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Exercício de ARESAS E PERIMETROS
por LUCAS OLIVEIRA (FIXAS DO PROFESSOR TOMAZ ( E.T.R ESCOLA TECNICA REGIONAL REVISÃO PRA ENTRA NO ASSUNTO DE MATEMATICA FINACEIRA) CRUSO LOGISTICA)
Normal Tuesday, February 17th, 2009
Exercício:
UM LOSANGO E UM QUADRADO TEM ÁREAS IGUAIS. UMA DAS DIAGONAIS DO LOSANGO É O TRIPLO DA OUTRA.O PERIMETRO DO QUADRADO MEDE 24M. CALCULE O COMPRIMENTO DA DIAGONAL MAIOR?
primeiro, vamos denominar os lados e diagonais dos poligonos.
a diagonal menor do losango é x, e a maior 3x pois é o triplo da outra, e como todos os lados do quadrado são iguais, chamemos todos de y.
o perimetro do quadrado, que é a soma do seus lados nada mais é do que 4y=24, portanto y=6. Se o lado equivale a 6m, a área do quadrado é 36m².
Se a área do quadrado equivale a do losango podemos dizer que a área do losango é 36m² também. Ou seja, A=36m².
A fórmula de área do losango é A=Diagonal maior x diagonal menor/2.
Sabendo que A=36, que a diagonal maior do losango é 3x e a menor x, então 36=4x/2, a resposta é x=18m. Sabemos, então, a resposta da diagonal menor, se a diagonal maior é o triplo dela, portanto seu valor é 54m.
Oii.. por favor, alguem poderia me dizer o que foi feito do 4??
Pois, sendo 36=4x/2 vai ficar 4x=36/2 4x=18 e pra ficar x=18.. onde foi o 4?? hehe
Agradeço desde já!
vai ficar 36*2/4 é só inverte
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