Materi : Lingkaran
∠ABC = ½ × ∠( refleks dari AOC )
130° = ½ × ( 360° - ∠AOC )
260° = 360° - ∠AOC
∠AOC = 360° - 260°
∠AOC = 100° { d. }
L = ∠AOC/360 × π × r²
L = 60°/360° × 22/7 × 14²
L = ⅙ × 616 cm²
L = ⅓ × 308 cm²
L = 102⅔ cm² { a. }
Cari OB
OB² = OA² + AB²
OB² = 9² + 12²
OB² = 81 + 144
OB² = 225
OB = √225
OB = √(15²)
OB = 15 cm
Langsung Saja
½AC × OB = OA × AB
½AC × 15 = 9 × 12
½AC = ( 3³ × 4 )/( 3 × 5 )
½AC = ( 3² × 4 )/5
½AC = 36/5
½AC = 7,2
AC = 14,4 cm { c. }
Semoga bisa membantu
[tex] \boxed{ \colorbox{darkblue}{ \sf{ \color{lightblue}{ answered\:by\: BLUEBRAXGEOMETRY}}}} [/tex]
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Materi : Lingkaran
Soal Nomor 11
∠ABC = ½ × ∠( refleks dari AOC )
130° = ½ × ( 360° - ∠AOC )
260° = 360° - ∠AOC
∠AOC = 360° - 260°
∠AOC = 100° { d. }
Soal Nomor 15
L = ∠AOC/360 × π × r²
L = 60°/360° × 22/7 × 14²
L = ⅙ × 616 cm²
L = ⅓ × 308 cm²
L = 102⅔ cm² { a. }
Soal Nomor 16
Cari OB
OB² = OA² + AB²
OB² = 9² + 12²
OB² = 81 + 144
OB² = 225
OB = √225
OB = √(15²)
OB = 15 cm
Langsung Saja
½AC × OB = OA × AB
½AC × 15 = 9 × 12
½AC = ( 3³ × 4 )/( 3 × 5 )
½AC = ( 3² × 4 )/5
½AC = 36/5
½AC = 7,2
AC = 14,4 cm { c. }
Semoga bisa membantu
[tex] \boxed{ \colorbox{darkblue}{ \sf{ \color{lightblue}{ answered\:by\: BLUEBRAXGEOMETRY}}}} [/tex]