Area of Polygons | IT

Question 2

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?

Solution

Understand the Question

Perimeter measures the 1-dimensional boundary length around a closed region (in linear units like $\text{cm}$ or $\text{m}$ ).
Area measures the 2-dimensional surface enclosed inside the boundary (by counting unit squares, in units like $\text{cm}^2$ or $\text{m}^2$ ).
Perimeter cannot be a measure of area because two regions can have the exact same perimeter while enclosing entirely different amounts of area.

Step 1 · Define Perimeter and Area

Perimeter and area measure two completely different geometric properties:

Perimeter: The total distance around the outer boundary of a shape, measured in linear units (such as $\text{cm}$ or $\text{m}$ ).
Area: The total surface or region covered inside the boundary, measured by the number of unit squares that fit inside (such as $\text{cm}^2$ or $\text{m}^2$ ).

Step 2 · Compare Rectangles with the Same Perimeter

Consider two different rectangles to compare their perimeter and area:

Rectangle 1: Length $L_1 = 6\text{ units}$ , Width $W_1 = 2\text{ units}$ Diagram 2

\begin{aligned} P_1 &= 2 \times (L_1 + W_1) = 2 \times (6 + 2) = 2 \times 8 = 16\text{ units} \\[0.6em] A_1 &= L_1 \times W_1 = 6 \times 2 = 12\text{ square units} \end{aligned}

Rectangle 2: Length $L_2 = 5\text{ units}$ , Width $W_2 = 3\text{ units}$ Diagram 3

\begin{aligned} P_2 &= 2 \times (L_2 + W_2) = 2 \times (5 + 3) = 2 \times 8 = 16\text{ units} \\[0.6em] A_2 &= L_2 \times W_2 = 5 \times 3 = 15\text{ square units} \end{aligned}

Both rectangles have the same perimeter ( $16\text{ units}$ ), but Rectangle 1 covers $12\text{ square units}$ while Rectangle 2 covers $15\text{ square units}$ . Thus, perimeter cannot determine or measure area.

Answer

No, perimeter cannot be a measure of area. Two different shapes can have the exact same perimeter but enclose completely different areas (e.g., a $6 \times 2$ rectangle and a $5 \times 3$ rectangle both have a perimeter of $16\text{ units}$ , but their areas are $12\text{ square units}$ and $15\text{ square units}$ , respectively).

Common Mistakes

Dimension Confusion: Confusing a 1-dimensional boundary length (measured in linear units) with a 2-dimensional enclosed space (measured in square units).
Constant Perimeter Assumption: Assuming that shapes with equal perimeters must enclose equal areas. In reality, area depends on the dimensions/shape, not just the boundary length.

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, $161.29 \text{ cm}^2 = \dfrac{161.29}{6.4516} \text{ in}^2$ .