(a\), \(b\), and \(c\) (where \(c\) is the longest side), it is a right triangle if \(a^2 + b^2 = c^2\).
We will calculate the side lengths for the triangles using the given areas of the squares, and then check if they satisfy the Pythagorean theorem.
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Triangle with areas 6, 10, 8:
- Side lengths: \( \sqrt{6} \), \( \sqrt{10} \), \( \sqrt{8} \)
- Check: \[ \sqrt{6}^2 + \sqrt{8}^2 = 6 + 8 = 14 \] \[ \sqrt{10}^2 = 10 \]
- Since \(14 \neq 10\), this is not a right triangle.
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Triangle with areas 9, 25, 16:
- Side lengths: \( \sqrt{9} \), \( \sqrt{25} \), \( \sqrt{16} \)
- Check: \[ 3^2 + 4^2 = 9 + 16 = 25 \] \[ 5^2 = 25 \]
- Since \(25 = 25\), this is a right triangle.
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Triangle with areas 25, 169, 144:
- Side lengths: \( \sqrt{25} \), \( \sqrt{169} \), \( \sqrt{144} \)
- Check: \[ 5^2 + 12^2 = 25 + 144 = 169 \] \[ 13^2 = 169 \]
- Since \(169 = 169\), this is a right triangle.
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Triangle with areas 7, 25, 24:
- Side lengths: \( \sqrt{7} \), \( \sqrt{25} \), \( \sqrt{24} \)
- Check: \[ \sqrt{7}^2 + \sqrt{24}^2 = 7 + 24 = 31 \] \[ \sqrt{25}^2 = 25 \]
- Since \(31 \neq 25\), this is not a right triangle.
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Triangle with areas 64, 225, 289:
- Side lengths: \( \sqrt{64} \), \( \sqrt{225} \), \( \sqrt{289} \)
- Check: \[ 8^2 + 15^2 = 64 + 225 = 289 \] \[ 17^2 = 289 \]
- Since \(289 = 289\), this is a right triangle.
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Triangle with areas 10, 26, 24:
- Side lengths: \( \sqrt{10} \), \( \sqrt{26} \), \( \sqrt{24} \)
- Check: \[ \sqrt{10}^2 + \sqrt{24}^2 = 10 + 24 = 34 \] \[ \sqrt{26}^2 = 26 \]
- Since \(34 \neq 26\), this is not a right triangle.
Thus, the triangles that are right triangles are:
- The triangle with areas 9, 25, and 16.
- The triangle with areas 25, 169, and 144.
- The triangle with areas 64, 225, and 289.