QuestionJune 18, 2025

Use the converse of the Pythagorean Theorem to determine whether the triangle with the given vertices is a right triangle. (1,4),(4,2),(6,6) a. Yes,it is a right triangle. b. No, it is not a right triangle. Please select the best answer from the choices provided A B

Use the converse of the Pythagorean Theorem to determine whether the triangle with the given vertices is a right triangle. (1,4),(4,2),(6,6) a. Yes,it is a right triangle. b. No, it is not a right triangle. Please select the best answer from the choices provided A B
Use the converse of the Pythagorean Theorem to determine whether the triangle with the given vertices is a right triangle. (1,4),(4,2),(6,6) a. Yes,it is a right triangle. b. No, it is not a right triangle. Please select the best answer from the choices provided A B

Solution
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Answer

b. No, it is not a right triangle. Explanation 1. Calculate the distances between points Use the distance formula d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} for each pair of points. - Between (1,4) and (4,2): d = \sqrt{(4-1)^2 + (2-4)^2} = \sqrt{9 + 4} = \sqrt{13} - Between (4,2) and (6,6): d = \sqrt{(6-4)^2 + (6-2)^2} = \sqrt{4 + 16} = \sqrt{20} - Between (1,4) and (6,6): d = \sqrt{(6-1)^2 + (6-4)^2} = \sqrt{25 + 4} = \sqrt{29} 2. Apply the converse of the Pythagorean Theorem Check if \sqrt{13}^2 + \sqrt{20}^2 = \sqrt{29}^2 or any permutation. - \sqrt{13}^2 + \sqrt{20}^2 = 13 + 20 = 33 - \sqrt{29}^2 = 29 Since 33 \neq 29, the triangle is not a right triangle.

Explanation

1. Calculate the distances between points<br /> Use the distance formula $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ for each pair of points.<br />- Between $(1,4)$ and $(4,2)$: $d = \sqrt{(4-1)^2 + (2-4)^2} = \sqrt{9 + 4} = \sqrt{13}$<br />- Between $(4,2)$ and $(6,6)$: $d = \sqrt{(6-4)^2 + (6-2)^2} = \sqrt{4 + 16} = \sqrt{20}$<br />- Between $(1,4)$ and $(6,6)$: $d = \sqrt{(6-1)^2 + (6-4)^2} = \sqrt{25 + 4} = \sqrt{29}$<br /><br />2. Apply the converse of the Pythagorean Theorem<br /> Check if $\sqrt{13}^2 + \sqrt{20}^2 = \sqrt{29}^2$ or any permutation.<br />- $\sqrt{13}^2 + \sqrt{20}^2 = 13 + 20 = 33$<br />- $\sqrt{29}^2 = 29$<br /><br /> Since $33 \neq 29$, the triangle is not a right triangle.
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