3 years ago

PLS HELP ASAP :DDDDDDDDDDDDDDDDDDD

1 answer:

6 0

Pythagoras in HBC : $HC=\sqrt{HB^2-BC^2}=\sqrt{20^2-12^2}=16in$

Pythagoras in HDC : $HD=\sqrt{HC^2-CD^2}=\sqrt{16^2-14^2}\approx\boxed{7.7 in}$

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