Question
Given is the right triangle ABC with altitude CD. Apply the Pythagorean Theorem and similar triangles as needed to find the length of AD with length 6 at CD and length 8 at DB
Answers
GPT-4o mini
To find the length of segment \(AD\) in right triangle \(ABC\) with altitude \(CD\), we can use the Pythagorean theorem and properties of similar triangles.
1. **Label the triangle and variables**:
- Let \(AB = c\) (the hypotenuse)
- \(AC = b\)
- \(BC = a\)
- The right angle is at \(C\)
- \(AD = x\)
- \(DB = 8\)
- \(CD = 6\)
Since \(D\) is the foot of the altitude from \(C\) to hypotenuse \(AB\), we have:
\[
AD + DB = AB \implies x + 8 = c
\]
2. **Using the property of similar triangles**:
Triangles \(CDA\) and \(CDB\) are similar to triangle \(CAB\) (both being right triangles). According to the similarity:
- For triangle \(CDA\):
\[
\frac{CD}{AC} = \frac{AD}{AB} \implies \frac{6}{b} = \frac{x}{c}
\]
- For triangle \(CDB\):
\[
\frac{CD}{BC} = \frac{DB}{AB} \implies \frac{6}{a} = \frac{8}{c}
\]
3. **Express \(a\) in terms of \(c\)**:
From the second similarity:
\[
\frac{6}{a} = \frac{8}{c} \implies 6c = 8a \implies a = \frac{6c}{8} = \frac{3c}{4}
\]
4. **Express \(b\) in terms of \(c\)**:
Now, we can substitute \(b\) back using the first similarity:
\[
\frac{6}{b} = \frac{x}{c} \implies 6c = bx \implies b = \frac{6c}{x}
\]
5. **Use Pythagorean theorem**:
In triangle \(ABC\):
\[
AB^2 = AC^2 + BC^2 \implies c^2 = b^2 + a^2
\]
Substituting \(b\) and \(a\):
\[
c^2 = \left(\frac{6c}{x}\right)^2 + \left(\frac{3c}{4}\right)^2
\]
6. **Calculate squares**:
\[
c^2 = \frac{36c^2}{x^2} + \frac{9c^2}{16}
\]
Multiply through by \(16x^2\) to eliminate denominators:
\[
16x^2c^2 = 576c^2 + 9c^2x^2
\]
Rearranging gives:
\[
16x^2c^2 - 9c^2x^2 - 576c^2 = 0
\]
Factoring out \(c^2\):
\[
c^2(16x^2 - 9x^2 - 576) = 0
\]
\[
c^2(7x^2 - 576) = 0
\]
Since \(c\) cannot be zero, set the remaining factor to zero:
\[
7x^2 - 576 = 0
\]
\[
7x^2 = 576
\]
\[
x^2 = \frac{576}{7}
\]
\[
x = \sqrt{\frac{576}{7}} = \frac{24}{\sqrt{7}} \approx 9.06
\]
Thus, the length of segment \(AD\) is approximately \(9.06\) or more precisely:
\[
AD = \frac{24}{\sqrt{7}}.
\]
1. **Label the triangle and variables**:
- Let \(AB = c\) (the hypotenuse)
- \(AC = b\)
- \(BC = a\)
- The right angle is at \(C\)
- \(AD = x\)
- \(DB = 8\)
- \(CD = 6\)
Since \(D\) is the foot of the altitude from \(C\) to hypotenuse \(AB\), we have:
\[
AD + DB = AB \implies x + 8 = c
\]
2. **Using the property of similar triangles**:
Triangles \(CDA\) and \(CDB\) are similar to triangle \(CAB\) (both being right triangles). According to the similarity:
- For triangle \(CDA\):
\[
\frac{CD}{AC} = \frac{AD}{AB} \implies \frac{6}{b} = \frac{x}{c}
\]
- For triangle \(CDB\):
\[
\frac{CD}{BC} = \frac{DB}{AB} \implies \frac{6}{a} = \frac{8}{c}
\]
3. **Express \(a\) in terms of \(c\)**:
From the second similarity:
\[
\frac{6}{a} = \frac{8}{c} \implies 6c = 8a \implies a = \frac{6c}{8} = \frac{3c}{4}
\]
4. **Express \(b\) in terms of \(c\)**:
Now, we can substitute \(b\) back using the first similarity:
\[
\frac{6}{b} = \frac{x}{c} \implies 6c = bx \implies b = \frac{6c}{x}
\]
5. **Use Pythagorean theorem**:
In triangle \(ABC\):
\[
AB^2 = AC^2 + BC^2 \implies c^2 = b^2 + a^2
\]
Substituting \(b\) and \(a\):
\[
c^2 = \left(\frac{6c}{x}\right)^2 + \left(\frac{3c}{4}\right)^2
\]
6. **Calculate squares**:
\[
c^2 = \frac{36c^2}{x^2} + \frac{9c^2}{16}
\]
Multiply through by \(16x^2\) to eliminate denominators:
\[
16x^2c^2 = 576c^2 + 9c^2x^2
\]
Rearranging gives:
\[
16x^2c^2 - 9c^2x^2 - 576c^2 = 0
\]
Factoring out \(c^2\):
\[
c^2(16x^2 - 9x^2 - 576) = 0
\]
\[
c^2(7x^2 - 576) = 0
\]
Since \(c\) cannot be zero, set the remaining factor to zero:
\[
7x^2 - 576 = 0
\]
\[
7x^2 = 576
\]
\[
x^2 = \frac{576}{7}
\]
\[
x = \sqrt{\frac{576}{7}} = \frac{24}{\sqrt{7}} \approx 9.06
\]
Thus, the length of segment \(AD\) is approximately \(9.06\) or more precisely:
\[
AD = \frac{24}{\sqrt{7}}.
\]