To apply the Pythagorean Theorem to the isosceles triangle \( \triangle ABC \) where \( AB = BC = 12 \) and \( BH \) is the altitude to the base \( AC \), we need to analyze the triangle further.
Since \( BH \) is an altitude, it divides the triangle into two right triangles: \( \triangle ABH \) and \( \triangle BCH \).
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Identify Components:
- Let \( AH = CH = x \).
- Since \( AC = AB + BC \), we have \( AC = 2x \).
- In \( \triangle ABH \), we have:
- The hypotenuse \( AB = 12 \)
- One leg \( AH = x \)
- The other leg \( BH = h \) (which we are determining for the altitude).
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Apply the Pythagorean Theorem: In \( \triangle ABH \): \[ AB^2 = AH^2 + BH^2 \] Substituting the known values: \[ 12^2 = x^2 + h^2 \] This becomes: \[ 144 = x^2 + h^2 \quad \text{(1)} \]
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Expression for \( h \): From the isosceles properties, \( h \) can also be expressed in terms of the legs. The triangle can be broken down further. The base \( AC \) will be half of \( AC = 2x \), thus: \[ \left(\frac{AC}{2}\right)^2 + h^2 = AB^2 \] In our case, this means: \[ \left(\frac{12}{2}\right)^2 + h^2 = 12^2 \] This results in: \[ 6^2 + h^2 = 12^2 \implies 36 + h^2 = 144 \implies h^2 = 108 \] So, \( h = \sqrt{108} = 6\sqrt{3} \).
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Substituting for x: The triangle property gives: \[ h^2 + x^2 = 144 \]
Putting \( h^2 = 108 \) into \( 144 = x^2 + 108 \): \[ x^2 = 144 - 108 = 36 \implies x = 6 \]
Thus, the correct response is: \[ \boxed{x = 6} \]
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