BH=V 25^2-15^2=20cm MH=V 17^2-15^2=8cm AM=BM=28cm
AB=28*2=56cm A=56*15/2=420cm2 h=420*2/25=33,6cm
AC = 25 cm
CM = 17 cm
altezza CH = 15 cm
AH = 5√5^2-3^2 = 20 cm
HM = √17^2-15^2 = 8 cm
AM = AH+HM = 20+8 = 28 cm
AB = 2AM = 28*2 = 56 cm
area A = AB*CH/2 = 28*15 = 420 cm
altezza BH = 2A/AC = 840/25 = 168/5 cm (33,60)
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Applica il teorema di Pitagora come segue:
$\small AH= \sqrt{(AC)^2-(CH)^2} = \sqrt{25^2-15^2} = 20\,cm;$
$\small HM= \sqrt{(CM)^2-(CH)^2} = \sqrt{17^2-15^2} = 8\,cm;$
per cui:
base $\small AB= 2(AH+HM) = 2(20+8) = 2×28 = 56\,cm;$
area $\small A_{ABC}= \dfrac{b×h}{2} = \dfrac{AB×CH}{2} = \dfrac{56×15}{2} = 420\,cm^2;$
altezza relativa al lato AC $\small h_{rel.AC}= \dfrac{2×A_{ABC}}{AC} = \dfrac{2×420}{25} = 33,6\,cm;$