61,5 cm²
Penjelasan dengan langkah-langkah:
diketahui ∆ sama sisi
panjang sisi = 10 cm = a
Keliling = 3a = 3(10) = 30 cm
s = ½K = ½(30) = 15
Luas ∆ =
[tex] = \sqrt{s(s - 10 {)}^{3} } \\ = \sqrt{15(15 - 10 {)}^{3} } \\ = \sqrt{15(5 {)}^{3} } \\ = \sqrt{15 \times 125} \\ = \sqrt{1875} \\ = 25 \sqrt{3} \: \: cm {}^{2} [/tex]
jari - jari lingkaran =
[tex] = \frac{ {a}^{3} }{4 \: luas \: \: segitiga} \\ = \frac{ {10}^{3} }{4 \times 25 \sqrt{3} } \\ = \frac{1000}{100 \sqrt{3} } \\ = \frac{10}{ \sqrt{3} } [/tex]
Luas daerah yg diarsir
= Luas Lingkaran - luas ∆
= (πr² ) - 25√3
= (22/7 x (10/√3)²) - 25√3
= (22/7 x 100/3) - 25√3
= (2200/21) - 25√3
= 104,8 - 43,3
= 61,5 cm²
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61,5 cm²
Penjelasan dengan langkah-langkah:
diketahui ∆ sama sisi
panjang sisi = 10 cm = a
Keliling = 3a = 3(10) = 30 cm
s = ½K = ½(30) = 15
Luas ∆ =
[tex] = \sqrt{s(s - 10 {)}^{3} } \\ = \sqrt{15(15 - 10 {)}^{3} } \\ = \sqrt{15(5 {)}^{3} } \\ = \sqrt{15 \times 125} \\ = \sqrt{1875} \\ = 25 \sqrt{3} \: \: cm {}^{2} [/tex]
jari - jari lingkaran =
[tex] = \frac{ {a}^{3} }{4 \: luas \: \: segitiga} \\ = \frac{ {10}^{3} }{4 \times 25 \sqrt{3} } \\ = \frac{1000}{100 \sqrt{3} } \\ = \frac{10}{ \sqrt{3} } [/tex]
Luas daerah yg diarsir
= Luas Lingkaran - luas ∆
= (πr² ) - 25√3
= (22/7 x (10/√3)²) - 25√3
= (22/7 x 100/3) - 25√3
= (2200/21) - 25√3
= 104,8 - 43,3
= 61,5 cm²