Question

Fie ABCD un trapez in care AB||CD , AB=40,BC=15,CD=15 si AD=20.Inaltimea trapezului are lungimea: a) 12cm . b) 8cm. c)14cm

Answer 1

ducem DN si CM inaltimi pe AB
notam AN=x    si BM=y
x+y=AB-CD=40-15=25
DN²=20²-x²
CM²=15²-y²

20²-x²=15²-y²
x²-y²=400-225
(x+y)(x-y)=175
x-y=175/25    
x-y=7
x+y=25
2x=25+7
x=32/2=16 cm
y=25-16=9 cm
DN=√(20²-16²)=√(400-256)=√144=14 cm

raspuns corect a) 12 cm