Question

Math 2, Trigonometry and Triangles on ALEKS See photo below

Answer 1

Answer:

FC = 28

Step-by-step explanation:

The parallel lines creates two similar triangles:

ΔBDC similar to ΔEDF

Similar triangle means corresponding lengths are equal.

BD corresponds to ED

BC corresponds to EF

DC corresponds to DF

Now, we want to find FC, we let that length be "x",

thus we can write:

DE similar DF

DB similar DC

So,

$\frac{DE}{DF}=\frac{DB}{DC}\\\frac{9}{21}=\frac{DE+EB}{DF+FC}\\\frac{9}{21}=\frac{9+12}{21+x}$

Now, we use algebra, cross multiplication and solve for x, which is FC:

$\frac{9}{21}=\frac{9+12}{21+x}\\\frac{9}{21}=\frac{21}{21+x}\\9(21+x)=21*21\\189+9x=441\\9x=252\\x=28$

FC = 28