[ BIG POINT ] a. hitung luas daerah yang diarsir ! b. hitung keliling daerah yang diarsir !

Mapel : Matematika
Bab : Lingkaran

A) Luas yng di-Arsir

Cari Luas juring besar

L Juring = 60/360 × 3,14 × r^2
L Juring = 1/6 × 3,14 × 24 × 24
L Juring = 301,44 cm^2

Cari Luas Juting kecil

L juring = 60/360 × 3,14 × 12 × 12
L juring = 75,36 cm^2

L yang diarsir = 301,44 - 75,36
L yang Diarsir = 226,08 cm^2

B) Keliling yang diarsir

Busur Besar = 60/360 × 3,14 × d
Busur Besar = 1/6 × 3,14 × 48
Busur Besar = 25,12 cm

Busur Kecil = 60/360 × 3,14 × d
Busur Kecil = 1/6 × 3,14 × 24
Busur kecil = 12,56 cm

K Lingkaran Besar = 24 + 24 + 25,12 = 73,12 cm

K lingkaran Kecil = 12 + 12 + 12,56 = 36,56 cm

Keliling yang diarsir = 73,12 - 36,56 = 36,56 cm

Semoga Membantu :)

diketahui :
- r1 = 24cm
- r2 = 12cm
- besar sudut = 60°

ditanya :
a. luas daerah arsir?
b. kll daerah arsir?

a.)
luas juring besar :

[tex] \frac{60}{360} = \frac{lj}{\pi \times {r}^{2} } \\ \frac{60}{360} = \frac{lj}{3.14 \times 24 \times 24} \\ \: \: \: \: \frac{1}{6} = \frac{lj}{1808.64} \\ \: \: \: \: lj = 301.44 \: {cm}^{2} [/tex]

luas juring kecil :

[tex] \frac{60}{360} = \frac{lj}{\pi \times {r}^{2} } \\ \frac{60}{360} = \frac{lj}{3.14 \times 12 \times 12} \\ \: \: \: \: \frac{1}{6} = \frac{lj}{452.16} \\ \: \: \: \: lj = 75.36 \: {cm}^{2} [/tex]

luas juring arsir :
= 301,44cm² - 75,36cm²
= 226,08cm²

b.)
panjang busur besar :

[tex] \frac{60}{360} = \frac{pb}{2 \times \pi \times r} \\ \frac{60}{360} = \frac{pb}{2 \times 3.14 \times 24} \\ \: \: \: \: \frac{1}{6} = \frac{pb}{150.72} \\ \: \: \: pb = 25.12cm [/tex]

panjang busur kecil :

[tex] \frac{60}{360} = \frac{pb}{2 \times \pi \times r} \\ \frac{60}{360} = \frac{pb}{2 \times 3.14 \times 12} \\ \: \: \: \: \frac{1}{6} = \frac{pb}{75.36} \\ \: \: \: pb = 12.56 \: cm[/tex]

kll daerah arsir :
= 25,12cm + 12,56cm + 12cm + 12cm
= 61,68cm

semoga membantu