luas irisan = 9,99188 satuan luas.

keliling irisan = 11,9875 satuan keliling.

Pembahasan :

mencari titik potong.

L1 dan L2 dieliminasi/substitusi dihasilkan

x = - ⅔√14 atau x = ⅔√14

y = 5/3

titik potong:

(-⅔√14, 5/3) dan (⅔√14, 5/3)

L2 = x² + y² - 9 = 0

y² = 9 - x²

y = (9 - x²)¹´²

L1 = x² + y² - 6y + 1 = 0

y² - 6y +(1+x²) = 0

rumus abc

y = (6 ± √(36-4(1+x²))/2

y = 3 ± ½√(32-4x²)

y = 3 ± ½√(4(8-x²))

y = 3 ± (8-x²)¹´²

Luas irisan =

x antara x = - ⅔√14 sampai x = ⅔√14

∫ (y2 - y1) dx =

∫ ( (9- x²)¹´² - (3 ± (8-x²)¹´²) dx =

9,99188

mencari keliling irisan?

jarak titik potong = CD =

(-⅔√14, 5/3) dan (⅔√14, 5/3) adalah:

= √((⅔√14 - -⅔√14)² + (5/3-5/3)²)

= (4/3 √14)²

= 16/9 × 14

= 224/9

L1 = x² + y² - 6y + 1 = 0

(x-0)² + (y-3)²- 9 + 1 =0

(x-0)² + (y-3)² = 8

r = √8

cos CAD = (2r² - CD²)/(2r²)

= (2(√8)²- 224/9)/((2(√8)²)

= -0,555

CAD = 123,75⁰

keliling L1 = 123,75⁰/360⁰ × 2πr

= 123,75/360 × 2(3,14)(√8)

= 6,1

L2 : x² + y² - 9 = 0

x² + y² = 9

r = 3

cos CBD = (2r² - CD²)/(2r²)

= (2(3)²- 224/9)/(2(3)²)

= -0,383

CBD = 112,5⁰

keliling L2 = 112,5/360 × 2(3,14)(3)

= 5,8875

jadi keliling irisan =

6,1 + 5,8875 =

11,9875