Question
page 15 of 15 Solving Right Triangles Quick Check 4 of 54 of 5 Items Question Use the image to answer the 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=62–√ x equals 6 Start Root 2 End Root x=122–√ x equals 12 Start Root 2 End Root x=6 x equals 6 x=72 x equals 72 Skip to navigation
Answers
GPT-4o mini
To find the correct application of the Pythagorean Theorem for the isosceles triangle \( \triangle ABC \), given that \( AB = BC = 12 \) and \( BH = AH = CH = x \), we can analyze the situation as follows:
1. First, recognize that dropping the altitude \( BH \) from \( B \) to \( AC \) will create two right triangles, \( \triangle ABH \) and \( \triangle BCH \).
2. Since it is an isosceles triangle, \( AC \) will be split into two equal sections, so we can denote \( AH = CH = x \) and each half of \( AC \) will be \( \frac{AC}{2} \).
3. The length of each half of the base \( AC \) can be derived from the fact that the whole base can be calculated using the Pythagorean theorem with \( AB \), \( AH \), and \( BH \).
4. The relationship can be expressed as \( (AH)^2 + (BH)^2 = (AB)^2 \).
Using our given lengths:
- \( AH = \frac{AC}{2} = 6 \) (since \( AB = 12 \)).
- \( BH = x \).
- \( AB = 12 \).
We can set up the equation as follows:
\[
x^2 + 6^2 = 12^2
\]
Now, simplifying:
\[
x^2 + 36 = 144
\]
\[
x^2 = 144 - 36
\]
\[
x^2 = 108
\]
\[
x = \sqrt{108} = \sqrt{36 \cdot 3} = 6\sqrt{3}
\]
However, the correct expression you are looking for applies the Pythagorean Theorem correctly without simplifying to numeric form, based on dimensions before solving.
From the responses you provided, while none directly represent the complete statement of the original Pythagorean theorem application, the closest one that applies is:
- \( x = 12\sqrt{2} \)
However, the appropriate setup would have shown the use of:
\[
x = \sqrt{12^2 - 6^2}
\]
If you seek the right derived expression but express it fully, it could eventually be represented as part of further simplifications leading to other forms.
Make sure to confirm if there's a misunderstanding in any dimensions, or provide additional context if required.
1. First, recognize that dropping the altitude \( BH \) from \( B \) to \( AC \) will create two right triangles, \( \triangle ABH \) and \( \triangle BCH \).
2. Since it is an isosceles triangle, \( AC \) will be split into two equal sections, so we can denote \( AH = CH = x \) and each half of \( AC \) will be \( \frac{AC}{2} \).
3. The length of each half of the base \( AC \) can be derived from the fact that the whole base can be calculated using the Pythagorean theorem with \( AB \), \( AH \), and \( BH \).
4. The relationship can be expressed as \( (AH)^2 + (BH)^2 = (AB)^2 \).
Using our given lengths:
- \( AH = \frac{AC}{2} = 6 \) (since \( AB = 12 \)).
- \( BH = x \).
- \( AB = 12 \).
We can set up the equation as follows:
\[
x^2 + 6^2 = 12^2
\]
Now, simplifying:
\[
x^2 + 36 = 144
\]
\[
x^2 = 144 - 36
\]
\[
x^2 = 108
\]
\[
x = \sqrt{108} = \sqrt{36 \cdot 3} = 6\sqrt{3}
\]
However, the correct expression you are looking for applies the Pythagorean Theorem correctly without simplifying to numeric form, based on dimensions before solving.
From the responses you provided, while none directly represent the complete statement of the original Pythagorean theorem application, the closest one that applies is:
- \( x = 12\sqrt{2} \)
However, the appropriate setup would have shown the use of:
\[
x = \sqrt{12^2 - 6^2}
\]
If you seek the right derived expression but express it fully, it could eventually be represented as part of further simplifications leading to other forms.
Make sure to confirm if there's a misunderstanding in any dimensions, or provide additional context if required.