¿calcula el area de un triangulo isosceles cuyos lados iguales miden 8 cm y el angulo desigual mide 45 grados?

Lo divido desde la mitad de la base al angulo superior, en 2 triángulos rectángulos, Separo unoa. A=19° B=71° C=90° a=? (mitad de la base) b=?

(altura) c=16dm Area de nuestro triángulo=(base x altura)/2 a=c sen A a=16 x 0.3256 a=5.2dm b=Raiz de c^2-a^2 b=Raiz de 256-27.04 b=Raiz de 228.96 b=15.13dm Area=(5.2 x 15.13)/2 Area=78.676/2 Area=39.338dm^2 como son 2 triángulos: Area del triángulo isosceles=78.676dm^2 Perimetro=10.4+16+16=42.4dm.

Los ángulos iguales miden: 180º - 38º = 142º entre los dos. 142º/2 = 71º cada uno. Por la fórmula del Seno: sen(38º) / 16 = sen(71º) / x Despejando x = 16por sen71 / sen36º x = 25.74 El perímetro: 16 + 25.74 + 25.74 = 67.48 Para el área: la altura del triángulo (por Pitágoras): altura h^2 = 25.74^2 - 8^2 h^2 = 598.5 h = Raíz cuadrada (598.5 h = 24.46 Área = (16 por 24.46) / 2 Área = 195.72 dm^2.

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