Răspuns:
Explicație pas cu pas:
a) Trasăm CE⊥AB, DF⊥AB. Atunci CE=h, inălțimea trapezului, EF=CD=14.
Atunci AF+BE=24-14=10
AF²=6²-h², ⇒AF=√(36-h²); BE²=8²-h², ⇒BE=√(64-h²). Deci, obtinem:
√(36-h²)+√(64-h²)=10 |^2 ⇒36-h²+64-h²+2√[(36-h²)(64-h²)]=100 ⇒
2√[(36-h²)(64-h²)]=2h² ⇒√[(36-h²)(64-h²)]=h² |^2 ⇒(36-h²)(64-h²)=h⁴ ⇒36·64-100h²+h⁴=h⁴ ⇒36·64=100h² ⇒h²=(36·64)/100 , deci h=4,8cm.
b) Aria(ABCD)=(24+14)·4,8/2=91,2 cm².
c) ΔSAB~ΔSDC ⇒AB/DC=SM/SN, unde SM si SN sunt inaltimi in aceste triunghiuri.
SM=SN+4,8. SM/SN=(SN+4,8)/SN=SN/SN+4,8/SN=1+4,8/SN.
Deci 24/14=1+4,8/SN ⇒12/7 -1=4,8/SN ⇒5/7=4,8/SN ⇒SN=7·4,8/5=6,72cm.
Atunci Aria(SCD)=(1/2)·CD·SN=(1/2)·14·6,72=7·6,72=47,04cm².
