Luas daerah yang dibatasi oleh grafik persamaan x![x^{2} + y^{2} = |x| + |y| adalah\\ [tex]A. \pi + \sqrt{2} \\B. \pi + 2\\C. \pi + \sqrt[2]{2} \\D. 2\pi + \sqrt{2}\\E. 2\pi + \sqrt[2]{2}](https://tex.z-dn.net/?f=x%5E%7B2%7D%20%2B%20y%5E%7B2%7D%20%3D%20%7Cx%7C%20%2B%20%7Cy%7C%20adalah%5C%5C%20%5Btex%5DA.%20%5Cpi%20%2B%20%5Csqrt%7B2%7D%20%5C%5CB.%20%5Cpi%20%2B%202%5C%5CC.%20%5Cpi%20%2B%20%5Csqrt%5B2%5D%7B2%7D%20%5C%5CD.%202%5Cpi%20%2B%20%5Csqrt%7B2%7D%5C%5CE.%202%5Cpi%20%2B%20%5Csqrt%5B2%5D%7B2%7D "x^{2} + y^{2} = |x| + |y| adalah\\ [tex]A. \pi + \sqrt{2} \\B. \pi + 2\\C. \pi + \sqrt[2]{2} \\D. 2\pi + \sqrt{2}\\E. 2\pi + \sqrt[2]{2}")
Pakai jalann ya kak
Pakai jalann ya kak
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Luas daerah yang dibatasi oleh grafik
x²+y²=|x|+|y|
Dari grafik, dapat dicari nilai s
s = √1²+1² = √2
Luas daerahnya
L = 4•½•π•s² + s²
= 2π(√2)² + (√2)²
= 4π+2 satuan luas
L = 4.(2.L tembereng + L persegi)
= 4.(2.(L 1/4 Lingkaran - L segitiga) + 1²)
= 4.(2.(1/4 . πr² - 1/2 . a.b) + 1) <= b = 1, a = 1/2, r = 1/2 √2, r² = 1/2
= 4.(2.(1/4 . 1/2 . π - 1/2 . 1/2 . 1) + 1)
= 4.(2.(1/8 π - 1/4) + 1)
= 4(2.1/4.(1/2 π - 1) + 1)
= 4.(1/2 . (1/2 π - 1) + 1)
= 2(1/2 π - 1) + 4
= π - 2 + 4
Luas = π + 2
1/4 lingkaran ya karena memang bisa di bagi jadi 1/4
1/4 lingkaran sudutnya 90°