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10-06-2015
Mathematics

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Hippalectryon Hippalectryon · Answer 1 · 2015-06-10T16:35:16-04:00

Hippalectryon Hippalectryon

10-06-2015

Pythagoras in HBC : [tex]HC=\sqrt{HB^2-BC^2}=\sqrt{20^2-12^2}=16in[/tex]

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

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