Question

pre - calculus friqz #2: pythagorean thm, parallel & perpendicular lines name: zadin casanova find the length of each missing side. 1. 10 cm 20 cm x cm state x in radical form (ex. √57): __ cm state x as a decimal approximation: cm state x in simplified radical form (ex. 3√15): cm 2. 40 cm 24 cm __ cm

Explanation:

Step1: Apply Pythagorean theorem for first triangle

According to the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), where \(c\) is the hypotenuse and \(a\) and \(b\) are the legs of the right - triangle. For the first triangle with legs \(a = 10\) and \(b=x\) and hypotenuse \(c = 20\), we have \(10^{2}+x^{2}=20^{2}\).

Step2: Solve for \(x\) in the first triangle

\[

$$\begin{align*} x^{2}&=20^{2}-10^{2}\\ x^{2}&=400 - 100\\ x^{2}&=300\\ x&=\sqrt{300} \end{align*}$$

\]

Step3: Simplify the radical for the first triangle

\(x=\sqrt{300}=\sqrt{100\times3}=10\sqrt{3}\)

Step4: Find the decimal approximation for the first triangle

\(x = 10\sqrt{3}\approx10\times1.732 = 17.32\)

Step5: Apply Pythagorean theorem for second triangle

For the second triangle with legs \(a = 24\) and \(b\) (unknown) and hypotenuse \(c = 40\), we have \(b^{2}=40^{2}-24^{2}\)

Step6: Solve for the missing side in the second triangle

\[

$$\begin{align*} b^{2}&=40^{2}-24^{2}\\ b^{2}&=(40 + 24)(40 - 24)\\ b^{2}&=64\times16\\ b^{2}&=1024\\ b&=\sqrt{1024}=32 \end{align*}$$

\]

Answer:

For the first triangle:
State \(x\) in radical form: \(\sqrt{300}\text{ cm}\)
State \(x\) as a decimal approximation: \(17.32\text{ cm}\)
State \(x\) in simplified radical form: \(10\sqrt{3}\text{ cm}\)
For the second triangle: \(32\text{ cm}\)