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Two triangles are similar if at least two angles of one triangle are equal to any two angles of the other triangle; this is known as Angle-Angle (AA) similarity.

Two triangles are similar if any two sides of the triangles are in proportion and the angles between those sides are equal; this is known as Side-Angle-Side (SAS) similarity.

According to the question we get the required figure

Let A be the place where her father is standing

B is the position of boat

C is the place where girl is standing

D is the position of watercraft

E is the position of swimmer

Here we have to find the distance ED

Here, BC = x meter; AB = (x-10) meter

Consider ∆ABC and ∆DBE

$\angle ABC = \angle DBE$ (Vertically opposite angles)

$\angle BAC = \angle BDE$ (Alternate angles)

Using AA similarity, $\Delta ABC \sim \Delta DBE$

$\dfrac{{AB}}{{DB}} = \dfrac{{BC}}{{BE}} = \dfrac{{AC}}{{ED}}$

So,$\dfrac{{AB}}{{DB}} = \dfrac{{BC}}{{BE}}$

From the figure

$\begin{gathered}

\dfrac{{x - 10}}{{98}} = \dfrac{x}{{126}} \\

126x - 1260 = 98x \\

28x = 1260 \\

x = 45m \\

\end{gathered} $

∴BC = 45 m

Also, $\dfrac{{BC}}{{BE}} = \dfrac{{AC}}{{ED}}$

$\begin{gathered}

ED = \dfrac{{AC \times BE}}{{BC}} \\

ED = \dfrac{{50 \times 126}}{{45}} \\

ED = 140m \\

\end{gathered} $

∴Man has to travel 140 m to rescue the swimmer