a) In trapezul dat, ducem CF _I_ pe AB (CF = AD = h trapezului)
Stim ca intr-un trapez dreptunghic ortodiagonal (ortodiagonal = cu diagonalele perpendiculare) :
h = √(B×b) = √48×12 = √576 =24 cm
deci AD = CF = 24 cm
A= (B+b) × h / 2 = (48+12) ×24 /2 = 720 cm²
b) Distanta de la A la BC o notam cu AE si este _I_ pe BC.
In Δ ACD, dreptunghic in D, (AC= diagonala trapez si ipotenuza in Δ):
AC² = AD² +CD²
AC² = 24² +12²
AC² = 720
AC = 12√5 cm
In Δ BCF, dreptunghic in F :
FB = AB-CD = 48-12 = 36
BC² = BF² +CF²
BC² = 36² +24²
BC² = 1872
BC = 12√13 cm
Prelungim laturile BC si AD care se vor intalni, sa zicem, in pct.V
Δ VAB ~ ΔVDC
⇒ DC/AB =VD/VA = VC/VB
VA = VD +AD
VB = VC +BC
DC/AB =VD/VA ⇒ DC (VD+AD) = AB ×VD
DC×VD -AB×VD = - DC×AD
VD(DC - AB) = - DC ×AD
VD = (DC ×AD) / (AB-DC) =12×24 / (48-12) = 8 cm
DC/AB =VC/VB ⇒ DC (VC +BC) =AB ×VC
VC (DC-AB) = -DC ×BC
VC = (DC×BC) / (AB-DC) = 12 ×12√13 / 36 = 144√13 /36 = 4√13 cm
VA = VD+AD = 8 +24 = 32 cm
VB = VC +BC = 4√13 +12√13 = 16√13 cm
Aria Δ VAB = VA×AB / 2 (semiprodusul catetelor)
dar si
AE × VB / 2 (adica baza × h / 2)
⇒VA×AB/2 = AE ×VB /2 ⇒ AE = VA×AB / VB
AE = 32 × 48 / 16√13 = 1536 / 16√13 = 96 /√13 = 96√13 /13 cm
