Question

the triangle below is isosceles. find the length of side ( x ) to the nearest tenth.

Explanation:

Step1: Use Pythagorean theorem

For a right - angled triangle, \(a^{2}+b^{2}=c^{2}\). In an isosceles right - angled triangle \(a = b=\sqrt{12}\), and \(c=x\).
So, \(x^{2}=(\sqrt{12})^{2}+(\sqrt{12})^{2}\).

Step2: Simplify the equation

\(x^{2}=12 + 12\).
\(x^{2}=24\).

Step3: Solve for \(x\)

\(x=\sqrt{24}\).
\(x = 2\sqrt{6}\approx4.9\).

Answer:

\(4.9\)