Area of Polygons | FIO

Question 19

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

Solution

Understand the Question

Joining the midpoints of the consecutive sides of any quadrilateral forms an inner quadrilateral (known as a Varignon parallelogram).
Using the Midpoint Theorem, each corner triangle has an area equal to $\dfrac{1}{4}$ of the triangle formed by the quadrilateral's diagonal.
Subtracting the sum of the four corner triangles from the total area proves that the inner quadrilateral's area is exactly half of the original quadrilateral's area.

Step 1 · Construct the Midpoint Quadrilateral

Let $ABCD$ be any given quadrilateral.

Mark the midpoints of its sides:

$P$ is the midpoint of $AB$
$Q$ is the midpoint of $BC$
$R$ is the midpoint of $CD$
$S$ is the midpoint of $DA$

Connect the midpoints in order ( $P \to Q \to R \to S \to P$ ) to form the quadrilateral $PQRS$ . Diagram 1

Step 2 · Apply Midpoint Theorem to Sides

Draw diagonal $AC$ .

In $\Delta ABC$ , $P$ and $Q$ are the midpoints of $AB$ and $BC$ respectively. By the Midpoint Theorem:

\begin{aligned} PQ &\parallel AC \\[0.6em] PQ &= \dfrac{1}{2} AC \end{aligned}

In $\Delta ADC$ , $S$ and $R$ are the midpoints of $DA$ and $CD$ respectively. By the Midpoint Theorem:

\begin{aligned} SR &\parallel AC \\[0.6em] SR &= \dfrac{1}{2} AC \end{aligned}

Since $PQ \parallel SR$ and $PQ = SR$ , quadrilateral $PQRS$ is a parallelogram.

Step 3 · Compare Areas of Corner Triangles

In $\Delta ABC$ , since $P$ and $Q$ are midpoints, $\Delta PBQ \sim \Delta ABC$ with a side ratio of $1 : 2$ .

The ratio of their areas is the square of their side ratio:

\begin{aligned} \text{Area}(PBQ) &= \left(\dfrac{1}{2}\right)^2 \times \text{Area}(ABC) \\[0.6em] &= \dfrac{1}{4} \times \text{Area}(ABC) \end{aligned}

Similarly, for the remaining three corner triangles: $\text{Area}(QCR) = \dfrac{1}{4} \times \text{Area}(BCD)$

$\text{Area}(SDR) = \dfrac{1}{4} \times \text{Area}(CDA)$

$\text{Area}(SAP) = \dfrac{1}{4} \times \text{Area}(DAB)$

Step 4 · Calculate Area of Quadrilateral PQRS

The area of $PQRS$ is obtained by subtracting the areas of the four corner triangles from $\text{Area}(ABCD)$ : $\text{Area}(PQRS) = \text{Area}(ABCD) - (\text{Area}(PBQ) + \text{Area}(QCR) + \text{Area}(SDR) + \text{Area}(SAP))$

Substitute the corner triangle areas: $\text{Area}(PQRS) = \text{Area}(ABCD) - \dfrac{1}{4} (\text{Area}(ABC) + \text{Area}(BCD) + \text{Area}(CDA) + \text{Area}(DAB))$

Express the quadrilateral area using both diagonals: $\text{Area}(ABCD) = \text{Area}(ABC) + \text{Area}(CDA)$ $\text{Area}(ABCD) = \text{Area}(BCD) + \text{Area}(DAB)$

Summing both equations gives: $\text{Area}(ABC) + \text{Area}(BCD) + \text{Area}(CDA) + \text{Area}(DAB) = 2 \times \text{Area}(ABCD)$

Substitute this back:

\begin{aligned} \text{Area}(PQRS) &= \text{Area}(ABCD) - \dfrac{1}{4} (2 \times \text{Area}(ABCD)) \\[0.6em] &= \text{Area}(ABCD) - \dfrac{2}{4} \times \text{Area}(ABCD) \\[0.6em] &= \text{Area}(ABCD) - \dfrac{1}{2} \times \text{Area}(ABCD) \\[0.6em] &= \dfrac{1}{2} \times \text{Area}(ABCD) \end{aligned}

Answer

Connect the midpoints of the four consecutive sides of the given quadrilateral $ABCD$ in order ( $P, Q, R, S$ ) to form quadrilateral $PQRS$ .

$\text{Area}(PQRS) = \dfrac{1}{2} \times \text{Area}(ABCD)$

Common Mistakes

Area vs Side Scaling: Confusing side ratios with area ratios. The side ratio of $\Delta PBQ$ to $\Delta ABC$ is $\dfrac{1}{2}$ , but its area ratio is $\left(\dfrac{1}{2}\right)^2 = \dfrac{1}{4}$ , not $\dfrac{1}{2}$ .
Generality Misconception: Assuming this method only applies to regular quadrilaterals like squares or rectangles. The midpoint property holds true for any general convex quadrilateral.

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 — $\text{Area of a rectangle} = \text{length} \times \text{width}$ .

[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 $14\text{ m}$ and $12\text{ m}$ , 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 $\text{BY}$ .

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

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 $\Delta\text{XYZ}$ is the area of $\Delta\text{XMN}$ ? [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 $AC = 22\text{ cm}$ , $BM = 3\text{ cm}$ , $DN = 3\text{ cm}$ , $BM$ 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 $\text{QN}$ .

Q23

Consider a rectangle and a parallelogram of the same sidelengths: $5\text{ cm}$ and $4\text{ cm}$ . 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 $\Delta \text{ADB}$ and $\Delta \text{ADC}$ 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 $20\text{ cm}$ and $15\text{ cm}$ .

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 $\Delta AHI \cong \Delta DGI$ and $\Delta BEJ \cong \Delta CFJ$ , 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 $ZY \parallel WX$ . $A$ is the midpoint of $XY$ . Show that the area of the trapezium ZYXW is equal to the area of $\triangle ZWB$ .