Area of Polygons | IT

Question 13

Area of any Polygon

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

Question diagram 1
Check your answer with HomiSolve it yourself, then let Homi check your steps and spot mistakes.
Solution

To find the area of any quadrilateral, we can split it into two triangles using a diagonal.

Step 1 — Required Measurements

Let us draw a diagonal, for example, AC, to divide the quadrilateral ABCD into two triangles. These triangles are triangle ABC and triangle ADC. We need to measure the length of this diagonal AC. Let us call this length dd. We also need to measure the perpendicular height from vertex B to the diagonal AC. Let us call this height h1h_1. Finally, we need the perpendicular height from vertex D to the diagonal AC. Let us call this height h2h_2.

Measurements needed: diagonal length d, and perpendicular heights h1,h2\boxed{\text{Measurements needed: diagonal length } d \text{, and perpendicular heights } h_1, h_2}

Diagram 1

Step 2 — Calculating the Area

The area of a triangle is found by multiplying half of its base by its height. For triangle ABC, the base is AC and the height is h1h_1. The area of triangle ABC is:

AreaABC=12×AC×h1\text{Area}_{\text{ABC}} = \frac{1}{2} \times \text{AC} \times h_1

=12×d×h1= \frac{1}{2} \times d \times h_1

For triangle ADC, the base is AC and the height is h2h_2. The area of triangle ADC is:

AreaADC=12×AC×h2\text{Area}_{\text{ADC}} = \frac{1}{2} \times \text{AC} \times h_2

=12×d×h2= \frac{1}{2} \times d \times h_2

The total area of the quadrilateral ABCD is the sum of the areas of these two triangles.

AreaABCD=AreaABC+AreaADC\text{Area}_{\text{ABCD}} = \text{Area}_{\text{ABC}} + \text{Area}_{\text{ADC}}

=(12×d×h1)+(12×d×h2)= \left( \frac{1}{2} \times d \times h_1 \right) + \left( \frac{1}{2} \times d \times h_2 \right)

We can take out the common factor 12×d\frac{1}{2} \times d.

=12×d×(h1+h2)= \frac{1}{2} \times d \times (h_1 + h_2)

Area=12×diagonal length×(sum of perpendicular heights)\boxed{\text{Area} = \frac{1}{2} \times \text{diagonal length} \times (\text{sum of perpendicular heights})}

Answer

(i) We need the length of one diagonal and the perpendicular heights from the other two vertices to that diagonal. (ii) The area is calculated by multiplying half the diagonal length by the sum of the two perpendicular heights.

More questions in IT

Q1

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

Q2

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?

Q3

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.

Q4

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.

Q5

In the given figure, which triangle has a greater area: Δ\DeltaXDC or Δ\DeltaYDC, if both the rectangles are identical?

Q6

In the given figure, which triangle has a greater area: Δ\DeltaXDC or Δ\DeltaYBC, if both the rectangles are identical?

Q7

Find the area of Δ\DeltaXDC.

Q8

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

Q9

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

(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 ΔAXD\Delta\text{AXD} and ABCX\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 ΔADB\Delta\text{ADB} and ΔCDB\Delta\text{CDB}. What formula does this give us?

Q21

Context: The area of rhombus ABCD can be written as: Area of rhombus ABCD=Area(ΔADB)+Area(ΔCDB)=12×AO×BD+12×CO×BD\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 cm2\text{cm}^2 is 1 in21\text{ in}^2?

Q32

Context: Convert 161.29 cm2161.29\text{ cm}^2 to in2\text{in}^2. Every 6.4516 cm26.4516\text{ cm}^2 gives an in2\text{in}^2. Hence, 161.29 cm2=161.296.4516 in2161.29\text{ cm}^2 = \frac{161.29}{6.4516}\text{ in}^2.

Q. Evaluate the quotient.

Q33

What do you think is the area of your classroom?

Q34

How many in2\text{in}^2 is 1 ft21\text{ ft}^2?

Q35

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

Q36

Find out the local unit of area measurement in your region.

Q37

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

Q38

How many m2\text{m}^2 is a km2\text{km}^2?

Q39

How many times is your village/town/city bigger than your school?

Q40

Find the city with the largest area in:

(i) India

(ii) the world

Q41

Find the city with the smallest area in:

(i) India

(ii) the world

← Back to Area of Polygons