In right triangle ABC, Angle C = 90 degrees, D is a point on AB, and CD is perpendicular to AB. AB = 20 and AD = 5. Find AC
I tried formulas like soacahtoa and Pythagorean theorem but I am unable to figure this out.
2 answers
nevermind! i had an epiphany
Let AD = x
The Geometric Mean Theorem states that
AD*DB = CD^2
x(20-x) = 25
x = 5(2+√3)
Since right triangles ADC and ACB are similar (Angle A is common to both),
AC/x = 20/AC
AC^2 = 20x = 100(2+√3)
AC = 10√(2+√3) = 5(√2+√6)
The Geometric Mean Theorem states that
AD*DB = CD^2
x(20-x) = 25
x = 5(2+√3)
Since right triangles ADC and ACB are similar (Angle A is common to both),
AC/x = 20/AC
AC^2 = 20x = 100(2+√3)
AC = 10√(2+√3) = 5(√2+√6)
1 answer