Area of Polygons | IT

Question 17

Can ΔAXD\Delta\text{AXD} and ABCX\text{ABCX} fit together, as shown in the figure, to get a rectangle?

Question diagram 1
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Solution

For the combined shape to be a rectangle, its opposite sides must be equal in length and all its angles must be 90 degrees.

Step 1 — Understand the shapes and transformation

Let us look at the original figure ABCD. A line segment AX is drawn from vertex A, perpendicular to the base DC. This means AX is the height of the figure. The original figure is divided into a right-angled triangle AXD and a quadrilateral ABCX. The figure shows that triangle AXD is cut off from the left side. The remaining part is the quadrilateral ABCX. The triangle AXD is then moved to the right side of ABCX. The arrow indicates that the side AX of the triangle is placed vertically at the right end of the figure. The base DX of the triangle is placed along the line DC, next to point C.

Step 2 — Analyze the properties of the resulting figure

Let us examine the resulting figure, which is shown as a rectangle. The top side of this new figure is AB. The left side of this new figure is AX. The right side of this new figure is the moved segment AX (from the triangle). The bottom side of this new figure is formed by combining the segment XC and the moved segment DX. So, the total length of the bottom side is XC+DXXC + DX. We know that the entire base DC of the original quadrilateral is DX+XCDX + XC. Therefore, the length of the bottom side of the new figure is equal to DC.

Step 3 — Check conditions for a rectangle

For a figure to be a rectangle, its opposite sides must be equal in length. The top side of the new figure is AB. The bottom side of the new figure is DC. For the new figure to be a rectangle, its top side must be equal to its bottom side. So, we must have AB = DC. Also, for a rectangle, all angles must be 90 degrees. The original AX is perpendicular to DC, so the left angles are 90 degrees. When triangle AXD is moved, its side AX is placed perpendicular to the line DC. So, the right angles are also 90 degrees. Thus, the main condition for the combined figure to be a rectangle is that AB must be equal to DC.

Step 4 — Conclusion

The original figure ABCD is a quadrilateral. If AB is parallel to DC (which is implied by AX being a height for a trapezoid) and AB = DC, then ABCD is a parallelogram. If ABCD is a parallelogram, then the transformation shown will indeed form a rectangle. However, the diagram of ABCD shows a general trapezoid, where AB is not necessarily equal to DC. If AB is not equal to DC, the resulting figure will have unequal top and bottom sides (AB and DC). In that case, the resulting figure would be a parallelogram, but not a rectangle. Since the question asks if they can fit together to get a rectangle as shown, and the figure shows a rectangle, it implies that the conditions for it to be a rectangle must be met. This means the original quadrilateral ABCD must be a parallelogram.

Answer

No, not for every quadrilateral ABCD. It can only fit together to form a rectangle if the original quadrilateral ABCD is a parallelogram (meaning AB is parallel to DC and AB = DC).

Diagram 1

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 l∥BCl \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

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