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Uploaded by Javier Armando Choque Sansuste

AREA DE UN PRISMA OCTOGONAL

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AREA DE UN PRISMA OCTOGONAL
En nuestro caso:
β„Ž = 15 π‘π‘š 𝑒𝑛 π‘™π‘’π‘”π‘Žπ‘Ÿ 𝑑𝑒 8π‘π‘š
𝑙 = 6π‘π‘š 𝑒𝑛 π‘™π‘’π‘”π‘Žπ‘Ÿ 𝑑𝑒 4π‘π‘š
π‘Žπ‘ = π‘Žπ‘π‘œπ‘‘π‘’π‘šπ‘Ž = 7.24 π‘π‘š 𝑒𝑛 π‘™π‘’π‘”π‘Žπ‘Ÿ 𝑑𝑒 3π‘π‘š
𝑃 = π‘ƒπ‘’π‘Ÿíπ‘šπ‘’π‘‘π‘Ÿπ‘œ 𝑑𝑒𝑙 𝑂𝑐𝑑áπ‘”π‘œπ‘›π‘œ = 8𝑙 = 8π‘₯6π‘π‘š = 48 π‘π‘š
𝐴. π΅π‘Žπ‘ π‘’ =
𝑃 π‘₯ π‘Žπ‘ 48 π‘π‘š π‘₯ 7.24 π‘π‘š
=
= 173.76 π‘π‘š2 π‘₯2 = 347.52 π‘π‘š2
2
2
𝐴. 𝐢. πΏπ‘Žπ‘‘π‘’π‘Ÿπ‘Žπ‘™ = 𝑏 π‘₯ β„Ž = 6 π‘₯ 15 = 90 π‘π‘š2
Como son 8 caras laterales:
𝐴. 𝐢. πΏπ‘Žπ‘‘π‘’π‘Ÿπ‘Žπ‘™π‘’π‘  = 8π‘₯90π‘π‘š2 = 720π‘π‘š2
Finalmente, el área total será:
π΄π‘π‘Ÿπ‘–π‘ π‘šπ‘Ž = 347.52 π‘π‘š2 + 720 π‘π‘š2
π‘¨π’‘π’“π’Šπ’”π’Žπ’‚ = πŸπŸŽπŸ”πŸ•. πŸ“πŸ π’„π’ŽπŸ
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