Question

An isosceles triangle, △ ABC , has the measurements AB=BC=12. The segment BH is the altitude of △ ABC , and BH=AH=CH=x. Which of the following correctly applies the Pythagorean Theorem to find x?

Answer 1

In the isosceles triangle \( \triangle ABC \) with \( AB = BC = 12 \) and \( C \) being the vertex opposite the base \( AB \), we can denote \( H \) as the foot of the altitude from vertex \( B \) to base \( AC \).

Since \( H \) is the midpoint of \( AC \) (due to the properties of isosceles triangles), we can assign the following:

- \( AH = CH = x \)
- The base \( AC = AH + HC = x + x = 2x \)

To apply the Pythagorean theorem, we can consider the right triangle \( \triangle ABH \) or \( \triangle BCH \). Let's focus on \( \triangle ABH \):

1. \( AB = 12 \) (the hypotenuse)
2. \( AH = x \) (one leg)
3. \( BH \) is the other leg, which we can find using \( BH \) as the height of the triangle from vertex \( B \) to base \( AC \).

According to the properties of isosceles triangles and right triangles formed by the altitude:

In \( \triangle ABH \), where \( AH = x \) and \( BH \) is the altitude, we have:
\[
AB^2 = AH^2 + BH^2
\]
Substituting the known values:
\[
12^2 = x^2 + BH^2
\]

To find \( BH \), we note that \( BH = x \) (as given in the problem). Hence we rewrite:
\[
12^2 = x^2 + x^2
\]
This simplifies to:
\[
144 = 2x^2
\]
Thus, we can solve for \( x \):
\[
2x^2 = 144 \implies x^2 = 72 \implies x = \sqrt{72} = 6\sqrt{2}
\]

Thus, the correct application of the Pythagorean theorem in this isosceles triangle is:
\[
12^2 = x^2 + x^2 \quad \text{(where \( BH = x \))}
\]