Question

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Answer 1

Explicação passo-a-passo:

Usando o Teorema de Tales:

Calculando x:

$\dfrac{x}{21} =\dfrac{4}{4+3}\\\\ \dfrac{x}{21} =\dfrac{4}{7}\\\\7x=21.4\\\\7x=84\\\\x=\dfrac{84}{7}\\\\\boxed{x=12}$

Calculando y:

$\dfrac{y}{21} =\dfrac{3}{4+3}\\\\ \dfrac{y}{21} =\dfrac{3}{7}\\\\7y=21.3\\\\7y=63\\\\y=\dfrac{63}{7}\\\\\boxed{y=9}$

Calculando $x^{2} -y^{2}$:

$x^{2} -y^{2} =12^{2} -9^{2} \\\\x^{2} -y^{2} =144-81\\\\\boxed{\boxed{x^{2} -y^{2} =63}}$

Answer 2

Explicação passo-a-passo:

=> Valor de x

Pelo Teorema de Tales:

$\sf \dfrac{x}{x+y}=\dfrac{4}{4+3}$

$\sf \dfrac{x}{21}=\dfrac{4}{7}$

$\sf 7\cdot x=21\cdot4$

$\sf 7\cdot x=84$

$\sf x=\dfrac{84}{7}$

$\sf \red{x=12}$

=> Valor de y

Pelo Teorema de Tales:

$\sf \dfrac{y}{x+y}=\dfrac{3}{4+3}$

$\sf \dfrac{y}{21}=\dfrac{3}{7}$

$\sf 7\cdot y=21\cdot3$

$\sf 7\cdot y=63$

$\sf y=\dfrac{63}{7}$

$\sf \red{y=9}$

Assim:

$\sf x^2-y^2=(x+y)\cdot(x-y)$

$\sf x^2-y^2=(12+9)\cdot(12-9)$

$\sf x^2-y^2=21\cdot3$

$\sf \large\red{x^2-y^2=63}$