Area of Polygons | FIO

Question 5

In this figure, if the sidelength of the square is doubled, what is the increase in the areas of the regions 1, 2 and 3? Give reasons.

Solution

We will first find the original areas of the three regions. Then, we will calculate their new areas when the square's side length is doubled. Finally, we will determine how much each area increases.

Step 1 — Define the square and its regions

Let the side length of the original square be 'a' units. Let us name the vertices of the square. Let the top-left vertex be A, top-right be B, bottom-right be C, and bottom-left be D. The diagonal shown in the figure is BD. The line segment from vertex A meets the diagonal BD at point E. The marks on the diagonal BD tell us that point E is exactly in the middle of BD. This means E is the midpoint of BD. Also, the line segment AE is perpendicular to BD. Region 1 is the triangle AED. Region 2 is the triangle AEB. Region 3 is the triangle BCD.

Diagram 1

Step 2 — Calculate the original areas of the regions

Let us find the area of each region when the side length of the square is 'a'.

For Region 3 (Triangle BCD): This is a right-angled triangle. Its base is BC, and its height is CD. Both BC and CD are sides of the square, so their length is 'a'. Area of a triangle is half times base times height. $\text{Area of Region 3} = \frac{1}{2} \times \text{base} \times \text{height}$ $= \frac{1}{2} \times a \times a$ $= \frac{a^2}{2}$

$\boxed{\text{Original Area of Region 3} = \frac{a^2}{2} \text{ square units}}$

For Region 1 (Triangle AED) and Region 2 (Triangle AEB): First, we need to find the length of the diagonal BD. In the right-angled triangle BCD, we can use the Pythagoras theorem. $BD^2 = BC^2 + CD^2$ $BD^2 = a^2 + a^2$ $BD^2 = 2a^2$ To find BD, we take the square root of both sides. $BD = \sqrt{2a^2}$ $BD = a\sqrt{2}$ Since E is the midpoint of BD, we can find the lengths of DE and EB. $DE = EB = \frac{BD}{2}$ $DE = EB = \frac{a\sqrt{2}}{2}$ We can simplify this by writing $\sqrt{2}$ in the denominator. $DE = EB = \frac{a}{\sqrt{2}}$ Now, we need the length of AE. In a square, the diagonal divides it into two isosceles right-angled triangles (like triangle ABD). The line AE is the altitude from A to the hypotenuse BD. In such a triangle, the altitude to the hypotenuse is half the hypotenuse. So, AE is also half of BD. $AE = \frac{BD}{2}$ $AE = \frac{a\sqrt{2}}{2}$ $AE = \frac{a}{\sqrt{2}}$ Now we can calculate the area of Region 1 (Triangle AED). $\text{Area of Region 1} = \frac{1}{2} \times \text{base DE} \times \text{height AE}$ $= \frac{1}{2} \times \frac{a}{\sqrt{2}} \times \frac{a}{\sqrt{2}}$ $= \frac{1}{2} \times \frac{a^2}{2}$ $= \frac{a^2}{4}$

$\boxed{\text{Original Area of Region 1} = \frac{a^2}{4} \text{ square units}}$ Next, we calculate the area of Region 2 (Triangle AEB). $\text{Area of Region 2} = \frac{1}{2} \times \text{base EB} \times \text{height AE}$ $= \frac{1}{2} \times \frac{a}{\sqrt{2}} \times \frac{a}{\sqrt{2}}$ $= \frac{1}{2} \times \frac{a^2}{2}$ $= \frac{a^2}{4}$ $\boxed{\text{Original Area of Region 2} = \frac{a^2}{4} \text{ square units}}$

Step 3 — Calculate the new areas when the side length is doubled

The side length of the square is now doubled. So, the new side length is $2a$ . We will replace 'a' with '2a' in our area formulas from Step 2.

For Region 3: $\text{New Area of Region 3} = \frac{(2a)^2}{2}$ $= \frac{4a^2}{2}$ $= 2a^2$

$\boxed{\text{New Area of Region 3} = 2a^2 \text{ square units}}$

For Region 1: $\text{New Area of Region 1} = \frac{(2a)^2}{4}$ $= \frac{4a^2}{4}$ $= a^2$

$\boxed{\text{New Area of Region 1} = a^2 \text{ square units}}$

For Region 2: $\text{New Area of Region 2} = \frac{(2a)^2}{4}$ $= \frac{4a^2}{4}$ $= a^2$

$\boxed{\text{New Area of Region 2} = a^2 \text{ square units}}$

Step 4 — Determine the increase in area for each region

To find how much the area increased, we divide the new area by the original area.

For Region 3: $\text{Increase in Region 3} = \frac{\text{New Area of Region 3}}{\text{Original Area of Region 3}}$ $= \frac{2a^2}{\frac{a^2}{2}}$ $= 2a^2 \times \frac{2}{a^2}$ $= 4$

$\boxed{\text{Increase in Region 3} = 4 \text{ times}}$

For Region 1: $\text{Increase in Region 1} = \frac{\text{New Area of Region 1}}{\text{Original Area of Region 1}}$ $= \frac{a^2}{\frac{a^2}{4}}$ $= a^2 \times \frac{4}{a^2}$ $= 4$

$\boxed{\text{Increase in Region 1} = 4 \text{ times}}$

For Region 2: $\text{Increase in Region 2} = \frac{\text{New Area of Region 2}}{\text{Original Area of Region 2}}$ $= \frac{a^2}{\frac{a^2}{4}}$ $= a^2 \times \frac{4}{a^2}$ $= 4$

$\boxed{\text{Increase in Region 2} = 4 \text{ times}}$

Answer

The area of each region increases by 4 times.

