Answer:
Given, E is the mid-point of AD \[\therefore AE=\frac{40}{2}=20\,m\] \[\angle A=90{}^\circ \] [Angle of a rectangle] \[\therefore \,\text{In}\,\,\text{rt}.\,\,\Delta \,BAE,\] \[E{{B}^{2}}=A{{B}^{2}}+A{{E}^{2}}\] [Pythagoras? theorem] \[={{(48)}^{2}}+{{(20)}^{2}}\] \[=2304+400=2704\] \[EB=\sqrt{2704}=52\,\,m\]
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