Quadrilaterals | IT

Question 5

Context: Let us check what quadrilateral we get if we draw the two diagonals such that their lengths are equal, they bisect each other and have an arbitrary angle, say $60^\circ$ , between them as shown in the figure to the right.

Q. Can you find all the remaining angles?

Solution

Understand the Question

A quadrilateral whose diagonals are equal in length and bisect each other is a rectangle.
The diagonals of a rectangle are equal and bisect each other into four equal segments ( $AO = BO = CO = DO$ ), forming four isosceles triangles with the sides of the rectangle.
Using properties of linear pairs, vertically opposite angles, and the angle sum property of triangles ( $180^\circ$ ), all remaining angles can be found.

Step 1 · Identify the Quadrilateral

Let the quadrilateral be $ABCD$ with diagonals $AC$ and $BD$ intersecting at point $O$ . Diagram 1

Given that the diagonals are equal and bisect each other: $AC = BD$ $AO = OC \quad \text{and} \quad BO = OD$

A quadrilateral with diagonals that bisect each other is a parallelogram, and a parallelogram with equal diagonals is a rectangle. Therefore, $ABCD$ is a rectangle.

Step 2 · Find Angles at Intersection Point $O$

Given $\angle AOB = 60^\circ$ .

Vertically opposite angles are equal:

\begin{aligned} \angle COD &= \angle AOB \\[0.6em] &= 60^\circ \end{aligned}

Angles $\angle AOB$ and $\angle BOC$ form a linear pair on line segment $AC$ :

\begin{aligned} \angle BOC &= 180^\circ - \angle AOB \\[0.6em] &= 180^\circ - 60^\circ \\[0.6em] &= 120^\circ \end{aligned}

Vertically opposite to $\angle BOC$ :

\begin{aligned} \angle DOA &= \angle BOC \\[0.6em] &= 120^\circ \end{aligned}

Step 3 · Find Angles in Triangles Formed by Diagonals

Since $AO = OC = BO = OD$ , the four triangles formed by the diagonals are isosceles triangles.

In $\triangle AOB$ ( $AO = BO$ ):

\begin{aligned} \angle OAB + \angle OBA + \angle AOB &= 180^\circ \\[0.6em] 2 \times \angle OAB + 60^\circ &= 180^\circ \\[0.6em] 2 \times \angle OAB &= 180^\circ - 60^\circ \\[0.6em] 2 \times \angle OAB &= 120^\circ \\[0.6em] \angle OAB &= \dfrac{120^\circ}{2} = 60^\circ \end{aligned}

$\angle OAB = \angle OBA = 60^\circ$

In $\triangle COD$ ( $CO = DO$ ):

\begin{aligned} \angle OCD + \angle ODC + \angle COD &= 180^\circ \\[0.6em] 2 \times \angle OCD + 60^\circ &= 180^\circ \\[0.6em] 2 \times \angle OCD &= 180^\circ - 60^\circ \\[0.6em] 2 \times \angle OCD &= 120^\circ \\[0.6em] \angle OCD &= \dfrac{120^\circ}{2} = 60^\circ \end{aligned}

$\angle OCD = \angle ODC = 60^\circ$

In $\triangle BOC$ ( $BO = CO$ ):

\begin{aligned} \angle OBC + \angle OCB + \angle BOC &= 180^\circ \\[0.6em] 2 \times \angle OBC + 120^\circ &= 180^\circ \\[0.6em] 2 \times \angle OBC &= 180^\circ - 120^\circ \\[0.6em] 2 \times \angle OBC &= 60^\circ \\[0.6em] \angle OBC &= \dfrac{60^\circ}{2} = 30^\circ \end{aligned}

$\angle OBC = \angle OCB = 30^\circ$

In $\triangle DOA$ ( $DO = AO$ ):

\begin{aligned} \angle ODA + \angle OAD + \angle DOA &= 180^\circ \\[0.6em] 2 \times \angle ODA + 120^\circ &= 180^\circ \\[0.6em] 2 \times \angle ODA &= 180^\circ - 120^\circ \\[0.6em] 2 \times \angle ODA &= 60^\circ \\[0.6em] \angle ODA &= \dfrac{60^\circ}{2} = 30^\circ \end{aligned}

