Oblicz pole trapezu, którego podstawy mają długość 16 cm i 44 cm a ramiona 17cm i 25 cm
h + x = 17 h + (28-x) = 25 h + x = 289 h + 784 - 56 x + x = 625 h + x = 289 h - 56 x + x = -159 56 x = 448 \ :56 x = 8
h + 8 = 17 h = 289 - 64 h = 225 h = 15
p = (a+b) × h p = (16+44) × 15 p = 30 × 15 p = 450 cm²
Odp. Pole wynosi 450 cm kwadratowych.
h²+x²=17²
h²+x²=289h+(28-x)=25 h+784-56x+ x=625h²+x²=289h-56x+x=-15956x=448 /: 56x=8
h+8=17 h = 15
P=a+b*h /2P=16+44 * 15/2P=60*15/2
P=900/2P=450 cm²
Odp: Pole wynosi 450 cm²
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h + x = 17
h + (28-x) = 25
h + x = 289
h + 784 - 56 x + x = 625
h + x = 289
h - 56 x + x = -159
56 x = 448 \ :56
x = 8
h + 8 = 17
h = 289 - 64
h = 225
h = 15
p = (a+b) × h
p = (16+44) × 15
p = 30 × 15
p = 450 cm²
Odp. Pole wynosi 450 cm kwadratowych.
h²+x²=17²
h²+x²=289
h+(28-x)=25
h+784-56x+ x=625
h²+x²=289
h-56x+x=-159
56x=448 /: 56
x=8
h+8=17
h = 15
P=a+b*h /2
P=16+44 * 15/2
P=60*15/2
P=900/2
P=450 cm²
Odp: Pole wynosi 450 cm²