Area of Polygons | IT

Question 35

What do you think is the area of your school? Make an estimate and compare it with the actual data.

Solution

We will estimate the school's total area by adding the areas of its main parts.

Step 1 — Estimate School Area

Let us imagine a typical school layout. It has a main building, a playground, and other areas. We will treat these as simple rectangles. Let the main building be a rectangle. Its length is $L_B = 80 \text{ m}$ . Its width is $W_B = 30 \text{ m}$ . Let the playground be a rectangle. Its length is $L_P = 100 \text{ m}$ . Its width is $W_P = 50 \text{ m}$ . Let other areas be a rectangle. These include gardens, parking, and pathways. Its length is $L_O = 50 \text{ m}$ . Its width is $W_O = 20 \text{ m}$ .

The area of the main building, $A_B$ , is: $A_B = L_B \times W_B$ $= 80 \text{ m} \times 30 \text{ m}$ $= \mathbf{2400 \text{ m}^2}$

The area of the playground, $A_P$ , is: $A_P = L_P \times W_P$ $= 100 \text{ m} \times 50 \text{ m}$ $= \mathbf{5000 \text{ m}^2}$

The area of other areas, $A_O$ , is: $A_O = L_O \times W_O$ $= 50 \text{ m} \times 20 \text{ m}$ $= \mathbf{1000 \text{ m}^2}$

The total estimated area of the school, $A_{\text{estimate}}$ , is the sum of these areas: $A_{\text{estimate}} = A_B + A_P + A_O$ $= 2400 \text{ m}^2 + 5000 \text{ m}^2 + 1000 \text{ m}^2$

$\boxed{\mathbf{8400 \text{ m}^2}}$

Diagram 1

Step 2 — Compare with Actual Data

Let us compare our estimate with actual data. Suppose the actual area of the school is known. Actual data often comes from official records. Let the actual area of the main building be $A_{B, \text{actual}} = 2500 \text{ m}^2$ . Let the actual area of the playground be $A_{P, \text{actual}} = 5200 \text{ m}^2$ . Let the actual area of other parts be $A_{O, \text{actual}} = 1200 \text{ m}^2$ .

The total actual area of the school, $A_{\text{actual}}$ , is: $A_{\text{actual}} = A_{B, \text{actual}} + A_{P, \text{actual}} + A_{O, \text{actual}}$ $= 2500 \text{ m}^2 + 5200 \text{ m}^2 + 1200 \text{ m}^2$

$\boxed{\mathbf{8900 \text{ m}^2}}$

Now, we find the difference between the areas. $\text{Difference} = A_{\text{actual}} - A_{\text{estimate}}$ $= 8900 \text{ m}^2 - 8400 \text{ m}^2$

$\boxed{\mathbf{500 \text{ m}^2}}$

Our estimate is less than the actual area. We can also find the percentage difference. $\text{Percentage Difference} = \frac{\text{Difference}}{A_{\text{actual}}} \times 100\%$ $= \frac{500 \text{ m}^2}{8900 \text{ m}^2} \times 100\%$ $= \frac{500}{89} \%$ $\approx \mathbf{5.62\%}$

Answer

(i) The estimated area of the school is 8400 m $^2$ . (ii) The actual area of the school is 8900 m $^2$ . (iii) Our estimate is 500 m $^2$ less than the actual area, which is approximately 5.62% of the actual area.

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 Perimeter of Region 1 > Perimeter of Region 2, but Area of Region 1 < 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: $\Delta$ XDC or $\Delta$ YDC, if both the rectangles are identical?

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

Find the area of $\Delta$ 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 BC as the base?

Q11

Line $l \parallel \text{BC}$ . Consider the different triangles that have 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: $\text{Area of rhombus ABCD} = \text{Area}(\Delta\text{ADB}) + \text{Area}(\Delta\text{CDB}) = \frac{1}{2} \times \text{AO} \times \text{BD} + \frac{1}{2} \times \text{CO} \times \text{BD}$

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 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 cm and 29.7 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 in

(ii) 7.4 in

Q30

Express the following lengths in inches:

(i) 5.08 cm

(ii) 11.43 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, $161.29\text{ cm}^2 = \frac{161.29}{6.4516}\text{ in}^2$ .