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.
- 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: \[ AD + DB = AB \implies x + 8 = c \]
- Using the property of similar triangles: Triangles \(CDA\) and \(CDB\) are similar to triangle \(CAB\) (both being right triangles). According to the similarity:
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For triangle \(CDA\): \[ \frac{CD}{AC} = \frac{AD}{AB} \implies \frac{6}{b} = \frac{x}{c} \]
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For triangle \(CDB\): \[ \frac{CD}{BC} = \frac{DB}{AB} \implies \frac{6}{a} = \frac{8}{c} \]
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Express \(a\) in terms of \(c\): From the second similarity: \[ \frac{6}{a} = \frac{8}{c} \implies 6c = 8a \implies a = \frac{6c}{8} = \frac{3c}{4} \]
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Express \(b\) in terms of \(c\): Now, we can substitute \(b\) back using the first similarity: \[ \frac{6}{b} = \frac{x}{c} \implies 6c = bx \implies b = \frac{6c}{x} \]
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Use Pythagorean theorem: In triangle \(ABC\): \[ AB^2 = AC^2 + BC^2 \implies c^2 = b^2 + a^2 \]
Substituting \(b\) and \(a\): \[ c^2 = \left(\frac{6c}{x}\right)^2 + \left(\frac{3c}{4}\right)^2 \]
- 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: \[ AD = \frac{24}{\sqrt{7}}. \]