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Bbrainly101
Bbrainly101 Bbrainly101
  • 10-06-2015
  • Mathematics
Answered

PLS HELP ASAP :DDDDDDDDDDDDDDDDDDD

PLS HELP ASAP :DDDDDDDDDDDDDDDDDDD class=

Verified Answer :

Hippalectryon
Hippalectryon Hippalectryon
  • 10-06-2015

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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