tgc=3/4⇔ab/ac=3/2
dar ac=24 din astea doua=>ab/24=3/4=>ab=24×3/4=>ab=6×3=>ab=18
bc²=ab²+ac²
bc²=18²+24²
bc²=324+576
bc²=900=>bc=√900=>bc=30
daca m mijlocullui bc si ∡a=90°=>am este mediana intr-un tiunghi dreptunghic=>am=1/2bc=>am=30/2=>am=15
aplicam teorema catetei:
ab²=bd x bc
18²=bd×30
324=bd×30=>bd=324/30(simplific cu 2)=>bd=162/15 (simplific cu 3)=>bd=54/5
ac²=cd × bc
24²= cd × 30
576=cd ×30=>cd=576/30 (simplific cu 3)=>cd=192/10 (simplific cu 2)=> cd=96/5
aplic teorema inaltimii
ad²=cd × db
ad²=96/5 × 54/5
ad²=5184/25=>ad=√5184/25=>ad=72/5
in Δadm:
ad²+dm²=am²
(72/5)²+dm²=15²
dm²=225-5184/25
dm²=225×25/25-5184/25
dm²=5626/25-5184/25
dm²=441/25=>dm=√441/25=>dm=21/5
aria Δabc=(ad ×bc)/2=(72/5 ×30)/2=(72 ×6)/2=72×3=216cm²
ariaΔadm=?
inΔadm duc dx⊥am unde x∈am
aplic teorema catetei
ad²=ax ori am
(72/5)²=ax ori 15
5184/25=ax ori 15=>ax=5184/25:15=>ax=5184/25 ori 1/15=ax=5184/375 care se simplifica cu 3=>ax=1728/125
ax+xm=am
1728/125+xm=15
xm=15-1728/125
xm=15 ori 125/125-1728/125
xm=1875/125-1728/125
xm=147/125
aplic teorema inaltimii
dx²=ax ori xm
dx²=1728/125×147/125
dx²=254016/125²
dx²=√254016/125²
dx=504/125
arie Δadm=dx ori am/2=(504/125 ori 15)/2=(504/25 ori 3)/2=1512/25/2=1512/25 ori1/2=756/25=30,24 cm²
daca aria abc=216 cm² si aria adm=30,2 cm²=> ...de aici chiar ca nu mai stiu...
dar vezi daca te ajuta ce am rezolvat..daca nu sterge
