Question

w trojkacie abc mamy dane :

|AC| = 3 pierwiastek2

|BC| = 5

| KĄT BAC| = 45 STOPNI

oblicz Pole tego trojkata i promien okregu wpisanego

daje naj !

Answer 1

z tw. cosinusów

 

5² = (3√2)² + x² - 2 * 3√2 * x * cos45*

25  = 18 + x² - 6√2x * √2/2

25 - 18 = x² - 6x

x² - 6x - 7 = 0

Δ = 36 + 28 = 64

√Δ = 8

x1 = (6 - 8)/2 = -2/2 = -1  ∉ D

x2 = (6 + 8)/2 = 14/2 = 7

 

|AB| = x = 7

 

obliczamy wysokosć tego trójkata:

a² + h² = (3√2)²

(7 - a)² + h² = 5²

 

h² = 18 - a²

49 - 14a + a² + 18 - a² = 25

67 - 14a = 25

-14a = 25 - 67

-14a = -42

a = 3

 

h² = 18 - 3²

h² = 18 - 9

h² = 9

h = 3

 

P = 1/2 * 7 * 3 = 10,5    -------- pole trójkąta

 

r = 2P / (a + b + c)

r = 2 * 10,5 / (7 + 5 + 3√2)

r = 21 / (12 + 3√2)

r = 21(12 - 3√2) / (12 + 3√2) (12 - 3√2)

r = 21(12 - 3√2) / (12² - (3√2)²)

r = 21(12 - 3√2) / (144 - 18)

r = 21(12 - 3√2) / 126

r = (12 - 3√2) / 6

r = 3(4 - √2) / 6

r = (4 - √2) / 2    --------- promień okegu wpisanego w ten trójkąt