Area of Polygons | FIO

Question 19

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

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Solution

FIO-19

Chapter: AREA OF POLYGONS
Class: 8 (Class 8)
Category: figure_it_out

Question

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

We can find a quadrilateral with half the area of a given one by connecting the middle points of its sides.

Step 1 — Setting up our quadrilateral

Let's start with any four-sided shape, called a quadrilateral. We will call its corners A, B, C, and D. Now, let's find the middle point of each side. The middle point of side AB is P. The middle point of side BC is Q. The middle point of side CD is R. The middle point of side DA is S. Next, we connect these middle points in order: P to Q, Q to R, R to S, and S back to P. This creates a new quadrilateral inside, called PQRS.

Diagram 1

Step 2 — Understanding the inner shape

Let's draw a line connecting corners A and C. This line is called a diagonal. Look at the triangle ABC. P is the midpoint of AB, and Q is the midpoint of BC. There's a special rule called the Midpoint Theorem. It says that the line connecting the midpoints of two sides of a triangle is parallel to the third side. It also says this connecting line is half the length of the third side. So, PQ is parallel to AC. The length of PQ is half the length of AC. PQ∥ACPQ \parallel AC PQ=12ACPQ = \frac{1}{2} AC Now look at triangle ADC. S is the midpoint of DA, and R is the midpoint of CD. Using the Midpoint Theorem again, SR is parallel to AC. The length of SR is half the length of AC. SR∥ACSR \parallel AC SR=12ACSR = \frac{1}{2} AC Since both PQ and SR are parallel to AC, they must be parallel to each other. Also, since both PQ and SR are half the length of AC, they must have the same length. So, PQ and SR are parallel and equal in length. A quadrilateral with opposite sides that are parallel and equal is a parallelogram. Therefore, the quadrilateral PQRS is a parallelogram.

Step 3 — Comparing triangle areas

Let's focus on triangle ABC again. P and Q are midpoints. This means triangle PBQ is similar to triangle ABC. They have the same angles, and their sides are in proportion. The ratio of their sides is 1:2. For example, PB is half of AB, and BQ is half of BC. Similar triangles have a special area relationship. The ratio of their areas is the square of their side ratio. So, the area of triangle PBQ is (1/2) squared, or 1/4, of the area of triangle ABC. Area(PBQ)=(12)2×Area(ABC)\text{Area(PBQ)} = \left(\frac{1}{2}\right)^2 \times \text{Area(ABC)} =14×Area(ABC)= \frac{1}{4} \times \text{Area(ABC)} We can do the same for the other three corner triangles: For triangle QCR, its area is 1/4 of the area of triangle BCD. Area(QCR)=14×Area(BCD)\text{Area(QCR)} = \frac{1}{4} \times \text{Area(BCD)} For triangle SDR, its area is 1/4 of the area of triangle CDA. Area(SDR)=14×Area(CDA)\text{Area(SDR)} = \frac{1}{4} \times \text{Area(CDA)} For triangle SAP, its area is 1/4 of the area of triangle DAB. Area(SAP)=14×Area(DAB)\text{Area(SAP)} = \frac{1}{4} \times \text{Area(DAB)}

Step 4 — Finding the area of PQRS

The area of PQRS is the area of ABCD. We subtract the areas of the four corner triangles. Area(PQRS)=Area(ABCD)−(Area(PBQ)+Area(QCR)+Area(SDR)+Area(SAP))\text{Area(PQRS)} = \text{Area(ABCD)} - (\text{Area(PBQ)} + \text{Area(QCR)} + \text{Area(SDR)} + \text{Area(SAP)}) Now, let's substitute the area relationships we found in Step 3: Area(PQRS)=Area(ABCD)−(14Area(ABC)+14Area(BCD)+14Area(CDA)+14Area(DAB))\text{Area(PQRS)} = \text{Area(ABCD)} - \left(\frac{1}{4} \text{Area(ABC)} + \frac{1}{4} \text{Area(BCD)} + \frac{1}{4} \text{Area(CDA)} + \frac{1}{4} \text{Area(DAB)}\right) We can take out the common factor of 1/4: Area(PQRS)=Area(ABCD)−14(Area(ABC)+Area(BCD)+Area(CDA)+Area(DAB))\text{Area(PQRS)} = \text{Area(ABCD)} - \frac{1}{4} (\text{Area(ABC)} + \text{Area(BCD)} + \text{Area(CDA)} + \text{Area(DAB)}) Let's look at the sum inside the bracket. The area of ABCD can be split in two ways. We use its diagonals to do this. Using diagonal AC, Area(ABCD) = Area(ABC) + Area(CDA). Using diagonal BD, Area(ABCD) = Area(BCD) + Area(DAB). The sum inside the bracket is (Area(ABC) + Area(CDA)). Then we add (Area(BCD) + Area(DAB)). This sum is Area(ABCD) plus Area(ABCD). So, the total sum is 2 times Area(ABCD). Area(ABC)+Area(BCD)+Area(CDA)+Area(DAB)=2×Area(ABCD)\text{Area(ABC)} + \text{Area(BCD)} + \text{Area(CDA)} + \text{Area(DAB)} = 2 \times \text{Area(ABCD)} Now, let's put this back into our equation for Area(PQRS): Area(PQRS)=Area(ABCD)−14(2×Area(ABCD))\text{Area(PQRS)} = \text{Area(ABCD)} - \frac{1}{4} (2 \times \text{Area(ABCD)}) =Area(ABCD)−24×Area(ABCD)= \text{Area(ABCD)} - \frac{2}{4} \times \text{Area(ABCD)} =Area(ABCD)−12×Area(ABCD)= \text{Area(ABCD)} - \frac{1}{2} \times \text{Area(ABCD)} =12×Area(ABCD)= \frac{1}{2} \times \text{Area(ABCD)}

Area(PQRS)=12×Area(ABCD)\boxed{\text{Area(PQRS)} = \frac{1}{2} \times \text{Area(ABCD)}}

Answer

(i) Start with the given quadrilateral, let's call it ABCD. (ii) Find the midpoint of each side: P on AB, Q on BC, R on CD, and S on DA. (iii) Connect these midpoints in order (P to Q, Q to R, R to S, S to P) to form a new quadrilateral PQRS. This quadrilateral PQRS will have an area exactly half of the area of ABCD.

More questions in FIO

Q1

Identify the missing sidelengths.

Q2

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×widthArea\ 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?

Q3

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.

Q4

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?

Q5

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.

Q6

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.

Q7

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Q8

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Q9

Find the area of Δ\DeltaSUB, given that it is isosceles, SE is perpendicular to UB, and the area of Δ\DeltaSEB 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 Δ\DeltaXYZ is the area of Δ\DeltaXMN? [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 cm2144\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 ΔZWB\Delta\text{ZWB}.

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