To find the length of segment \( BD \), we first need the lengths of segments \( AC \) and the radius of the semicircle.
- Find the length of \( AC \) (the hypotenuse):
Since triangle \( ABC \) is a right triangle with vertices on a semicircle, we can use the Pythagorean theorem:
\[ AC^2 = AB^2 + BC^2 \]
Given that \( AB = 8 \) cm and \( BC = 12 \) cm:
\[ AC^2 = 8^2 + 12^2 \] \[ AC^2 = 64 + 144 \] \[ AC^2 = 208 \] \[ AC = \sqrt{208} \approx 14.42 \text{ cm} \]
- Find the radius \( r \) of the semicircle:
The diameter of the semicircle is equal to the hypotenuse \( AC \) of triangle \( ABC \), so the radius \( r \) is:
\[ r = \frac{AC}{2} = \frac{14.42}{2} \approx 7.21 \text{ cm} \]
- Find the length of \( BD \):
In the right triangle \( ABD \), we can now consider \( BD \) as part of triangle \( ABD \) where \( AD \) (the radius) is perpendicular to the hypotenuse. Using the Pythagorean theorem again:
In triangle \( ADB \):
\[ AD^2 = AB^2 + BD^2 \]
Substituting \( AD = r \), \( AB = 8 \):
\[ (7.21)^2 = 8^2 + BD^2 \] \[ 52.0641 = 64 + BD^2 \] \[ BD^2 = 52.0641 - 64 \] \[ BD^2 = -11.9359 \text{ (this means we made a mistake, let’s recalculate the correct way)} \]
Now, instead we should consider the coordinates or find a perpendicular way that uses simpler ratios.
- Calculating using area:
We can also find \( BD \) using ratios, where the area of triangle \( ABC \) can also be derived as:
Area \( = \frac{1}{2} \times AB \times BC \) \[ = \frac{1}{2} \times 8 \times 12 = 48 \text{ cm}^2 \]
From the area, relating it with segment BD:
\[ Area = \frac{1}{2} \times AC \times BD \rightarrow 48 = \frac{1}{2} \times 14.42 \times BD \]
Now solve for \( BD \):
\[ 96 = 14.42 \times BD \] \[ BD = \frac{96}{14.42} \approx 6.65 \text{ cm} \]
- Final Result:
Rounding to the nearest tenth gives \( BD \approx 6.7 \) cm.
Thus, the length of segment \( BD \) is approximately \( 6.7 \) cm.
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