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Geometry and Trigonometry / Right triangles and trigonometry Difficulty: Hard

An isosceles right triangle has a hypotenuse of length $58$ inches. What is the perimeter, in inches, of this triangle?

$29 StartRoot 2 EndRoot$

$58 StartRoot 2 EndRoot$

$58 plus 58 StartRoot 2 EndRoot$

$58 plus 116 StartRoot 2 EndRoot$

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Explanation

Choice C is correct. Since the triangle is an isosceles right triangle, the two sides that form the right angle must be the same length. Let $x$ be the length, in inches, of each of those sides. The Pythagorean theorem states that in a right triangle, $a^{2} + b^{2} = c^{2}$ , where $c$ is the length of the hypotenuse and $a$ and $b$ are the lengths of the other two sides. Substituting $x$ for $a$ , $x$ for $b$ , and $58$ for $c$ in this equation yields $x^{2} + x^{2} = 58^{2}$ , or $2 x squared equals 58 squared$ . Dividing each side of this equation by $2$ yields $x^{2} = \frac{58^{2}}{2}$ , or $x^{2} = \frac{2 \cdot 58^{2}}{4}$ . Taking the square root of each side of this equation yields two solutions: $x = \frac{58 \sqrt{2}}{2}$ and $x = - \frac{58 \sqrt{2}}{2}$ . The value of $x$ must be positive because it represents a side length. Therefore, $x = \frac{58 \sqrt{2}}{2}$ , or $x equals 29 StartRoot 2 EndRoot$ . The perimeter, in inches, of the triangle is $58 + x + x$ , or $58 plus 2 x$ . Substituting $29 StartRoot 2 EndRoot$ for $x$ in this expression gives a perimeter, in inches, of $58 plus 2 left parenthesis 29 StartRoot 2 EndRoot right parenthesis$ , or $58 plus 58 StartRoot 2 EndRoot$ .

Choice A is incorrect. This is the length, in inches, of each of the congruent sides of the triangle, not the perimeter, in inches, of the triangle.

Choice B is incorrect. This is the sum of the lengths, in inches, of the congruent sides of the triangle, not the perimeter, in inches, of the triangle.

Choice D is incorrect and may result from conceptual or calculation errors.