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Zina [86]
3 years ago
12

Tienes un triángulo equilátero con lados de 10cm, lo divides en dos partes de manera horizontal tal que el área de ambos figuras

resultantes es igual, ¿cuánto vale la distancia del lado x?

Mathematics
1 answer:
Tanya [424]3 years ago
6 0

Answer:

x = 10/√2 ≈ 7.07

Step-by-step explanation:

Comenzaremos por dividir el triángulo en dos partes y definir H, como en la figura adjunta.

Aplicando el teorema de Tales, sabemos que:

\dfrac{l/2}{H}=\dfrac{x/2}{h}\\\\\\\dfrac{l}{H}=\dfrac{x}{h}

También sabemos que, dado que el tirángulo menor es la mitad que el triángulo mayor, la relación entre áreas es:

\dfrac{A}{A_x}=\dfrac{lH/2}{xh/2}=\dfrac{lH}{xh}=2

Dado que formamos dos triángulos rectángulos, podemos despejar el valor de H como:

(l/2)^2+H^2=l^2\\\\H^2=l^2-(l/2)^2=10^2-5^2=100-25=75\\\\H^2=\sqrt{75}

Podemos entonces despejar x de la siguiente manera:

h=\dfrac{H}{l}\cdot x=\dfrac{lH}{2x}\\\\\\2x^2\dfrac{H}{l}=lH\\\\\\2x^2=l^2\\\\\\x=\dfrac{l}{\sqrt{2}}=\dfrac{10}{\sqrt{2}}\approx7.07

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Step-by-step explanation:

Given data

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5 0
2 years ago
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C + a = 132.....a = 132 - c
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5.10c + 9.60(132 - c) = 898.20
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