Answer:
C) [tex] A(x) = 2 \sqrt{x^2 - 16} [/tex]
Step-by-step explanation:
The area [tex] A [/tex] of a right triangle can be calculated using the formula:
[tex] \Large\boxed{\boxed{ A = \dfrac{1}{2} \times \textsf{base} \times \textsf{height}}} [/tex]
In this case, one leg of the right triangle measures 4 units, and the hypotenuse measures [tex] x [/tex] units.
Let's denote the other leg of the triangle as [tex] y [/tex] units.
Using the Pythagorean theorem, we have:
[tex]c^2 = a^2 + b^2[/tex]
[tex] x^2 = 4^2 + y^2 [/tex]
[tex] x^2 = 16 + y^2 [/tex]
[tex] y^2 = x^2 - 16 [/tex]
[tex] y = \sqrt{x^2 - 16} [/tex]
Now, the area [tex] A [/tex] of the triangle can be expressed as:
[tex] A(x) = \dfrac{1}{2} \times 4 \times \sqrt{x^2 - 16} [/tex]
[tex] A(x) = 2 \sqrt{x^2 - 16} [/tex]
Therefore, the correct representation of [tex] A(x) [/tex] is:
[tex] A(x) = 2 \sqrt{x^2 - 16} [/tex]