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? (1 point) Responses x=72 x equals 72 x=122–√ x equals 12 Start Root 2 End Root x=6 x equals 6 x=62–√
Answers
Answered by
GPT-4o mini
Given an isosceles triangle \( \triangle ABC \) with \( AB = BC = 12 \) and \( BH \) being the altitude, we can denote the length of segment \( AC \) as \( 2x \) because \( H \) is the midpoint of \( AC \).
Using the Pythagorean Theorem in triangle \( ABH \):
\[
AB^2 = AH^2 + BH^2
\]
This can be written in terms of \( x \):
\[
12^2 = x^2 + x^2
\]
This simplifies to:
\[
144 = 2x^2
\]
Now, solving for \( x^2 \):
\[
x^2 = \frac{144}{2} = 72
\]
Taking the square root gives:
\[
x = \sqrt{72} = 6\sqrt{2}
\]
Thus, the correct application of the Pythagorean theorem leading to the value of \( x \) is:
\[
x = 6\sqrt{2}
\]
So the answer from the choices provided is:
\[
\text{x} = 6\sqrt{2}
\]
Using the Pythagorean Theorem in triangle \( ABH \):
\[
AB^2 = AH^2 + BH^2
\]
This can be written in terms of \( x \):
\[
12^2 = x^2 + x^2
\]
This simplifies to:
\[
144 = 2x^2
\]
Now, solving for \( x^2 \):
\[
x^2 = \frac{144}{2} = 72
\]
Taking the square root gives:
\[
x = \sqrt{72} = 6\sqrt{2}
\]
Thus, the correct application of the Pythagorean theorem leading to the value of \( x \) is:
\[
x = 6\sqrt{2}
\]
So the answer from the choices provided is:
\[
\text{x} = 6\sqrt{2}
\]