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

When the diagonals of a quadrilateral are equal in length and bisect each other, the quadrilateral is a rectangle.

Step 1 — Identify the quadrilateral

Let the quadrilateral be ABCD. Its diagonals are AC and BD, intersecting at point O. We are given that the diagonals are equal, so AC = BD. We are given that the diagonals bisect each other, so AO = OC and BO = OD. A quadrilateral with diagonals that bisect each other is a parallelogram. A parallelogram with equal diagonals is a rectangle. So, the quadrilateral ABCD is a rectangle.

Diagram 1

Step 2 — Find angles at the intersection of diagonals

We are given that the angle between the diagonals is $60^\circ$ . So, $\angle AOB = 60^\circ$ . Angles $\angle AOB$ and $\angle COD$ are vertically opposite angles. Vertically opposite angles are equal. $\angle COD = \angle AOB$ $= 60^\circ$ Angles $\angle AOB$ and $\angle BOC$ form a linear pair on the straight line AC. A linear pair of angles adds up to $180^\circ$ . $\angle BOC = 180^\circ - \angle AOB$ $= 180^\circ - 60^\circ$

$\boxed{\angle BOC = 120^\circ}$ Angles $\angle BOC$ and $\angle DOA$ are vertically opposite angles. $\angle DOA = \angle BOC$ $= 120^\circ$

Step 3 — Find angles within the triangles formed by the diagonals

In a rectangle, the diagonals are equal and bisect each other. This means that AO = OC = BO = OD. So, the four triangles formed by the diagonals ( $\triangle AOB, \triangle BOC, \triangle COD, \triangle DOA$ ) are isosceles triangles.

Let us find the angles in $\triangle AOB$ . We know AO = BO and $\angle AOB = 60^\circ$ . Since AO = BO, $\triangle AOB$ is an isosceles triangle. The base angles $\angle OAB$ and $\angle OBA$ are equal. The sum of angles in a triangle is $180^\circ$ . $\angle OAB + \angle OBA + \angle AOB = 180^\circ$ $2 \times \angle OAB + 60^\circ = 180^\circ$ $2 \times \angle OAB = 180^\circ - 60^\circ$ $2 \times \angle OAB = 120^\circ$ $\angle OAB = \frac{120^\circ}{2}$

$\boxed{\angle OAB = 60^\circ}$ Since $\angle OAB = \angle OBA$ , we also have $\angle OBA = 60^\circ$ . Because all three angles are $60^\circ$ , $\triangle AOB$ is an equilateral triangle.

Let us find the angles in $\triangle COD$ . We know CO = DO and $\angle COD = 60^\circ$ . Since CO = DO, $\triangle COD$ is an isosceles triangle. The base angles $\angle OCD$ and $\angle ODC$ are equal. The sum of angles in a triangle is $180^\circ$ . $\angle OCD + \angle ODC + \angle COD = 180^\circ$ $2 \times \angle OCD + 60^\circ = 180^\circ$ $2 \times \angle OCD = 120^\circ$ $\angle OCD = \frac{120^\circ}{2}$

$\boxed{\angle OCD = 60^\circ}$ Since $\angle OCD = \angle ODC$ , we also have $\angle ODC = 60^\circ$ . Because all three angles are $60^\circ$ , $\triangle COD$ is an equilateral triangle.

Let us find the angles in $\triangle BOC$ . We know BO = CO and $\angle BOC = 120^\circ$ . Since BO = CO, $\triangle BOC$ is an isosceles triangle. The base angles $\angle OBC$ and $\angle OCB$ are equal. The sum of angles in a triangle is $180^\circ$ . $\angle OBC + \angle OCB + \angle BOC = 180^\circ$ $2 \times \angle OBC + 120^\circ = 180^\circ$ $2 \times \angle OBC = 180^\circ - 120^\circ$ $2 \times \angle OBC = 60^\circ$ $\angle OBC = \frac{60^\circ}{2}$

$\boxed{\angle OBC = 30^\circ}$ Since $\angle OBC = \angle OCB$ , we also have $\angle OCB = 30^\circ$ .

Let us find the angles in $\triangle DOA$ . We know DO = AO and $\angle DOA = 120^\circ$ . Since DO = AO, $\triangle DOA$ is an isosceles triangle. The base angles $\angle ODA$ and $\angle OAD$ are equal. The sum of angles in a triangle is $180^\circ$ . $\angle ODA + \angle OAD + \angle DOA = 180^\circ$ $2 \times \angle ODA + 120^\circ = 180^\circ$ $2 \times \angle ODA = 60^\circ$ $\angle ODA = \frac{60^\circ}{2}$

$\boxed{\angle ODA = 30^\circ}$ Since $\angle ODA = \angle OAD$ , we also have $\angle OAD = 30^\circ$ .

Step 4 — Find the angles of the rectangle ABCD

Since ABCD is a rectangle, all its internal angles are $90^\circ$ . Let us verify this using the angles we found. $\angle DAB = \angle OAD + \angle OAB$ $= 30^\circ + 60^\circ$

$\boxed{\angle DAB = 90^\circ}$ $\angle ABC = \angle OBA + \angle OBC$ $= 60^\circ + 30^\circ$ $\boxed{\angle ABC = 90^\circ}$ $\angle BCD = \angle OCB + \angle OCD$ $= 30^\circ + 60^\circ$ $\boxed{\angle BCD = 90^\circ}$ $\angle CDA = \angle ODC + \angle ODA$ $= 60^\circ + 30^\circ$ $\boxed{\angle CDA = 90^\circ}$

Answer

The remaining angles are:

(i) Angles at the intersection O: $\angle COD = \mathbf{60^\circ}$ $\angle BOC = \mathbf{120^\circ}$ $\angle DOA = \mathbf{120^\circ}$

(ii) Angles within the triangles formed by the diagonals: $\angle OAB = \mathbf{60^\circ}$ $\angle OBA = \mathbf{60^\circ}$ $\angle OCD = \mathbf{60^\circ}$ $\angle ODC = \mathbf{60^\circ}$ $\angle OBC = \mathbf{30^\circ}$ $\angle OCB = \mathbf{30^\circ}$ $\angle ODA = \mathbf{30^\circ}$ $\angle OAD = \mathbf{30^\circ}$

(iii) Angles of the rectangle ABCD: $\angle DAB = \mathbf{90^\circ}$ $\angle ABC = \mathbf{90^\circ}$ $\angle BCD = \mathbf{90^\circ}$ $\angle CDA = \mathbf{90^\circ}$

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° (30° + 60°).

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 $x, x, 180 - x,$ and $180 - x.$

Q. Can you find the other angles?

Q11

Context: Since we know that $\Delta AOB$ 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°.

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°. Would we be wrong if we just define a rectangle as a quadrilateral in which all the angles are 90°?

Q14

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

Are you able to construct such a quadrilateral?

Q15

Is it wrong to write ΔBAD ≅ ΔCDB? Why?

Q16

Can you similarly show that AB is parallel to DC (AB || 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 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° and the fourth angle not equal to 90°?

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?