Reason: The area of any two-dimensional shape depends on the square of its side lengths. If we double the side length, the area will become $2 \times 2 = 4$ times larger.

More questions in FIO

Identify the missing sidelengths.

The figure shows a path (the shaded portion) laid around a rectangular park EFGH.

(i) What measurements do you need to find the area of the path? Once you identify the lengths to be measured, assign possible values of your choice to these measurements and find the area of the path. Give a formula for the area.

An example of a formula — $Area\ of\ a\ rectangle = length \times width$ .

[Hint: There is a relation between the areas of EFGH, the path, and ABCD.**]

(ii) If the width of the path along each side is given, can you find its area? If not, what other measurements do you need? Assign values of your choice to these measurements and find the area of the path. Give a formula for the area using these measurements.

[Hint: Break the path into rectangles.**]

(iii) Does the area of the path change when the outer rectangle is moved while keeping the inner rectangular park EFGH inside it, as shown?

The figure shows a plot with sides 14m and 12m, and with a crosspath. What other measurements do you need to find the area of the crosspath? Once you identify the lengths to be measured, assign some possible values of your choice and find the area of the path. Give a formula for the area based on the measurements you choose.

Find the area of the spiral tube shown in the figure. The tube has the same width throughout.

[Hint: There are different ways of finding the area. Here is one method.]

What should be the length of the straight tube if it is to have the same area as the bent tube on the left?

In this figure, if the sidelength of the square is doubled, what is the increase in the areas of the regions 1, 2 and 3? Give reasons.

Divide a square into 4 parts by drawing two perpendicular lines inside the square as shown in the figure.

Rearrange the pieces to get a larger square, with a hole inside.

You can try this activity by constructing the square using cardboard, thick chart paper, or similar materials.

Find the areas of the following triangles:

Find the length of the altitude BY.

Find the area of $\Delta$ SUB, given that it is isosceles, SE is perpendicular to UB, and the area of $\Delta$ SEB is 24 sq. units.

Q10

[Śulba-Sūtras] Give a method to transform a rectangle into a triangle of equal area.

Q11

[Śulba-Sūtras] Give a method to transform a triangle into a rectangle of equal area.

Q12

ABCD, BCEF, and BFGH are identical squares.

(i) If the area of the red region is 49 sq. units, then what is the area of the blue region?

(ii) In another version of this figure, if the total area enclosed by the blue and red regions is 180 sq. units, then what is the area of each square?

Q13

If M and N are the midpoints of XY and XZ, what fraction of the area of $\Delta$ XYZ is the area of $\Delta$ XMN? [Hint: Join NY]

Q14

Gopal needs to carry water from the river to his water tank. He starts from his house. What is the shortest path he can take from his house to the river and then to the water tank? Roughly recreate the map in your notebook and trace the shortest path.

Q15

Find the area of the quadrilateral ABCD given that AC = 22 cm, BM = 3 cm, DN = 3 cm, BM is perpendicular to AC, and DN is perpendicular to AC.

Q16

Find the area of the shaded region given that ABCD is a rectangle.

Q17

What measurements would you need to find the area of a regular hexagon?

Q18

What fraction of the total area of the rectangle is the area of the blue region?

Q19

Give a method to obtain a quadrilateral whose area is half that of a given quadrilateral.

Q20

Observe the parallelograms in the figure below.

(i) What can we say about the areas of all these parallelograms?

(ii) What can we say about their perimeters? Which figure appears to have the maximum perimeter, and which has the minimum perimeter?

Q21

Find the areas of the following parallelograms:

Q22

Find QN.

Q23

Consider a rectangle and a parallelogram of the same sidelengths: 5 cm and 4 cm. Which has the greater area? [Hint: Imagine constructing them on the same base.]

Q24

Give a method to obtain a rectangle whose area is twice that of a given triangle. What are the different methods that you can think of?

Q25

[Śulba-Sūtras] Give a method to obtain a rectangle of the same area as a given triangle.

Q26

[Śulba-Sūtras] An isosceles triangle can be converted into a rectangle by dissection in a simpler way. Can you find out how to do it?

[Hint: Show that triangles ΔADB and ΔADC can be made into halves of a rectangle. Figure out how they should be assembled to get a rectangle. Use cut-outs if necessary.]

Q27

[Śulba-Sūtras] Give a method to convert a rectangle into an isosceles triangle by dissection.

Q28

Which has greater area—an equilateral triangle or a square of the same sidelength as the triangle? Which has greater area—two identical equilateral triangles together or a square of the same sidelength as the triangle? Give reasons.

Q29

Find the area of a rhombus whose diagonals are 20 cm and 15 cm.

Q30

Give a method to convert a rectangle into a rhombus of equal area using dissection.

Q31

Find the areas of the following figures:

Q32

[Śulba-Sūtras] Give a method to convert an isosceles trapezium to a rectangle using dissection.

Q33

Here is one of the ways to convert trapezium ABCD into a rectangle EFGH of equal area —

Given the trapezium ABCD, how do we find the vertices of the rectangle EFGH?

[Hint: If ΔAHI ≅ ΔDGI and ΔBEJ ≅ ΔCFJ, then the trapezium and rectangle have equal areas.]

Q34

Using the idea of converting a trapezium into a rectangle of equal area, and vice versa, construct a trapezium of area $144\text{ cm}^2$ .

Q35

A regular hexagon is divided into a trapezium, an equilateral triangle, and a rhombus, as shown. Find the ratio of their areas.

Q36

ZYXW is a trapezium with ZY || WX. A is the midpoint of XY. Show that the area of the trapezium ZYXW is equal to the area of $\Delta\text{ZWB}$ .