To solve for \( x \), the altitude of a right triangle with one leg measuring \( 4 \) (the short leg) and the other leg measuring \( 16 \) (the long leg), we can use the formula for the area of the triangle in two different ways.
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Using the legs to find area: The area \( A \) of a right triangle can be calculated using the formula: \[ A = \frac{1}{2} \times \text{base} \times \text{height} \] Here, we can take \( 4 \) as the base and \( 16 \) as the height: \[ A = \frac{1}{2} \times 4 \times 16 = \frac{1}{2} \times 64 = 32 \]
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Using the altitude to find area: The area can also be expressed using the altitude \( x \) to the hypotenuse. If we denote the hypotenuse as \( h \), we then calculate the area as: \[ A = \frac{1}{2} \times \text{hypotenuse} \times \text{altitude} \] To find the hypotenuse \( h \), we can use the Pythagorean theorem: \[ h = \sqrt{4^2 + 16^2} = \sqrt{16 + 256} = \sqrt{272} = 4\sqrt{17} \] Now, using the hypotenuse \( h \) and the altitude \( x \): \[ A = \frac{1}{2} \times h \times x = \frac{1}{2} \times 4\sqrt{17} \times x \]
Setting both area equations equal to each other gives us: \[ 32 = \frac{1}{2} \times 4\sqrt{17} \times x \]
Now we can simplify: \[ 32 = 2\sqrt{17} x \] To isolate \( x \), divide both sides by \( 2\sqrt{17} \): \[ x = \frac{32}{2\sqrt{17}} = \frac{16}{\sqrt{17}} \]
To rationalize the denominator: \[ x = \frac{16\sqrt{17}}{17} \]
Thus, the altitude \( x \) is: \[ \boxed{\frac{16\sqrt{17}}{17}} \]