To find the correct application of the Pythagorean Theorem for triangle \( \triangle ABC \), we note that \( AB = BC = 12 \) and \( BH \) is the altitude from \( B \) to \( AC \), with \( AH = CH = x \).
Since \( \triangle ABC \) is isosceles, \( AC = AH + CH = x + x = 2x \).
Now applying the Pythagorean Theorem in triangle \( \triangle ABH \), we have:
\[ AB^2 = AH^2 + BH^2 \]
Plugging in the values:
\[ 12^2 = x^2 + BH^2 \]
Since \( BH = x \):
\[ 12^2 = x^2 + x^2 \]
This becomes:
\[ 144 = 2x^2 \]
Dividing by 2 gives:
\[ 72 = x^2 \]
Thus:
\[ x = \sqrt{72} = 6\sqrt{2} \]
Therefore, the correct response is:
x = 6√2 (x equals 6 Start Root 2 End Root).
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