1Cademy - Finding Equal Leg Lengths in a Right Triangle with a 10-Inch Hypotenuse

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Finding Equal Leg Lengths in a Right Triangle with a 10-Inch Hypotenuse

Apply the Pythagorean Theorem to a real-world situation in which both legs of a right triangle are equal, producing a quadratic equation that is solved using the Square Root Property and requires radical simplification.

Problem: Kelvin is building a gazebo and places a $10$ -inch piece of wood diagonally in each corner as a brace. If the brace's ends are the same distance from the corner, how far from the corner should each end be fastened? Round to the nearest tenth of an inch.

Read: A diagonal brace forms the hypotenuse of a right triangle whose two legs are equal.
Identify: The distance from the corner (the length of each leg).
Name: Let $x$ = the distance from the corner. Because both legs are equal, each leg has length $x$ .
Translate: Substitute into the Pythagorean Theorem:

$a^2 + b^2 = c^2$

$x^2 + x^2 = 10^2$

Solve: Combine the like terms on the left: $2x^2 = 100$ . Divide both sides by $2$ to isolate the squared variable: $x^2 = 50$ . Apply the Square Root Property: $x = \pm\sqrt{50}$ . Simplify the radical by extracting the largest perfect square factor: $\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}$ , so $x = \pm 5\sqrt{2}$ . Since $x$ represents a physical length, the negative solution $x = -5\sqrt{2}$ does not make sense and is discarded. Therefore $x = 5\sqrt{2} \approx 7.1$ .
Check: Verify: $(7.1)^2 + (7.1)^2 = 50.41 + 50.41 = 100.82 \approx 10^2$ $\checkmark$
Answer: Kelvin should fasten each piece of wood approximately $7.1$ inches from the corner.

This problem introduces several elements beyond simpler Pythagorean Theorem examples: (1) both legs share the same variable, so $x^2 + x^2$ combines into $2x^2$ , requiring division before the square root step; (2) applying the Square Root Property produces two solutions ( $\pm 5\sqrt{2}$ ), but the negative value must be discarded because a physical length cannot be negative; and (3) because $50$ is not a perfect square, $\sqrt{50}$ must be simplified to $5\sqrt{2}$ using the Product Property of Square Roots, and the result is then approximated to the nearest tenth.

Updated 2026-04-21

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