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=122–√ x equals 12 Start Root 2 End Root x=62–√ x equals 6 Start Root 2 End Root x=72 x equals 72 x=6
Answers
Answered by
GPT-4o mini
In the isosceles triangle \( \triangle ABC \), we have \( AB = BC = 12 \) and the altitude \( BH \) divides the triangle into two right triangles \( \triangle ABH \) and \( \triangle BCH \).
Since \( BH \) is the altitude, it also bisects \( AC \) at point \( H \); thus, \( AH = HC \). Let \( AH = HC = x \). Consequently, we have:
\[
AC = AH + HC = x + x = 2x.
\]
In \( \triangle ABH \), we can apply the Pythagorean Theorem as follows:
\[
AB^2 = AH^2 + BH^2.
\]
Substituting the known values:
\[
12^2 = x^2 + x^2,
\]
\[
144 = 2x^2.
\]
To solve for \( x^2 \):
\[
x^2 = \frac{144}{2} = 72.
\]
Thus,
\[
x = \sqrt{72} = 6\sqrt{2}.
\]
The suitable application of the Pythagorean theorem to find \( x \) is:
\[
x = 6\sqrt{2}.
\]
So, the correct response is:
**x = 6√2**.
Since \( BH \) is the altitude, it also bisects \( AC \) at point \( H \); thus, \( AH = HC \). Let \( AH = HC = x \). Consequently, we have:
\[
AC = AH + HC = x + x = 2x.
\]
In \( \triangle ABH \), we can apply the Pythagorean Theorem as follows:
\[
AB^2 = AH^2 + BH^2.
\]
Substituting the known values:
\[
12^2 = x^2 + x^2,
\]
\[
144 = 2x^2.
\]
To solve for \( x^2 \):
\[
x^2 = \frac{144}{2} = 72.
\]
Thus,
\[
x = \sqrt{72} = 6\sqrt{2}.
\]
The suitable application of the Pythagorean theorem to find \( x \) is:
\[
x = 6\sqrt{2}.
\]
So, the correct response is:
**x = 6√2**.