Context: We can compute the four angles between the diagonals to be and Q. Can you find the other angles?

The diagonals of a rectangle are equal in length and bisect each other, forming four isosceles triangles. Step 1 — Identify Isosceles Triangles Let O be the point where the diagonals AC and BD intersect. In a rectangle, the diagonals are equal in length. Also, the diagonals bisect each other. So, we have and . Since , it means that all four segments are equal: . This tells us that the four triangles formed by the diagonals are isosceles triangles. These triangles are , , , and . Step 2 — Find Angles in Triangles with Angle Consider . Since , is an isosceles triangle. The angle is given. The sum of angles in any triangle is . So, we can write: Because is isosceles, its base angles are equal: . Let us call these angles . Now, we solve for : > Similarly, for : (these are vertically opposite angles to ). Since , is also an isosceles triangle. So, its base angles are equal: . Following the same calculation: > Step 3 — Find Angles in Triangles with Angle Consider . Since , is an isosceles triangle. The angle is given. The sum of angles in any triangle is . So, we can write: Because is isosceles, its base angles are equal: . Let us call these angles . Now, we solve for : > Similarly, for : (these are vertically opposite angles to ). Since , is also an isosceles triangle. So, its base angles are equal: . Following the same calculation: > Step 4 — Find the Angles of the Rectangle The angles of the rectangle are formed by combining the angles we just found. For example, the angle at vertex A, , is the sum of and . Let us check the other angles: > Answer > (i) The angles formed by the diagonals with the sides, which are , , , and , are all equal to . > (ii) The other angles formed by the diagonals with the sides, which are , , , and , are all equal to . > (iii) The interior angles of the rectangle, , , , and , are all equal to .

Quadrilaterals | IT

Question 10

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?

Solution

The diagonals of a rectangle are equal in length and bisect each other, forming four isosceles triangles.

Step 1 — Identify Isosceles Triangles

Let O be the point where the diagonals AC and BD intersect. In a rectangle, the diagonals are equal in length. Also, the diagonals bisect each other. So, we have $AO = OC$ and $BO = OD$ . Since $AC = BD$ , it means that all four segments are equal: $AO = OB = OC = OD$ . This tells us that the four triangles formed by the diagonals are isosceles triangles. These triangles are $\triangle AOB$ , $\triangle BOC$ , $\triangle COD$ , and $\triangle DOA$ .

Diagram 1

Step 2 — Find Angles in Triangles with Angle $x$

Consider $\triangle AOB$ . Since $AO = OB$ , $\triangle AOB$ is an isosceles triangle. The angle $\angle AOB = x$ is given. The sum of angles in any triangle is $180^\circ$ . So, we can write: $\angle OAB + \angle OBA + \angle AOB = 180^\circ$ Because $\triangle AOB$ is isosceles, its base angles are equal: $\angle OAB = \angle OBA$ . Let us call these angles $\alpha$ . $2\alpha + x = 180^\circ$ Now, we solve for $\alpha$ : $2\alpha = 180^\circ - x$ $\alpha = \frac{180^\circ - x}{2}$ $= 90^\circ - \frac{x}{2}$

$\boxed{\angle OAB = \angle OBA = 90^\circ - \frac{x}{2}}$ Similarly, for $\triangle COD$ : $\angle COD = x$ (these are vertically opposite angles to $\angle AOB$ ). Since $OC = OD$ , $\triangle COD$ is also an isosceles triangle. So, its base angles are equal: $\angle OCD = \angle ODC$ . Following the same calculation: $\boxed{\angle OCD = \angle ODC = 90^\circ - \frac{x}{2}}$

Step 3 — Find Angles in Triangles with Angle $180^\circ - x$

Consider $\triangle BOC$ . Since $OB = OC$ , $\triangle BOC$ is an isosceles triangle. The angle $\angle BOC = 180^\circ - x$ is given. The sum of angles in any triangle is $180^\circ$ . So, we can write: $\angle OBC + \angle OCB + \angle BOC = 180^\circ$ Because $\triangle BOC$ is isosceles, its base angles are equal: $\angle OBC = \angle OCB$ . Let us call these angles $\beta$ . $2\beta + (180^\circ - x) = 180^\circ$ Now, we solve for $\beta$ : $2\beta = 180^\circ - (180^\circ - x)$ $2\beta = x$ $\beta = \frac{x}{2}$

$\boxed{\angle OBC = \angle OCB = \frac{x}{2}}$ Similarly, for $\triangle DOA$ : $\angle DOA = 180^\circ - x$ (these are vertically opposite angles to $\angle BOC$ ). Since $OD = OA$ , $\triangle DOA$ is also an isosceles triangle. So, its base angles are equal: $\angle ODA = \angle OAD$ . Following the same calculation: $\boxed{\angle ODA = \angle OAD = \frac{x}{2}}$

Step 4 — Find the Angles of the Rectangle

The angles of the rectangle are formed by combining the angles we just found. For example, the angle at vertex A, $\angle DAB$ , is the sum of $\angle OAD$ and $\angle OAB$ . $\angle DAB = \angle OAD + \angle OAB$ $= \frac{x}{2} + (90^\circ - \frac{x}{2})$ $= 90^\circ$ Let us check the other angles: $\angle ABC = \angle OBA + \angle OBC$ $= (90^\circ - \frac{x}{2}) + \frac{x}{2}$ $= 90^\circ$ $\angle BCD = \angle OCB + \angle OCD$ $= \frac{x}{2} + (90^\circ - \frac{x}{2})$ $= 90^\circ$ $\angle CDA = \angle ODC + \angle ODA$ $= (90^\circ - \frac{x}{2}) + \frac{x}{2}$ $= 90^\circ$

$\boxed{\text{Each interior angle of the rectangle is } 90^\circ}$

Answer

(i) The angles formed by the diagonals with the sides, which are $\angle OAD$ , $\angle ODA$ , $\angle OBC$ , and $\angle OCB$ , are all equal to $\frac{x}{2}$ . (ii) The other angles formed by the diagonals with the sides, which are $\angle OAB$ , $\angle OBA$ , $\angle OCD$ , and $\angle ODC$ , are all equal to $90^\circ - \frac{x}{2}$ . (iii) The interior angles of the rectangle, $\angle DAB$ , $\angle ABC$ , $\angle BCD$ , and $\angle CDA$ , are all equal to $90^\circ$ .

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)
$AD = CB$

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?

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?