Question

for the following right triangle, find the side length x. round your answer to the nearest hundredth.

Explanation:

Step1: Apply Pythagorean theorem

In a right - triangle, \(a^{2}+b^{2}=c^{2}\), where \(c\) is the hypotenuse. Here, \(c = 10\), \(a=8\), and \(b = x\). So, \(x^{2}+8^{2}=10^{2}\).

Step2: Simplify the equation

\(x^{2}+64 = 100\). Then, \(x^{2}=100 - 64\).

Step3: Calculate \(x^{2}\)

\(x^{2}=36\).

Step4: Solve for \(x\)

\(x=\sqrt{36}\) (since \(x>0\) as it is a side length of a triangle). But wait, no, we made a mistake. Wait, no, hold on. Wait, actually, if \(x\) is one of the legs (assuming the hypotenuse is \(10\)), then \(x=\sqrt{10^{2}-8^{2}}=\sqrt{100 - 64}=\sqrt{36}\) is wrong. Wait, no, no: \(x=\sqrt{10^{2}-8^{2}}=\sqrt{100 - 64}=\sqrt{36}\) is wrong. Wait, no, \(10^{2}-8^{2}=100 - 64 = 36\), \(x=\sqrt{36}=6\). But wait, let's check again.
Wait, formula: \(x=\sqrt{c^{2}-a^{2}}\), where \(c = 10\), \(a = 8\).
\(x=\sqrt{10^{2}-8^{2}}=\sqrt{(10 + 8)(10 - 8)}=\sqrt{18\times2}=\sqrt{36}=6.00\)

Answer:

\(6.00\)