Question

Alguém me ajuda a responder essas duas questões?!
Por favooooor

Answer 1

Por proporção:

$\frac{x}{10} = \frac{y-x}{17,5-10} \\\\ \frac{x}{10} = \frac{9}{7,5} \\\\ x = 10 \cdot \frac{9}{7,5} \\\\ x = \frac{90}{7,5} \\\\ \boxed{x = 12} \\\\\\ \Rightarrow y-x = 9 \\\\ y-12 = 9 \\\\ y = 9+12 \\\\ \boxed{y = 21}$


b) 
$y = x-15 \therefore x-y = 15 \\\\\\ \frac{y}{33,6} = \frac{x-y}{(18+33,6)-33,6} \\\\ \frac{y}{33,6} = \frac{15}{18} \\\\ y = \frac{33,6 \cdot 15}{18} \\\\ y = \frac{504}{18} \\\\ \boxed{y = 28} \\\\\\ \Rightarrow x-y = 15 \\\\ x-28 = 15 \\\\ x = 15+28 \\\\ \boxed{x = 43}$

Answer 2

Em ambos os exercícios, vamos aplicar o conceito de proporcionalidade do Teorema de Tales.

a) $y-x=9$ ⇒ $y=x+9$

$\dfrac{x}{y} = \dfrac{10}{17,5}$

$10y=17,5x$

$10.(x+9)=17,5x$

$10x+90=17,5x$

$10x-17,5x=-90$

$-7,5x=-90$

$x= \dfrac{-90}{-7,5}= \dfrac{90}{7,5}= \dfrac{900:25}{75:25}= \dfrac{36}{3}=12\text{ cm}$

$y=x+9=12+9=21\text{ cm}$

b) $y=x-15$

$\dfrac{x}{y} = \dfrac{18+33,6}{33,6}$

$\dfrac{x}{y} = \dfrac{51,6}{33,6}$

$51,6y=33,6x$

$51,6.(x-15)=33,6x$

$51,6x-774=33,6x$

$51,6x-33,6x=774$

$18x=774$

$x= \dfrac{774}{18} = \dfrac{387}{9} = \dfrac{129}{3} =43\text{ cm}$

$y=x-15=43-15=28\text{ cm}$