Question

In each of the figures given below, $ABCD$ is a rectangle. Find the values of $x$ and $y$ in each case.

Answer 1

(i) $ABCD$ is a rectangle.
The diagonals of a rectangle are congruent and bisect each other. Therefore, in​ $△AOB$ we have:
$OA=OB$
$\therefore \angle OAB=\angle OBA={35}^{\circ }$ [Angles opposite to equal sides are equal]
$\therefore x={90}^{\circ }-{35}^{\circ }={55}^{\circ }$ [Each angle of rectangle is ${90}^{\circ }$]
And $\angle AOB={180}^{\circ }-({35}^{\circ }+{35}^{\circ })={110}^{\circ }$ [By Angle Sum Property]
$\therefore y=AOB={110}^{\circ }$ ​[Vertically opposite angles]
Hence, $x={55}^{\circ }\text{and}y={110}^{\circ }$
(ii) In $△AOB$ we have:
$OA=OB$
Now, $\angle OAB=\angle OBA=\frac{1}{2}\times ({180}^{\circ }-{110}^{\circ })=\frac{1}{2}\times {70}^{\circ }={35}^{\circ }$
$\therefore y=\angle BAC={35}^{\circ }$ [Alternate interior angles]
Also, $x={90}^{\circ }-y$ [ $\because \angle C={90}^{\circ }=x+y$ ]
⇒​$x={90}^{\circ }-{35}^{\circ }={55}^{\circ }$
Hence, $x={55}^{\circ }$ and $y={35}^{\circ }$ ​