Question

an isosceles right triangle has leg lengths of 4 centimeters. what is the length of the altitude drawn from the right angle to the hypotenuse?

\\(2\text{ cm}\\)
\\(2\sqrt{2}\text{ cm}\\)
\\(4\text{ cm}\\)
\\(4\sqrt{2}\text{ cm}\\)

Explanation:

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<reasoning>

Calculate the hypotenuse length

\[

$$\begin{aligned} a &= 4\text{ cm},\quad b = 4\text{ cm}\\ c &= \sqrt{a^2 + b^2} = \sqrt{4^2 + 4^2} = \sqrt{32} = 4\sqrt{2}\text{ cm} \end{aligned}$$

\]

Calculate the altitude to the hypotenuse

\[

$$\begin{aligned} \text{Area} &= \frac{1}{2} \cdot a \cdot b = \frac{1}{2} \cdot 4 \cdot 4 = 8\text{ cm}^2\\ \text{Area} &= \frac{1}{2} \cdot c \cdot h \implies 8 = \frac{1}{2} \cdot 4\sqrt{2} \cdot h\\ h &= \frac{16}{4\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}\text{ cm} \end{aligned}$$

\]
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<answer>
<mcq-option>\(2\text{ cm}\)</mcq-option>
<mcq-correct>\(2\sqrt{2}\text{ cm}\)</mcq-correct>
<mcq-option>\(4\text{ cm}\)</mcq-option>
<mcq-option>\(4\sqrt{2}\text{ cm}\)</mcq-option>
</answer>

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Answer:

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<reasoning>

Calculate the hypotenuse length

\[

$$\begin{aligned} a &= 4\text{ cm},\quad b = 4\text{ cm}\\ c &= \sqrt{a^2 + b^2} = \sqrt{4^2 + 4^2} = \sqrt{32} = 4\sqrt{2}\text{ cm} \end{aligned}$$

\]

Calculate the altitude to the hypotenuse

\[

$$\begin{aligned} \text{Area} &= \frac{1}{2} \cdot a \cdot b = \frac{1}{2} \cdot 4 \cdot 4 = 8\text{ cm}^2\\ \text{Area} &= \frac{1}{2} \cdot c \cdot h \implies 8 = \frac{1}{2} \cdot 4\sqrt{2} \cdot h\\ h &= \frac{16}{4\sqrt{2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}\text{ cm} \end{aligned}$$

\]
</reasoning>

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<mcq-option>\(2\text{ cm}\)</mcq-option>
<mcq-correct>\(2\sqrt{2}\text{ cm}\)</mcq-correct>
<mcq-option>\(4\text{ cm}\)</mcq-option>
<mcq-option>\(4\sqrt{2}\text{ cm}\)</mcq-option>
</answer>

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