Consider rectangel CREB that is formed of squares COAB and OREA, where AE=4ft. Assume squares COAB and OREA are congruent. BO=3m-2nAR=7(m-1)-4.8nDetermine the values of m and n and the length of RB.

tinfoQ 2021-01-04 Answered

Consider rectangel CREB that is formed of squares COAB and OREA, where \(AE=4ft\). Assume squares COAB and OREA are congruent.
\(BO=3m-2n\)
\(AR=7(m-1)-4.8n\)
Determine the values of m and n and the length of RB.

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Dora
Answered 2021-01-05 Author has 26888 answers

The diagonal of each square divides it into two triangles with angles of \(\displaystyle{45}^{\circ},{45}^{\circ}{\quad\text{and}\quad}{90}^{\circ}\). The diagonals (BO and AR) are the triangles' hypotenuses. The hypotenuses is \(\displaystyle\sqrt{{2}}\) times the leg in such triangles. A leg is also a square side, which is 4ft, so
\(\displaystyle{B}{O}={A}{R}={4}\sqrt{{2}}\)
\(\displaystyle{3}{m}-{2}{n}={7}{\left({m}-{1}\right)}-{4.8}{n}={4}\sqrt{{2}}\)
\(\displaystyle{3}{m}-{2}{n}={4}\sqrt{{2}}\)
\(\displaystyle-{2}{n}=-{3}{n}+{4}\sqrt{{2}}\)
\(\displaystyle{n}={1.5}{m}-{2}\sqrt{{2}}\)
\(\displaystyle{7}{\left({m}-{1}\right)}-{4.8}{n}={4}\sqrt{{2}}\)
\(\displaystyle{7}{\left({m}-{1}\right)}-{4.8}{\left({1.5}{m}-{2}\sqrt{{2}}\right)}={4}\sqrt{{2}}\)
\(\displaystyle{7}{m}-{7}-{7.2}{m}+{9.6}\sqrt{{2}}={4}\sqrt{{2}}\)
\(\displaystyle-{0.2}{m}={4}\sqrt{{2}}+{7}-{9.6}\sqrt{{2}}\)
\(\displaystyle-{0.2}{m}=-{5.6}\sqrt{{2}}+{7}\)
\(\displaystyle{m}={28}\sqrt{{2}}-{35}\)
\(\displaystyle{n}={1.5}{m}-{2}\sqrt{{2}}\)
\(\displaystyle={1.5}{\left({28}\sqrt{{2}}-{35}\right)}-{2}\sqrt{{2}}\)
\(\displaystyle={42}\sqrt{{2}}-{52.5}-{2}\sqrt{{2}}\)
\(\displaystyle={40}\sqrt{{2}}-{52.5}\)
\(\displaystyle{R}{B}^{{2}}={B}{A}^{{2}}+{A}{E}^{{2}}\)
\(\displaystyle={8}^{{2}}+{4}^{{2}}\)
\(=64+16\)
\(=80\)
\(\displaystyle{R}{B}=\sqrt{{80}}\)
\(\displaystyle={4}\sqrt{{5}}\)

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