Answer:
[tex]\large\boxed{a=4\sqrt5,\ b=8\sqrt5,\ f=8}[/tex]
Step-by-step explanation:
ΔBDC and ΔCDA are similar (AA). Therefore the sides are in proportion:
[tex]\dfrac{BD}{DC}=\dfrac{DC}{DA}[/tex]
We have:
BD = 4, DA = 16 and DC = f. Substitute:
[tex]\dfrac{4}{f}=\dfrac{f}{16}[/tex] cross multiply
[tex]f^2=(4)(16)\\\\f^2=64\to f=\sqrt{64}\\\\\boxed{f=8}[/tex]
We can use the Pythagorean theorem:
[tex]leg^2+leg^2=hypotenuse^2[/tex]
ΔBDC:
leg = 4, leg = 8, hypotenuse = a. Substitute:
[tex]a^2=4^2+8^2\\\\a^2=16+64\\\\a^2=80\to a=\sqrt{80}\\\\a=\sqrt{16\cdot5}\\\\a=\sqrt{16}\cdot\sqrt5\\\\\boxed{a=4\sqrt5}[/tex]
ΔCDA:
leg = 8, leg = 16, hypotenuse = b. Substitute:
[tex]b^2=8^2+16^2\\\\b^2=64+256\\\\b^2=320\to b=\sqrt{320}\\\\b=\sqrt{16\cdot4\cdot5}\\\\b=\sqrt{16}\cdot\sqrt4\cdot\sqrt5\\\\b=4\cdot2\cdot\sqrt5\\\\\boxed{b=8\sqrt5}[/tex]
