Penjelasan dengan langkah-langkah:
PythaGoras & LpBR
Sisi miring
= √(12²+5²)
= √(144+25)
= √169
= 13 cm
Maka luas permukaan
= 2la + Jum.Ls
= 2(½×12×5) + 20(12+15+13)
= 60 + 20(40)
= 60 + 800
= 860 cm²
..
[tex]\begin{aligned} LP &= 2(LA) + t(K) \\&= (a × b) + t(a + b + \sqrt{a² + b²}) \\&= (12 × 5) + 20(12 + 15 + \sqrt{5² + 12²}) \\&= 60 + 20(27 + 13) \\&= 60 + 20(40) \\&= 60 + 800 \\&= 860~cm² \end{aligned}[/tex]
[tex]\begin{array}{lr}\texttt{Selamat Menunaikan Ibadah Puasa}\end{array}[/tex]
[tex]\boxed{\colorbox{ccddff}{Answered by Danial Alf'at | 03 - 04 - 2023}}[/tex]
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Penjelasan dengan langkah-langkah:
PythaGoras & LpBR
Sisi miring
= √(12²+5²)
= √(144+25)
= √169
= 13 cm
Maka luas permukaan
= 2la + Jum.Ls
= 2(½×12×5) + 20(12+15+13)
= 60 + 20(40)
= 60 + 800
= 860 cm²
Bangun Ruang Sisi Tegak
[Prisma Segitiga Siku-Siku]
..
[tex]\begin{aligned} LP &= 2(LA) + t(K) \\&= (a × b) + t(a + b + \sqrt{a² + b²}) \\&= (12 × 5) + 20(12 + 15 + \sqrt{5² + 12²}) \\&= 60 + 20(27 + 13) \\&= 60 + 20(40) \\&= 60 + 800 \\&= 860~cm² \end{aligned}[/tex]
[tex]\begin{array}{lr}\texttt{Selamat Menunaikan Ibadah Puasa}\end{array}[/tex]
[tex]\boxed{\colorbox{ccddff}{Answered by Danial Alf'at | 03 - 04 - 2023}}[/tex]