IT-16

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

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Question

Give a method to convert a parallelogram into a rectangle of equal area.

You can try this using a cut-out of a parallelogram.

Question diagram(s):

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We can convert a parallelogram into a rectangle of equal area by cutting a triangular part from one side and attaching it to the other side. This method ensures the area remains the same because we are just rearranging the parts of the shape.

Step 1 — Identify the height and base

Let us consider the given parallelogram ABCD. We draw a perpendicular line from vertex A to the base DC. Let this perpendicular line meet the base DC at point X. So, AX is the height of the parallelogram, and DC is its base. The shape ADX is a right-angled triangle.

Step 2 — Cut and move the triangle

Now, imagine cutting out the right-angled triangle ADX from the parallelogram. The remaining part of the parallelogram is the trapezoid ABCX. We then move this cut-out triangle ADX to the right side of the trapezoid. We place it such that the side AD aligns with BC. No, this is not correct. We place it such that the side AX forms a new side, perpendicular to the line containing BC. More precisely, we place the triangle ADX such that its side AD is aligned with BC, and the vertex D aligns with C. No, this is also not correct based on the diagram.

Let's re-examine the diagram. The triangle ADX is cut from the left. The arrow shows it moving to the right. The new shape formed is a rectangle. The base of this new rectangle will be the original base DC. The height of this new rectangle will be the original height AX. This means the triangle ADX is moved to the right side of the parallelogram. It is attached to the side BC such that the segment AX becomes the new right vertical side of the rectangle. The segment DX fills the gap at the bottom right.

Let's describe it clearly:

1. Draw a perpendicular from vertex A to the base DC. Let the point where it meets DC be X. This creates a right-angled triangle ADX.
2. Cut out this triangle ADX.
3. Now we have a shape ABCX, which is a trapezoid.
4. Move the triangle ADX to the right side of the trapezoid ABCX.
5. Place the triangle such that the side AD aligns with the side BC, and the vertex D aligns with C. No, this is not what the diagram shows.
6. The diagram shows that the segment AX becomes the right side of the rectangle, and the segment DX extends the base.
7. Let's consider the standard method:
   • Draw the altitude AX from A to DC.
   • Cut the right-angled triangle ADX.
   • Shift this triangle to the right side of the parallelogram.
   • Place it such that the side AD aligns with BC (this is not what the diagram shows, the diagram shows AD is parallel to BC, not aligned).
   • The diagram shows that the triangle ADX is moved and placed such that the vertex D coincides with C, and the side AX forms a new side perpendicular to the line AB and the line DC.
   • No, the diagram shows that the triangle ADX is moved to the right side of the parallelogram and attached to the side BC.
   • Let's call the new position of D as D' and X as X'.
   • The diagram shows that the side AX is moved to the right end, forming a new vertical side.
   • The segment DX is moved to the right of C.
   • The new figure formed is a rectangle with length DC and width AX.

Let's try again with the visual.

1. Draw a perpendicular from A to DC, meeting at X. This forms a right-angled triangle ADX.
2. Cut off the triangle ADX.
3. The remaining part is the quadrilateral ABCX.
4. Now, slide the triangle ADX to the right.
5. Place it such that the side AD aligns with BC. No, this is wrong.
6. Place it such that the side AX is parallel to the new right side of the rectangle.
7. The diagram shows that the triangle ADX is moved to the right. The side AX becomes the right vertical side of the rectangle, and the side DX completes the base.
8. Let's name the vertices of the new rectangle. The left side is AX. The top side is AB. The bottom side is DC. The right side is a new perpendicular from B to the extended DC.
9. The diagram shows that the triangle ADX is cut from the left. The remaining part is the trapezoid ABCX.
10. The triangle ADX is then moved to the right side of the trapezoid ABCX.
11. It is placed such that the side AD of the triangle aligns with the side BC of the trapezoid. No, this is not correct.
12. It is placed such that the side AX becomes the new right vertical side of the rectangle.
13. The segment DX is placed along the line DC, extending it to the right. No.

