Are you up for another challenging ACT Math problem? This one will require you to use concepts you’ve learned in both Algebra and Geometry.

The equation of the parabola in the figure below is *f*(*x*) = -12/25 *x*^{2} + 12. Point *A* is a vertex of the triangle and the *y*-intercept of the parabola. Points *B* and *C* are also vertices of the triangle and the *x*-intercepts of the parabola. What is the perimeter of triangle *ABC* ?

A. 24

B. 28

C. 30

D. 32

E. 36

Scroll down for the solution.

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Because the *y*-intercept of the parabola is 12, the coordinates of Point *A* are (0, 12). Points *B* and *C* are the *x*-intercepts of the parabola, so they occur at the points where *y* = 0. You can find the *x*-coordinates of these points by setting *y* equal to 0 in the equation and solving for *x*.

*y* = -12/25*x²* + 12

0 = -12/25*x*² + 12

0 – 12 = -12/25*x**² + *12 – 12

-12 = -12/25*x**² *

(25/-12)(-12) = (25/-12)(-12/25)*x**²*

25 =* x*²

– 5 or 5 = *x*

The coordinates of Point *B* are (-5, 0) and the coordinates of Point *C* are (5, 0).

You can now use either the Distance Formula or Pythagorean Theorem to find the lengths of segment *AC* and segment *AB*.

*a**² + b**² = c*²

5² + 12² = *c**²*

25 + 144 =* c**²*

169 =* c*²

13 = *c*

The lengths of both segment *AB* and segment *AC* are 13 units. The endpoints of segment *BC* are (-5, 0) and (5, 0), so the length of that segment is 10 units.

Therefore, the perimeter of triangle *ABC* is equal to 13 + 13 + 10 = 36 units.

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