Higher Order Thinking The side lengths of three triangles are given. Triangle 1: $$ \sqrt{62 ext { units, } 13 ext { units, } \sqrt{107} ext { units }} $$ 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.

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Higher Order Thinking The side lengths of three triangles are given. Triangle 1: $$ \sqrt{62 	ext { units, } 13 	ext { units, } \sqrt{107} 	ext { units }} $$ 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.

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**a. Determining which triangles are right triangles**

1. **Triangle 1:**  
   The side lengths are  
   $$
   \sqrt{62}, \quad 13, \quad \sqrt{107}.
   $$
   The largest number among these is $$13$$, so if Triangle 1 is a right triangle, $$13$$ must be the hypotenuse. We check using the Pythagorean theorem:
   $$
   (\sqrt{62})^2+(\sqrt{107})^2 = 62+107 = 169.
   $$
   Since
   $$
   13^2 = 169,
   $$
   Triangle 1 satisfies the Pythagorean theorem and is a right triangle.

2. **Triangle 2:**  
   The side lengths are  
   $$
   11, \quad \sqrt{48}, \quad \sqrt{52}.
   $$
   Evaluate the squares:
   $$
   11^2=121,\quad (\sqrt{48})^2=48,\quad (\sqrt{52})^2=52.
   $$
   The largest side is $$11$$ (since $$11 > \sqrt{48}$$ and $$11 > \sqrt{52}$$). Assuming $$11$$ is the hypotenuse, check:
   $$
   (\sqrt{48})^2+(\sqrt{52})^2 = 48+52 = 100,
   $$
   which is not equal to $$121$$. Therefore, Triangle 2 is not a right triangle.

3. **Triangle 3:**  
   The side lengths are  
   $$
   \sqrt{72}, \quad 12, \quad \sqrt{72}.
   $$
   Note that
   $$
   \sqrt{72} = 6\sqrt{2}.
   $$
   Here, the largest side is $$12$$, which serves as the hypotenuse if the triangle is right. Check:
   $$
   (6\sqrt{2})^2+(6\sqrt{2})^2 = 72+72 = 144,
   $$
   and
   $$
   12^2 = 144.
   $$
   Thus, Triangle 3 satisfies the Pythagorean theorem and is a right triangle.

**Answer for part (a):**  
Triangles 1 and 3 represent the side lengths of right triangles; Triangle 2 does not.

---

**b. Forming a right triangle from any two sides of a non-right triangle**

For Triangle 2, since it is not a right triangle as given, we can select any two sides and then construct a right triangle with those two sides as the legs. One natural choice is to use the sides  
$$
\sqrt{48} \quad \text{and} \quad \sqrt{52}.
$$
The hypotenuse $$c$$ of the new right triangle formed using these legs is found by:
$$
c = \sqrt{(\sqrt{48})^2 + (\sqrt{52})^2} = \sqrt{48 + 52} = \sqrt{100} = 10.
$$

**Answer for part (b):**  
Using $$\sqrt{48}$$ and $$\sqrt{52}$$ as the legs, the hypotenuse of the right triangle is $$10$$ units. Alternatively, one could choose the pair $$11$$ and $$\sqrt{48}$$ which gives:
$$
c = \sqrt{11^2+(\sqrt{48})^2} = \sqrt{121+48}=\sqrt{169}=13.
$$
However, the pair $$\sqrt{48}$$ and $$\sqrt{52}$$ produces a right triangle with a hypotenuse of $$10$$ units, which is an elegant result.
**a. Right Triangles:** - **Triangle 1:** Yes, it's a right triangle. - **Triangle 2:** No, it's not a right triangle. - **Triangle 3:** Yes, it's a right triangle. **b. Forming a Right Triangle:** - From Triangle 2, using sides $$\sqrt{48}$$ and $$\sqrt{52}$$, the hypotenuse is $$10$$ units.

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