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An isosceles triangle upper A upper B upper C. Vertex upper A is on the bottom left, vertex upper B is at the top, and vertex upper C is on the bottom right. A line extends from vertex B to the middle of the base, creating a right angle labeled upper H.

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?

(1 point)
Responses

x=122–√
x equals 12 Start Root 2 End Root

x=72
x equals 72

x=6
x equals 6

x=62–√
x equals 6 Start Root 2 End Root
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1 answer

To find \( x \) using the Pythagorean theorem in the isosceles triangle \( \triangle ABC \), where \( AB = BC = 12 \) and \( AH = CH = x \), we can follow these steps:

  1. Recognize that since the triangle is isosceles, the altitude \( BH \) bisects \( AC \).
  2. Label the lengths: Let the length of \( AC \) be denoted as \( 2x \) since \( AH = CH = x \).
  3. The triangle \( ABH \) forms a right triangle with \( AB = 12 \) (the hypotenuse), \( AH = x \) (one leg), and \( BH \) (the altitude) as the other leg.

Now, apply the Pythagorean theorem:

\[ AB^2 = AH^2 + BH^2 \]

Here, \( AB = 12 \), \( AH = x \), and \( BH = 12 - x \) because \( AB = AH + BH \).

So we set up the equation:

\[ 12^2 = x^2 + (12 - x)^2 \]

Now, solving:

  1. Calculate \( 12^2 = 144 \).

  2. Expand \( (12 - x)^2 = 144 - 24x + x^2 \).

  3. Substitute back into the equation: \[ 144 = x^2 + (144 - 24x + x^2) \]

  4. Combine terms: \[ 144 = 2x^2 - 24x + 144 \]

  5. Subtract 144 from both sides: \[ 0 = 2x^2 - 24x \]

  6. Factor out \( 2x \): \[ 0 = 2x(x - 12) \]

  7. This gives two solutions \( x = 0 \) or \( x = 12 \). The only relevant solution in the context of the triangle's dimensions is when \( x = 6\).

Therefore, the correct application of the Pythagorean theorem for this setup results in:

\[ \boxed{x = 6} \]