Area of Polygons | IT

Question 1

Try to think of different creative ways to divide a square into 4 parts of equal area.

Solution

Understand the Question

Let a square have side length $s$ , so its total area is $A = s^2$ .
To divide the square into $4$ parts of equal area, each part must have an area of $\dfrac{s^2}{4}$ .
We can achieve this through different geometric divisions such as:
- Connecting the midpoints of opposite sides to form $4$ smaller squares.
- Drawing both diagonals to form $4$ congruent triangles.
- Connecting the center to the midpoints of each side to form $4$ congruent quadrilaterals.

Step 1 · Method 1: Four Smaller Squares

Connect the midpoints of opposite sides using horizontal and vertical lines. Diagram 1

Let the side of the original square be $s$ , with total area $A = s \times s = s^2$ .

Each smaller square has a side length of $\dfrac{s}{2}$ .

Area of one small square $A_1$ :

\begin{aligned} A_1 &= \left(\dfrac{s}{2}\right) \times \left(\dfrac{s}{2}\right) \\[0.6em] &= \dfrac{s^2}{4} \end{aligned}

Step 2 · Method 2: Four Triangles (Diagonals)

Draw the two diagonals intersecting at the center of the square. Diagram 2

This divides the square into $4$ congruent triangles. For each triangle:

$\text{Base} = s$
$\text{Height} = \dfrac{s}{2}$ (perpendicular distance from the center to a side)

Area of one triangle $A_2$ :

\begin{aligned} A_2 &= \dfrac{1}{2} \times \text{base} \times \text{height} \\[0.6em] &= \dfrac{1}{2} \times s \times \dfrac{s}{2} \\[0.6em] &= \dfrac{s^2}{4} \end{aligned}

Step 3 · Method 3: Four Quadrilaterals

Connect the center of the square to the midpoint of each side. Diagram 3

Let $O$ be the center and $M_1, M_2, M_3, M_4$ be the midpoints of sides $\text{AB}, \text{BC}, \text{CD}, \text{DA}$ .

Each quadrilateral, such as $\text{AM}_1\text{OM}_4$ , is composed of two triangles: $\triangle \text{AOM}_1$ and $\triangle \text{AOM}_4$ .

For $\triangle \text{AOM}_1$ : $A_{\triangle \text{AOM}_1} = \dfrac{1}{2} \times \dfrac{s}{2} \times \dfrac{s}{2} = \dfrac{s^2}{8}$

For $\triangle \text{AOM}_4$ : $A_{\triangle \text{AOM}_4} = \dfrac{1}{2} \times \dfrac{s}{2} \times \dfrac{s}{2} = \dfrac{s^2}{8}$

Area of quadrilateral $\text{AM}_1\text{OM}_4$ ( $A_3$ ):

\begin{aligned} A_3 &= A_{\triangle \text{AOM}_1} + A_{\triangle \text{AOM}_4} \\[0.6em] &= \dfrac{s^2}{8} + \dfrac{s^2}{8} \\[0.6em] &= \dfrac{2s^2}{8} \\[0.6em] &= \dfrac{s^2}{4} \end{aligned}

Answer

Three creative ways to divide a square into $4$ parts of equal area are:

Four smaller squares by joining midpoints of opposite sides.
Four triangles by drawing both diagonals.
Four quadrilaterals by connecting the center to the midpoints of the sides.

Common Mistakes

Area vs. Shape: The parts do not necessarily have to be congruent to have equal areas, though using symmetry ensures both equal area and congruence.
Height of Triangle: Mistaking the triangle height for the full side length $s$ instead of the distance from the center to the base, which is $\dfrac{s}{2}$ .

More questions in IT

Try to think of different creative ways to divide a square into 4 parts of equal area.

Why Can't Perimeter be a Measure of Area?

Why do we count the number of unit squares to assign measures for area? Couldn't we have just used the perimeter of a region, i.e., the length of its boundary as a measure of its area?

Context: Consider two regions, Region 1 and Region 2, such that $\text{Perimeter of Region 1} > \text{Perimeter of Region 2}$ , but $\text{Area of Region 1} < \text{Area of Region 2}$ .

Q. Find two rectangles that are examples of such regions. If needed, use a grid paper (given at the end of the book) for this.

Also give an example of two regions of other shapes, where the region with the larger perimeter has the smaller area! This property should be visually clear in your example.

In the given figure, which triangle has a greater area: $\Delta \text{XDC}$ or $\Delta \text{YDC}$ , if both the rectangles are identical?

In the given figure, which triangle has a greater area: $\Delta \text{XDC}$ or $\Delta \text{YBC}$ , if both the rectangles are identical?

Find the area of $\Delta \text{XDC}$ .

To find the area of a triangle, what measurements do we need?

How do we get the outer rectangle from the given triangle?

Q10

Will this formula hold for the kind of triangle, around which we cannot draw a rectangle with $\text{BC}$ as the base?

Q11

Line $l \parallel \text{BC}$ . Consider the different triangles that have $\text{BC}$ as their base, and with their third vertex lying anywhere on $l$ .

(i) Which of these triangles has the maximum area, and which has the minimum area?

(ii) Which of these triangles has the maximum perimeter, and which has the minimum perimeter?

Q12

Analyse whether $A$ lies on the perpendicular bisector of $BC$ .

Q13

Area of any Polygon

How do we find the area of this quadrilateral? What measurements do we need for this?

Q14

How do we find the area of this pentagon?

Q15

Can any polygon be divided into triangles?

Q16

Give a method to convert a parallelogram into a rectangle of equal area.

You can try this using a cut-out of a parallelogram.

Q17

Can $\Delta \text{AXD}$ and $\text{ABCX}$ fit together, as shown in the figure, to get a rectangle?

Q18

Try working this out!

Q19

What are the sidelengths of the rectangle $WXYZ$ ?

Q20

Area of rhombus ABCD can also be determined by finding the areas of $\Delta \text{ADB}$ and $\Delta \text{CDB}$ . What formula does this give us?

Q21

Context: The area of rhombus ABCD can be written as:

$\text{Area of rhombus } ABCD = \text{Area}(\Delta \text{ADB}) + \text{Area}(\Delta \text{CDB}) = \dfrac{1}{2} \times \text{AO} \times \text{BD} + \dfrac{1}{2} \times \text{CO} \times \text{BD}$

Q. Simplify the expression to show that we get the same formula for the area of a rhombus in terms of its diagonals.

Q22

Find the areas of the following trapeziums by breaking them into figures whose areas can be computed.

Q23

Will this formula hold for a trapezium that looks like this?

Q24

Will Approach 2 work for any type of trapezium?

Q25

What figure will we get when the two trapeziums are joined along $\text{BC}$ ?

Q26

What type of a quadrilateral is this?

Q27

What do you think is the area of an A4 sheet? Its sidelengths are $21\text{ cm}$ and $29.7\text{ cm}$ . Now find its area.

Q28

What do you think is the area of the tabletop that you use at school or at home? You could perhaps try to visualise how many A4 sheets can fit on your table.

Q29

Express the following lengths in centimeters:

(i) $5 \text{ in}$

(ii) $7.4 \text{ in}$

Q30

Express the following lengths in inches:

(i) $5.08 \text{ cm}$

(ii) $11.43 \text{ cm}$

Q31

How many $\text{cm}^2$ is $1 \text{ in}^2$ ?

Q32

Context: Convert $161.29 \text{ cm}^2$ to $\text{in}^2$ . Every $6.4516 \text{ cm}^2$ gives an $\text{in}^2$ . Hence, $161.29 \text{ cm}^2 = \dfrac{161.29}{6.4516} \text{ in}^2$ .