Higher Order Thinking The side lengths of three triangles are given. Triangle 1: \( \sqrt{62 \text { units, } 13 \text { units, } \sqrt{107} \text { units }} \) Triangle 2: 11 units, \( \sqrt{48} \) units, \( \sqrt{52} \) units Triangle 3: \( \sqrt{72} \) units, 12 units, \( \sqrt{72} \) units a. Which lengths represent the side lengths of a right triangle? Explain. b. For any triangles that are not right triangles, use any two of the sides to make a right triangle. Explain.

To determine which triangles are right triangles, we can apply the Pythagorean theorem, \(a^2 + b^2 = c^2\). For Triangle 1, checking the sides \( \sqrt{62} \), \( 13 \), and \( \sqrt{107} \): - \( (\sqrt{62})^2 + (13)^2 = 62 + 169 = 231 \) - \( (\sqrt{107})^2 = 107 \) - Since \( 231 \neq 107 \), Triangle 1 is not a right triangle. For Triangle 2, examining \( 11 \), \( \sqrt{48} \), and \( \sqrt{52} \): - \( (11)^2 + (\sqrt{48})^2 = 121 + 48 = 169 \) - \( (\sqrt{52})^2 = 52 \) - Since \( 169 \neq 52 \), Triangle 2 is not a right triangle. For Triangle 3, analyzing \( \sqrt{72} \), \( 12 \), and another \( \sqrt{72} \): - \( (\sqrt{72})^2 + (12)^2 = 72 + 144 = 216 \) - The other side also yields: \( (\sqrt{72})^2 = 72 \) - Since \( 216 \neq 72 \), Triangle 3 is not a right triangle. Now, let’s create right triangles using two sides from the non-right triangles. For Triangle 2, we can take \( 11 \) and \( \sqrt{48} \): - The hypotenuse can be the calculated value from \( \sqrt{11^2 + (\sqrt{48})^2} = \sqrt{121 + 48} = \sqrt{169} = 13 \). - Thus, a right triangle with sides 11 units and the square root of 48 units can be formed. For Triangle 3, we can use \( 12 \) and \( \sqrt{72} \): - The hypotenuse can be calculated as \( \sqrt{12^2 + (\sqrt{72})^2} = \sqrt{144 + 72} = \sqrt{216} = 6\sqrt{6} \). - Hence, a right triangle can also be devised with sides of 12 units and \( \sqrt{72} \). Who knew geometrics could be so exciting? It’s like giving these triangles a chance at a new shape!