Question

find the length of side ( x ) to the nearest tenth.
(there is a right - angled isosceles triangle in the figure, with two acute angles of ( 45^{circ} ) each, one of the right - angled sides is ( sqrt{2} ), and the hypotenuse is ( x ))
answer attempt 1 out of 2
( x = ) (input box)
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Explanation:

Step1: Identify triangle type

This is a right isosceles triangle (two 45° angles, right angle). The given leg is $\sqrt{2}$.

Step2: Use Pythagorean theorem

For right triangle: $x^2 = (\sqrt{2})^2 + (\sqrt{2})^2$

Step3: Calculate right-hand side

$x^2 = 2 + 2 = 4$

Step4: Solve for x

$x = \sqrt{4} = 2$
(Alternatively, use sine: $\sin(45^\circ)=\frac{\sqrt{2}}{x}$, so $x=\frac{\sqrt{2}}{\sin(45^\circ)}=\frac{\sqrt{2}}{\frac{\sqrt{2}}{2}}=2$)

Answer:

$x = 2.0$