L = πr²
1.256 = 3,14.r²
r² = 1256/3,14
r² = 400
r = √400
r = 20 cm
d = r × 2
d = 20 × 2
d = 40 cm
[tex] \mathbb \color{aqua} \underbrace{JAWABAN}[/tex]
40 cm
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[tex] \mathbb \color{gold} \underbrace{RUMUS}[/tex]
Luas lingkaran :
[tex] \sf L = \pi \times {r}^{2} [/tex]
Keliling lingkaran :
[tex] \sf K = 2 \times \pi \times r[/tex]
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[tex] \mathbb \color{orange} \underbrace{PENYELESAIAN}[/tex]
[tex] \underline {\boxed{ \sf diketahui}}[/tex]
- Luas (L) = 1.256 cm²
- π = 3,14
[tex] \\ \underline {\boxed{ \sf ditanya}}[/tex]
- diameter (d)
[tex] \\ \underline {\boxed{ \sf jawab}}[/tex]
=> menentukan jari-jari (r) :
[tex] \sf L = \pi \times {r}^{2} [/tex]
[tex] \sf 1.256 = 3,14 \times r {}^{2} [/tex]
[tex] \sf \frac{1256}{3,14} = {r}^{2} [/tex]
[tex] \sf 400 = {r}^{2} [/tex]
[tex] \sf \sqrt{400} = r[/tex]
[tex] \sf \sqrt{ {20}^{2} } = r[/tex]
[tex] \sf20 \: cm = r \\ [/tex]
=> menentukan diameter (d) :
[tex] \sf d = 2 \times r[/tex]
[tex] \sf d = 2 \times 20[/tex]
[tex] \sf d = 40 \: cm[/tex]
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[tex] \mathbb \color{red} \underbrace{KESIMPULAN}[/tex]
Jadi, diameter lingkaran tersebut adalah 40 cm
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