Solve for x, a triangle with a short leg of 4, long leg of 16 and x as the altitude

1 answer

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.

  1. 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: \[ A = \frac{1}{2} \times 4 \times 16 = \frac{1}{2} \times 64 = 32 \]

  2. 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: \[ \boxed{\frac{16\sqrt{17}}{17}} \]