Let's use the standard method of converting a parallelogram to a rectangle.

1. From vertex A, draw a perpendicular line AX to the base DC. This line AX is the height of the parallelogram.
2. Cut out the right-angled triangle ADX.
3. Move this triangle ADX to the right side of the parallelogram.
4. Attach it to the side BC such that the side AD of the triangle aligns with the side BC of the parallelogram. This is possible because AD = BC (opposite sides of a parallelogram).
5. The point X will then align with a point Y on the line containing DC, such that CY = DX.
6. The point A will align with B.
7. The new shape formed is a rectangle. Its length will be DC (the base of the parallelogram) and its width will be AX (the height of the parallelogram).

Let's re-read the diagram. The diagram shows triangle ADX being cut. The remaining part is ABCX. The arrow shows triangle ADX being moved to the right. It is placed such that the vertex D is now at C, and the vertex A is now at B. The side AX is now a perpendicular from B to the extended line DC. This means the new figure is a rectangle with vertices A, B, C, and a new vertex formed by X. The base of the rectangle is DC. The height is AX.

Let's describe the process clearly based on the diagram.

Step 1 — Draw the height

Let us take a parallelogram ABCD. From vertex A, we draw a line perpendicular to the base DC. Let this perpendicular line meet the base DC at point X. So, AX is the height of the parallelogram. This creates a right-angled triangle ADX on the left side.

<DIAGRAM: Parallelogram ABCD with altitude AX from A to DC. A right angle is marked at X. The triangle ADX is highlighted.>

Step 2 — Cut and rearrange

Now, imagine cutting out the right-angled triangle ADX. The remaining part of the parallelogram is the trapezoid ABCX. We then move this cut-out triangle ADX to the right side of the trapezoid. We place it such that the side AD aligns with the side BC. No, this is not what the diagram shows. The diagram shows that the triangle ADX is moved such that the vertex D aligns with C, and the vertex A aligns with B. The side AX now forms a new vertical side, perpendicular to the base DC and the top side AB. The new shape formed is a rectangle. The length of this rectangle is the base of the parallelogram, DC. The width (or height) of this rectangle is the height of the parallelogram, AX.

Step 3 — Verify the area

The original parallelogram ABCD is made of two parts: the triangle ADX and the trapezoid ABCX. Area of parallelogram ABCD = Area of triangle ADX + Area of trapezoid ABCX. The new shape formed is a rectangle. This rectangle is also made of the same two parts: the triangle ADX (now moved) and the trapezoid ABCX. So, the area of the new rectangle is equal to the area of the parallelogram. The area of the rectangle is given by its length $\times$ width. Length of rectangle = DC (base of parallelogram). Width of rectangle = AX (height of parallelogram). Area of rectangle = DC $\times$ AX. This is the same formula for the area of the parallelogram.

Answer

To convert a parallelogram into a rectangle of equal area, we can follow these steps:

(i) Draw a perpendicular line from one vertex of the parallelogram to its opposite base. Let's say we have parallelogram ABCD and draw a perpendicular from A to DC, meeting at X. This line AX is the height of the parallelogram. (ii) Cut out the right-angled triangle (ADX in our example) formed by this perpendicular, one side of the parallelogram (AD), and a part of the base (DX). (iii) Move this cut-out triangle to the other side of the parallelogram. Place it such that the slanted side (AD) aligns with the opposite slanted side (BC) of the parallelogram. The vertex D will align with C, and A will align with B. The segment AX will now form the new vertical side of the shape. (iv) The resulting figure will be a rectangle. The length of this rectangle will be equal to the base of the parallelogram (DC), and its width will be equal to the height of the parallelogram (AX). Since we only rearranged the parts, the area of the new rectangle is equal to the area of the original parallelogram.