Question

Question 156
A playground is in the form of a rectangle ATEF. Two players are standing at the point F and B, where EF = EB. Find the values of x and y.

Answer 1

Given, a rectangle ATEF in which EF = EB. Then, $\Delta FEB$ is an isosceles triangle. Therefore, by the angle sum property of a triangle, we have

$\angle EFB+\angle EBF+\angle FEB={180}^{\circ }$ [angle sum property of triangle]

$\Rightarrow \angle EFB+\angle EBF+{90}^{\circ }={180}^{\circ }$ [$\because$ in a rectangle, each angle is of ${90}^{\circ }$]

$\Rightarrow 2\angle EFB={90}^{\circ }$ [$\because EB=EF\Rightarrow \angle EFB=\angle EBF$]

$\angle EFB={45}^{\circ }$ and $\angle EBF={45}^{\circ }$
Now, $\angle x={180}^{\circ }-{45}^{\circ }={135}^{\circ }$ [linear pair]

and $\angle EFB+\angle y={90}^{\circ }$ [$\because$ in a rectangle, each angle is of ${90}^{\circ }$]

$\Rightarrow \angle y={90}^{\circ }-{45}^{\circ }={45}^{\circ }$ .