Quadrilaterals | FIO

Question 1

Find all the other angles inside the following rectangles.

Solution

Understand the Question

In any rectangle:
- All four interior vertex angles are $90^\circ$ .
- Diagonals are equal in length and bisect each other ( $OA = OB = OC = OD$ ), creating four isosceles triangles around the center $O$ .
- The base angles of each isosceles triangle are equal.
- Opposite sides are parallel, so alternate interior angles made by the diagonals are equal.
- Angles forming a linear pair sum to $180^\circ$ , and vertically opposite angles are equal.

(i) Find all the other angles inside rectangle $\text{ABCD}$ .

Step 1 · Find the Angles in Rectangle ABCD

Diagram 1

Given $\angle 1 = \angle CAB = 30^\circ$ .

Since $\angle DAB = 90^\circ$

\begin{aligned} \angle 2 &= \angle DAB - \angle CAB \\ &= 90^\circ - 30^\circ = 60^\circ \end{aligned}

Since $AD \parallel BC$ , alternate interior angles are equal $\angle 6 = \angle DAC = 60^\circ$

Since $AB \parallel DC$ , alternate interior angles are equal $\angle 5 = \angle CAB = 30^\circ$

In $\triangle AOB$ , $OA = OB$ (diagonals of a rectangle are equal and bisect each other) $\angle 8 = \angle OAB = 30^\circ$

By angle sum property of $\triangle AOB$

\begin{aligned} \angle 9 &= 180^\circ - (\angle OAB + \angle OBA) \\ &= 180^\circ - (30^\circ + 30^\circ) \\ &= 180^\circ - 60^\circ = 120^\circ \end{aligned}

Vertically opposite angles are equal $\angle 11 = \angle AOB = 120^\circ$

Linear pair on diagonal $AC$ $\angle 10 = 180^\circ - 120^\circ = 60^\circ$

Vertically opposite angles are equal $\angle 12 = \angle BOC = 60^\circ$

In $\triangle BOC$ , $OB = OC$ $\angle 7 = \angle OCB = 60^\circ$

In $\triangle COD$ , $OC = OD$ $\angle 4 = \angle OCD = 30^\circ$

In $\triangle DOA$ , $OD = OA$ $\angle 3 = \angle OAD = 60^\circ$

Answer

(i) $\angle 1 = 30^\circ, \angle 2 = 60^\circ, \angle 3 = 60^\circ, \angle 4 = 30^\circ, \angle 5 = 30^\circ, \angle 6 = 60^\circ, \angle 7 = 60^\circ, \angle 8 = 30^\circ, \angle 9 = 120^\circ, \angle 10 = 60^\circ, \angle 11 = 120^\circ, \angle 12 = 60^\circ$

(ii) Find all the other angles inside rectangle $\text{PSRQ}$ .

Step 1 · Find the Angles in Rectangle PSRQ

Diagram 2

Given $\angle 9 = \angle QOR = 110^\circ$ .

Vertically opposite angles are equal $\angle 11 = \angle QOR = 110^\circ$

Linear pair on straight line $QS$ $\angle 10 = 180^\circ - 110^\circ = 70^\circ$

Vertically opposite angles are equal $\angle 12 = \angle QOP = 70^\circ$

In $\triangle POS$ , $OP = OS$ , so $\angle OPS = \angle OSP$

\begin{aligned} \angle OPS + \angle OSP + \angle POS &= 180^\circ \\ 2 \times \angle OPS + 110^\circ &= 180^\circ \\ 2 \times \angle OPS &= 180^\circ - 110^\circ \\ 2 \times \angle OPS &= 70^\circ \\ \angle OPS &= 35^\circ \end{aligned}

$\angle 1 = 35^\circ, \quad \angle 8 = \angle OPS = 35^\circ$

Since corner $\angle QPS = 90^\circ$

\begin{aligned} \angle 2 &= \angle QPS - \angle OPS \\ &= 90^\circ - 35^\circ = 55^\circ \end{aligned}

In $\triangle QOP$ , $OQ = OP$ $\angle 3 = \angle OPQ = 55^\circ$

Since corner $\angle PQR = 90^\circ$

\begin{aligned} \angle 4 &= \angle PQR - \angle PQS \\ &= 90^\circ - 55^\circ = 35^\circ \end{aligned}

In $\triangle QOR$ , $OQ = OR$ $\angle 5 = \angle OQR = 35^\circ$

Since corner $\angle QRS = 90^\circ$

\begin{aligned} \angle 6 &= \angle QRS - \angle ORQ \\ &= 90^\circ - 35^\circ = 55^\circ \end{aligned}

In $\triangle ROS$ , $OR = OS$ $\angle 7 = \angle ORS = 55^\circ$

Answer

(ii) $\angle 1 = 35^\circ, \angle 2 = 55^\circ, \angle 3 = 55^\circ, \angle 4 = 35^\circ, \angle 5 = 35^\circ, \angle 6 = 55^\circ, \angle 7 = 55^\circ, \angle 8 = 35^\circ, \angle 9 = 110^\circ, \angle 10 = 70^\circ, \angle 11 = 110^\circ, \angle 12 = 70^\circ$

Common Mistakes

Assuming Diagonals are Perpendicular: Diagonals of a rectangle are not perpendicular to each other ( $90^\circ$ ) unless the rectangle is a square.
Base Angles of Triangles: Forgetting that diagonals bisect each other into equal halves, which makes each of the four triangles isosceles with equal base angles.

Question 1

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