Question

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

Answer 1

Answer (i) ABCD is a rectangle.
The diagonals of a rectangle are congruent and bisect each other. Therefore, in​ $△AOB$ we have:
OA = OB
$\therefore \angle OAB=\angle OBA={35}^0$
$\therefore x={90}^0-{35}^0={55}^0$
And $\angle AOB={180}^0-({35}^0+{35}^0)={110}^0$
$\therefore y=AOB={110}^0$ ​
[Vertically opposite angles]
Hence, $x={55}^{0}andy={110}^0$
(ii) In $△AOB$ we have:
OA = OB
Now, $\angle OAB=\angle OBA=12\times {180}^0-{110}^0={35}^0$
$\therefore$ y = $\angle BAC={35}^0$
[Interior alternate angles]
Also, $x={90}^0-y$
[ $\therefore \angle C={90}^0=x+y$ ]
⇒​$x={90}^0-{35}^0={55}^0$
Hence, $x={55}^0$ and $y={35}^0$ ​