Question

Obwód trapezu równoramiennego jest równy 24 cm, a cosinus jego kąta ostrego jest równy 3/5. Jakie powinny być długości boków tego trapezu, aby jego pole było równe 28cm2?

Dzięki :)

Answer 1

L = 24 →L = a+b+2c→ a+b+2c = 24
cosα = 3/5 → cosα =(a-b)/2c→ (a-b)/2c = 3/5
P = 28 → P = 1/2(a+b)h → 1/2(a+b)h = 28

cos²α +sin²α = 1
sin²α = 1- 9/25
sin²α = 16/25
sinα = 4/5

z tw sinusów:
h/sinα = c/sin90
h/(4/5) = c
h = (4/5)c

a+b+2c = 24
(a-b)/2c = 3/5
(4/10)c*(a+b) = 28

a+b+2c = 24
5(a+b) = 6c
c(a+b) = 70

a+b = 24 - 2c
5(a+b) = 6c
c(24 - 2c) = 70→ c² - 12c+ 35 = 0
Δ = 144 - 140 = 4
√Δ = 2
c = (12 - 2)/2 = 5 ∨c = (12 - 2)/2 = 7
c= 5
a+b = 14
5(a- b) = 30

c = 5
a = 14 - b
5(14 - 2b) = 30

c = 5
a = 14 - b
70 - 10b = 30

c = 5
a = 14-b
-10b = -40

a = 10
b = 4
c = 5

lub:
c= 7
a+b= 10
5(a- b) = 42

c= 7
a = 10 - 5
5(10 - 2b) = 42

c= 7
a = 10 - 5
50 - 10b = 42

c= 7
a = 10 - b
b = 4/5

a = 46/5
b = 4/5
c = 7