Area of Polygons | FIO

Question 12

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?

Solution

We will use the side length of the squares to express the areas of the shaded regions and then solve for the unknown values.

Step 1 — Define the side length

Let us say that the side length of each identical square is 's' units. This means that AD = DC = CB = BA = s. Also, BC = CE = EF = FB = s. And BF = FG = GH = HB = s.

Step 2 — Calculate the area of the red region

The red region is a triangle with vertices D, C, and H. The base of this triangle is the side DC. The length of DC is s units. The height of this triangle is the perpendicular distance from H to the line containing DC. This height is the length of the segment HC. HC is made up of HB and BC. Since BFGH and BCEF are squares, HB = s and BC = s. So, the height HC = HB + BC = s + s = 2s units. The formula for the area of a triangle is (1/2) * base * height.

$\text{Area of red region} = \frac{1}{2} \times \text{DC} \times \text{HC}$

$= \frac{1}{2} \times s \times (2s)$

$= s^2$

$\boxed{\text{Area of red region} = s^2 \text{ sq. units}}$

Diagram 1

Step 3 — Calculate the area of the blue region

The blue region is a triangle with vertices A, D, and the point where the line segment DH intersects AB. Let's call this intersection point P. So, the blue region is triangle ADP. The base of this triangle is the side AD. The length of AD is s units. The height of this triangle is the perpendicular distance from P to the line containing AD. This height is the length of the segment AP. To find AP, we can use coordinates. Let D be at the origin (0,0). Then A is at (0,s) and B is at (s,s). H is at (s, 2s) because HB = s and BC = s. The line segment DH connects D(0,0) to H(s, 2s). The equation of the line DH is $y = 2x$ . The line segment AB is part of the line $y = s$ . The point P is where the line DH crosses the line AB. At point P, the y-coordinate is s. Substituting $y=s$ into the line equation $y=2x$ :

$s = 2x$

$x = \frac{s}{2}$

So, the coordinates of P are $(s/2, s)$ . The length of AP is the x-coordinate of P, which is s/2 units.

$\text{Area of blue region} = \frac{1}{2} \times \text{AD} \times \text{AP}$

$= \frac{1}{2} \times s \times \frac{s}{2}$

$= \frac{s^2}{4}$

$\boxed{\text{Area of blue region} = \frac{s^2}{4} \text{ sq. units}}$

Diagram 2

Step 4 — Solve part (i)

We are given that the area of the red region is 49 sq. units. From Step 2, we know that the area of the red region is $s^2$ .

$s^2 = 49$

To find 's', we take the square root of 49.

$s = \sqrt{49}$

$s = 7 \text{ units}$

Now we need to find the area of the blue region. From Step 3, we know that the area of the blue region is $s^2 / 4$ .

$\text{Area of blue region} = \frac{s^2}{4}$

$= \frac{49}{4}$

$= 12.25$

$\boxed{\text{Area of blue region} = 12.25 \text{ sq. units}}$

Step 5 — Solve part (ii)

We are given that the total area enclosed by the blue and red regions is 180 sq. units. From Step 2, the area of the red region is $s^2$ . From Step 3, the area of the blue region is $s^2 / 4$ .

$\text{Total area} = \text{Area of red region} + \text{Area of blue region}$

$= s^2 + \frac{s^2}{4}$

To add these, we find a common denominator.

$= \frac{4s^2}{4} + \frac{s^2}{4}$

$= \frac{4s^2 + s^2}{4}$

$= \frac{5s^2}{4}$

We are given that this total area is 180 sq. units.

$\frac{5s^2}{4} = 180$

To find $s^2$ , we multiply both sides by 4.

$5s^2 = 180 \times 4$

$5s^2 = 720$

Now, we divide both sides by 5.

$s^2 = \frac{720}{5}$

$s^2 = 144$

The question asks for the area of each square. The area of each square is $s^2$ .

$\boxed{\text{Area of each square} = 144 \text{ sq. units}}$

Answer

(i) The area of the blue region is 12.25 sq. units. (ii) The area of each square is 144 sq. units.

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.

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

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[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

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Q15

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Q16

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Q17

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Q18

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Q19

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Q20

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Q22

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Q29

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Q30

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Q31

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Q32

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

Q33

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[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

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