Question

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

X= 6 square root 2
X=12 square root 2
X=6
X=72

Answer 1

To solve for the length \( x \) (the altitude \( BH \) and the segments \( AH \) and \( CH \) in the isosceles triangle \( ABC \)), we can apply the Pythagorean theorem.

In triangle \( ABC \):
- \( AB = BC = 12 \) (the two equal sides)
- \( AH = CH = x \) (the segments from the vertex to the foot of the altitude at base \( AC \))
- \( BH \) is the altitude from vertex \( B \) to base \( AC \), so \( BH = x \).

We can denote \( A \) and \( C \) as points on a horizontal line at positions \( A \) and \( C \) such that \( AC = 2 \times AH = 2x \).

Now, since \( H \) (the foot of the altitude) is where \( BH \) meets \( AC \), and given \( AH \) and \( CH \) are equal, we know:
\[
AC = AH + CH = x + x = 2x.
\]

The horizontal distance from \( B \) to line \( AC \) forms a right triangle \( ABH \) where:
- \( AB = 12 \) (the hypotenuse)
- \( AH = x \) (one leg)
- \( BH = x \) (the other leg).

Using the Pythagorean theorem:
\[
AB^2 = AH^2 + BH^2.
\]
Substituting the known values:
\[
12^2 = x^2 + x^2.
\]
This simplifies to:
\[
144 = 2x^2.
\]
Now, dividing both sides by 2:
\[
72 = x^2.
\]
Taking the square root:
\[
x = \sqrt{72}.
\]
Simplifying \( \sqrt{72} \):
\[
x = \sqrt{36 \times 2} = 6\sqrt{2}.
\]

Thus, the correct answer is:
\[
\boxed{6\sqrt{2}}.
\]