OA: B

Given AB=6 ,BC = 8 , AC =10

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As can be seen from figure above, all four \(\triangle\) i.e. \(\triangle\)ABC ,\(\triangle\)ARQ ,\(\triangle\)QBP ,\(\triangle\) PSC are similar.

Because of AA similarity as their corresponding angles are equal in measure.

Considering \(\triangle\) ABC and \(\triangle\) QBP ,we get

\(\frac{BP}{BC}=\frac{BQ}{AB}=\frac{PQ}{AC} = k\) (a constant)

We get \(QP=10k\) and \(BQ=6k\)

As all sides of a square are equal,QR will be also equal to QP i.e \(QR =10k\).

Considering \(\triangle\) ABC and \(\triangle\) QBP ,we get

\(\frac{QR}{BC}=\frac{AR}{AB}=\frac{AQ}{AC} = p\) (a constant)

\(\frac{QR}{BC}=\frac{AQ}{AC}\)

\(\frac{{10k}}{8}=\frac{AQ}{10}\)

\(AQ =\frac{{100k}}{8}=\frac{{25k}}{2}\)

\(AB =BQ+AQ\)

\(6= 6k + \frac{{25k}}{2}\)

\(k = \frac{12}{37}\)

Side of Square\(= 10k = 10* \frac{12}{37} =\frac{120}{37}\)

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