MATH SOLVE

5 months ago

Q:
# Need help badly don’t have answer to this problem.

Accepted Solution

A:

there are a couple of ways to tackle this one, using the 45-45-90 rule or just using the pythagorean theore, let's use the pythagorean theorem.

the angle at A is 45°, and its opposite side is BC, the angle at C is 45° as well, and its opposite side is AB, well, the angles are the same, thus BC = AB.

hmmm le'ts call hmmm ohh hmmm say z, thus BC = AB = z.

[tex]\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2 \implies (6\sqrt{2})^2=z^2+z^2 \qquad \begin{cases} c=\stackrel{6\sqrt{2}}{hypotenuse}\\ a=\stackrel{z}{adjacent}\\ b=\stackrel{z}{opposite}\\ \end{cases}\\\\\\ 6^2(\sqrt{2})^2=2z^2\implies 36(2)=2z^2\implies \cfrac{36(2)}{2}=z^2\implies 36=z^2 \\\\\\ \sqrt{36}=z\implies 6=z[/tex]

the angle at A is 45°, and its opposite side is BC, the angle at C is 45° as well, and its opposite side is AB, well, the angles are the same, thus BC = AB.

hmmm le'ts call hmmm ohh hmmm say z, thus BC = AB = z.

[tex]\bf \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2 \implies (6\sqrt{2})^2=z^2+z^2 \qquad \begin{cases} c=\stackrel{6\sqrt{2}}{hypotenuse}\\ a=\stackrel{z}{adjacent}\\ b=\stackrel{z}{opposite}\\ \end{cases}\\\\\\ 6^2(\sqrt{2})^2=2z^2\implies 36(2)=2z^2\implies \cfrac{36(2)}{2}=z^2\implies 36=z^2 \\\\\\ \sqrt{36}=z\implies 6=z[/tex]