$\angle ODA = \angle OAD = 30^\circ$

Step 4 · Verify Internal Angles of Rectangle $ABCD$

Since $ABCD$ is a rectangle, each vertex angle is $90^\circ$ :

\begin{aligned} \angle DAB &= \angle OAD + \angle OAB = 30^\circ + 60^\circ = 90^\circ \\[0.6em] \angle ABC &= \angle OBA + \angle OBC = 60^\circ + 30^\circ = 90^\circ \\[0.6em] \angle BCD &= \angle OCB + \angle OCD = 30^\circ + 60^\circ = 90^\circ \\[0.6em] \angle CDA &= \angle ODC + \angle ODA = 60^\circ + 30^\circ = 90^\circ \end{aligned}

Answer

Angles at intersection $O$ : $\angle COD = 60^\circ, \quad \angle BOC = 120^\circ, \quad \angle DOA = 120^\circ$

Base angles of the triangles: $\angle OAB = \angle OBA = \angle OCD = \angle ODC = 60^\circ$ $\angle OBC = \angle OCB = \angle ODA = \angle OAD = 30^\circ$

Angles of rectangle $ABCD$ : $\angle DAB = \angle ABC = \angle BCD = \angle CDA = 90^\circ$

Common Mistakes

Assuming Perpendicular Diagonals: Diagonals intersecting at $90^\circ$ form a rhombus or square; here the angle is given as $60^\circ$ , which confirms it is a non-square rectangle.
Overlooking Equal Halves: Forgetting that in a rectangle, diagonals bisect each other into four equal segments ( $AO = BO = CO = DO$ ), meaning the opposite triangles are equilateral (angles $60^\circ, 60^\circ, 60^\circ$ ) and the adjacent triangles are isosceles (angles $120^\circ, 30^\circ, 30^\circ$ ).

More questions in IT

Observe the following figures.

Figs. (i), (ii), and (iii) are quadrilaterals, and the others are not. Why?

Are there other ways to define a rectangle?

A Carpenter's Problem

A carpenter needs to put together two thin strips of wood, as shown in Fig. 1, so that when a thread is passed through their endpoints, it forms a rectangle. She already has one 8 cm long strip. What should be the length of the other strip? Where should they both be joined?

Let us first model the structure that the carpenter has to make. The strips can be modelled as line segments. They are the diagonals of the quadrilateral formed by their endpoints. For the quadrilateral to be a rectangle, we need to answer the following questions —

What is the length of the other diagonal?
What is the point of intersection of the two diagonals?
What should the angle be between the diagonals?

Can the following equalities be used to establish that $\Delta AOD \cong \Delta COB$ ?

$AO = CO$ (proved above)
$\angle AOB = \angle COD$ (vertically opposite angles)
$AD = CB$

Q. Can you find all the remaining angles?

Context: In $\Delta AOB$ , since $OA = OB$ , the angles opposite them are equal, say $a$ .

Q. Can you find the value of $a$ ?

Can we now identify what type of quadrilateral ABCD is?

Notice that its angles all add up to $90^\circ$ ( $30^\circ + 60^\circ$ ).

What can we say about its sides?

Will $ABCD$ remain a rectangle if the angles between the diagonals are changed? Can we generalise this?

Take one of the angles between the diagonals as $x$ .

Q10

Context: We can compute the four angles between the diagonals to be $x$ , $x$ , $180^\circ - x$ , and $180^\circ - x$ .

Q. Can you find the other angles?

Q11

Context: Since we know that $\triangle AOB$ is isosceles, we can denote the measures of both of its base angles by $a$ .

Q. What is the value of $a$ (in degrees) in terms of $x$ ?

Q12

Context: Thus, all four angles of the quadrilateral $ABCD$ are $90^\circ$ .

Q. What can we say about $AB$ and $CD$ , and $AD$ and $BC$ ?

Q13

In the earlier definition, we stated that a rectangle has (a) opposite sides of equal length, and (b) all angles equal to $90^\circ$ . Would we be wrong if we just define a rectangle as a quadrilateral in which all the angles are $90^\circ$ ?

Q14

If you think that this definition is incomplete, try constructing a quadrilateral in which the angles are all $90^\circ$ but the opposite sides are not equal.

