Question

Bazele trapezului isoscel ABCD au lungimile AD=25cmsi BC=7cm ,iar inaltimea sa este de 12cm .Se noteaza cu S intersectia laturilor neparalele .

Determinati :aria si perimetrul trapezului ,sinusul masurii unghiului BAD, lungimea segmentlui (SA)si distanta d(D,AB)

Answer 1

ducem BE_I_AD
ducem CF_I_AD
AE=FD=(25-7)/2=9
AB=CD=√(12²+9²)=√(144+81)=√225=15 cm
A=(7+25)*12/2=192 cm²
sin (<BAD)=12/15=4/5
SB/SA=(BC/2)/(AD/2)=3,5/12,5 
SB/(SB+SA)=SB/(SB+15)=3,5/12,5=7/25
25SB=7SB+105
18SB=105
SB=105/18=5,8(3)
SA=15+5,8(3)=15+105/18=375/18=125/6=20,8(3) 
ducem DM _I_ AB
SD=SA=20,8(3)
notam AM=x
DM²=20,8(3)²- [20,8(3)-x]²
DM²=25²- x²
20,8(3)²- [20,8(3)-x]² =25²- x²
(125/6)²-(125/6)²+2*125/6 *x-x²=625-x²
125/3 *x=625
x=15
deci M coincide cu B
DB²=√(25²-15²)=√(625-225)=√400=20 cm