Here are the step-by-step solutions for finding the missing side lengths in these special right triangles.
1) 45°-45°-90° Triangle
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Rule: In a 45-45-90 triangle, the legs are equal ($a = b$), and the hypotenuse is the leg times $\sqrt{2}$ ($hypotenuse = leg \cdot \sqrt{2}$).
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Given: One leg is $2\sqrt{2}$.
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Find $b$: Since the legs are equal, $b = 2\sqrt{2}$.
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Find $a$ (hypotenuse): Multiply the leg by $\sqrt{2}$.
$$a = 2\sqrt{2} \cdot \sqrt{2} = 2 \cdot 2 = 4$$
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Answer: $a = 4$, $b = 2\sqrt{2}$
2) 45°-45°-90° Triangle
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Rule: Same as above. Hypotenuse = leg $\cdot \sqrt{2}$.
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Given: The hypotenuse is 4.
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Find $x$ and $y$ (legs): To find the leg from the hypotenuse, divide by $\sqrt{2}$.
$$leg = \frac{4}{\sqrt{2}}$$
Rationalize the denominator:
$$\frac{4}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$$
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Answer: $x = 2\sqrt{2}$, $y = 2\sqrt{2}$
3) 45°-45°-90° Triangle
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Rule: Legs are equal; Hypotenuse = leg $\cdot \sqrt{2}$.
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Given: One leg is $\frac{2\sqrt{2}}{2}$, which simplifies to just $\sqrt{2}$.
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Find $y$ (other leg): It is equal to the given leg. So, $y = \sqrt{2}$.
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Find $x$ (hypotenuse): Multiply the leg by $\sqrt{2}$.
$$x = \sqrt{2} \cdot \sqrt{2} = 2$$
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Answer: $x = 2$, $y = \sqrt{2}$
4) 45°-45°-90° Triangle
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Rule: Legs are equal; Hypotenuse = leg $\cdot \sqrt{2}$.
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Given: One leg is $3\sqrt{2}$.
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Find $y$ (other leg): It is equal to the given leg. So, $y = 3\sqrt{2}$.
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Find $x$ (hypotenuse): Multiply the leg by $\sqrt{2}$.
$$x = 3\sqrt{2} \cdot \sqrt{2} = 3 \cdot 2 = 6$$
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Answer: $x = 6$, $y = 3\sqrt{2}$
5) 45°-45°-90° Triangle
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Rule: Legs are equal; Hypotenuse = leg $\cdot \sqrt{2}$.
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Given: The hypotenuse is 6.
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Find $x$ and $y$ (legs): Divide the hypotenuse by $\sqrt{2}$.
$$leg = \frac{6}{\sqrt{2}}$$
Rationalize the denominator:
$$\frac{6}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{6\sqrt{2}}{2} = 3\sqrt{2}$$
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Answer: $x = 3\sqrt{2}$, $y = 3\sqrt{2}$
6) 45°-45°-90° Triangle
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Rule: Legs are equal; Hypotenuse = leg $\cdot \sqrt{2}$.
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Given: The hypotenuse is $2\sqrt{6}$.
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Find $x$ and $y$ (legs): Divide the hypotenuse by $\sqrt{2}$.
$$leg = \frac{2\sqrt{6}}{\sqrt{2}} = 2\sqrt{\frac{6}{2}} = 2\sqrt{3}$$
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Answer: $x = 2\sqrt{3}$, $y = 2\sqrt{3}$
7) 30°-60°-90° Triangle
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Rule:
* Side opposite 30° (shortest) = $s$
* Side opposite 60° (longer leg) = $s\sqrt{3}$
* Side opposite 90° (hypotenuse) = $2s$
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Given: The side opposite the 60° angle is 16.
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Find $s$ (side opposite 30°, labeled $x$):
$$s\sqrt{3} = 16 \rightarrow s = \frac{16}{\sqrt{3}}$$
Rationalize:
$$x = \frac{16\sqrt{3}}{3}$$
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Find hypotenuse (labeled $y$): Multiply $s$ by 2.
$$y = 2 \cdot \frac{16\sqrt{3}}{3} = \frac{32\sqrt{3}}{3}$$
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Answer: $x = \frac{16\sqrt{3}}{3}$, $y = \frac{32\sqrt{3}}{3}$
8) 30°-60°-90° Triangle
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Rule: Same ratios as #7.
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Given: The side opposite the 30° angle is 2. This is our base value $s$.
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Find $v$ (side opposite 60°): Multiply $s$ by $\sqrt{3}$.
$$v = 2\sqrt{3}$$
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Find $u$ (hypotenuse): Multiply $s$ by 2.
$$u = 2 \cdot 2 = 4$$
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Answer: $u = 4$, $v = 2\sqrt{3}$
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Final Answer:
1) $a = 4, b = 2\sqrt{2}$
2) $x = 2\sqrt{2}, y = 2\sqrt{2}$
3) $x = 2, y = \sqrt{2}$
4) $x = 6, y = 3\sqrt{2}$
5) $x = 3\sqrt{2}, y = 3\sqrt{2}$
6) $x = 2\sqrt{3}, y = 2\sqrt{3}$
7) $x = \frac{16\sqrt{3}}{3}, y = \frac{32\sqrt{3}}{3}$
8) $u = 4, v = 2\sqrt{3}$
Parent Tip: Review the logic above to help your child master the concept of special right triangles worksheet answer key with work.
