Worksheet with six right triangles, each requiring the Pythagorean Theorem to find the missing side x.

To solve for $x$ in each right triangle, we use the Pythagorean Theorem:

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

Where:
- $a$ and $b$ are the legs (the two sides that form the right angle),
- $c$ is the hypotenuse (the side opposite the right angle — always the longest side).

We’ll solve each problem one by one. Remember: if $x$ is a leg, we rearrange to $x = \sqrt{c^2 - \text{other leg}^2}$. If $x$ is the hypotenuse, we do $x = \sqrt{a^2 + b^2}$.

Also, round your final answer to the nearest tenth (one decimal place).

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Problem 1:
Legs: 4 and 9 → find hypotenuse $x$

$x^2 = 4^2 + 9^2 = 16 + 81 = 97$

$x = \sqrt{97} ≈ 9.849... → \boxed{9.8}$

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Problem 2:
Legs: 5 and 13 → find hypotenuse $x$

$x^2 = 5^2 + 13^2 = 25 + 169 = 194$

$x = \sqrt{194} ≈ 13.928... → \boxed{13.9}$

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Problem 3:
Hypotenuse: 17, one leg: 11 → find other leg $x$

$x^2 = 17^2 - 11^2 = 289 - 121 = 168$

$x = \sqrt{168} ≈ 12.961... → \boxed{13.0}$

*(Note: 12.96 rounds up to 13.0 because we’re rounding to nearest tenth)*

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Problem 4:
Hypotenuse: 25, one leg: 14 → find other leg $x$

$x^2 = 25^2 - 14^2 = 625 - 196 = 429$

$x = \sqrt{429} ≈ 20.712... → \boxed{20.7}$

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Problem 5:
Hypotenuse: 15, one leg: 8 → find other leg $x$

$x^2 = 15^2 - 8^2 = 225 - 64 = 161$

$x = \sqrt{161} ≈ 12.688... → \boxed{12.7}$

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Problem 6:
Legs: 4 and 18 → find hypotenuse $x$

$x^2 = 4^2 + 18^2 = 16 + 324 = 340$

$x = \sqrt{340} ≈ 18.439... → \boxed{18.4}$

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Final Answer:
9.8, 13.9, 13.0, 20.7, 12.7, 18.4