Are you able to construct such a quadrilateral?

Q15

Is it wrong to write $\Delta BAD \cong \Delta CDB$ ? Why?

Q16

Can you similarly show that $\text{AB}$ is parallel to $\text{DC}$ ( $\text{AB} \parallel \text{DC}$ )?

Q17

In the quadrilaterals below, are there any non-rectangles?

Q18

Let us consider the Carpenter's Problem again. If the wooden strips have to be placed such that the thread passing through their endpoints forms a square, what must be done?

Q19

What more needs to be done to get equal sidelengths as well? Can this be achieved by properly choosing the angle between the diagonals? See if you can reason and/or experiment to figure this out!

Q20

Can this be used to find the angles $\angle\text{BOA}$ and $\angle\text{BOC}$ formed by the diagonals?

Q21

Context: The diagonals of a square are of equal lengths and bisect each other at right angles.

Q. Using this fact, construct a square with a diagonal of length $8\text{ cm}$ .

Q22

Context: Since a square is a special type of rectangle, all the properties of a rectangle hold true for a square.

Q. Verify if this is true by going through geometric reasoning in Deduction 1 and Deduction 2, and see if they apply to a square as well.

Q23

Q. Similarly, find $\angle 2$ and $\angle 4$ .

Q24

4.2 Angles in a Quadrilateral

Is it possible to construct a quadrilateral with three angles equal to $90^\circ$ and the fourth angle not equal to $90^\circ$ ?

Q25

But why not?

Q26

Are there quadrilaterals that have parallel opposite sides that are not rectangles?

Q27

Construct such a figure by recalling how parallel lines can be constructed using a ruler and a set-square, or a compass and a ruler.

Q28

Context: Consider a parallelogram $ABCD$ with adjacent sides of lengths $4\text{ cm}$ and $5\text{ cm}$ , and an angle of $30^\circ$ between them.

Q. What are the remaining angles of the parallelogram? What are the lengths of the remaining sides? See if you can reason out and/or experiment to figure these out.

Q29

Deduction 7— What can we say about the sides of a parallelogram?

By looking at a parallelogram, it appears that the opposite sides are equal. Can we again use congruence to show this? Which two triangles can be considered for this?

Q30

Is it wrong to write $\Delta \text{ABD} \cong \Delta \text{CBD}$ ? Why?

Q31

Are the diagonals of a parallelogram always equal? Check with the parallelogram that you have constructed.

Q32

Context: We see that the diagonals of a parallelogram need not be equal.

Q. Do they bisect each other (do they intersect at their midpoints)? Reason and/or experiment to figure this out.

Q33

Is it wrong to write $\Delta \text{AOE} \cong \Delta \text{SOY}$ ? Why?

Q34

Do the diagonals of a parallelogram intersect at a particular angle?

Q35

What are the other angles of the rhombus $ABCD$ that we have constructed? Reason and/or experiment to figure this out.

Q36

It can be seen that $\Delta GAE \cong \Delta MAE$ (How?)

Q37

So a rhombus is a parallelogram, and a rectangle is also a parallelogram. How can this be represented using a Venn diagram?

Q38

Where will the set of squares occur in this diagram?

Q39

Are the diagonals of a rhombus equal?

Q40

Do the diagonals of a rhombus intersect at any particular angle? Reason out and/or experiment to figure this out!

Q41

In the rhombus GAME, we have $\Delta \text{GEO} \cong \Delta \text{MEO}$ (why?).

Q42

In the kite, show that the diagonal $BD$

(i) bisects $\angle ABC$ and $\angle ADC$ ,

(ii) bisects the diagonal $AC$ , that is, $AO = OC$ , and is perpendicular to it.

Hint: Is $\triangle AOB \cong \triangle COB$ ?

Q43

Construct a trapezium. Measure the base angles (marked in the figure).

Q44

Can you find the remaining angles without measuring them?

Q45

How do we construct an isosceles trapezium?

Q46

Construct an isosceles trapezium $UVWX$ , with $UV \parallel XW$ . Measure $\angle U$ .

Q47

Now, it can be shown that $\Delta \text{UXY} \cong \Delta \text{VWZ}$ . (How?)