Question

utilizando o Teorema de Pitágoras determine o valor de X nos triângulos acima:​

Answer 1

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$\rm c)\\\large\boxed{\begin{array}{l}\sf(x+1)^2=(\sqrt{7})^2+x^2\\\sf \diagup\!\!\!\!\!x^2+2x+1=7+\diagup\!\!\!\!\!x^2\\\sf 2x=7-1\\\sf 2x=6\\\sf x=\dfrac{6}{2}\\\huge\boxed{\boxed{\boxed{\boxed{\sf x=3}}}}\end{array}}\blue{\checkmark}$

$\rm d)\\\Large\boxed{\begin{array}{l}\sf x^2+x^2=(3\sqrt{2})^2\\\sf \diagup\!\!\!2x^2=9\cdot\diagup\!\!\!2\\\sf x=\sqrt{9}\\\boxed{\boxed{\boxed{\boxed{\sf x=3}}}}\end{array}}\blue{\checkmark}$