In the previous section we completed the discussion on 'unchanging areas of triangles between parallel lines'. In this section, we will see the relation between the areas of quadrilaterals and triangles.
• Fig.14.11(a) below, shows a quadrilateral ABCD
• In the fig.b, a diagonal BD is drawn. This diagonal splits the quadrilateral into two triangles: ΔABD and ΔCBD. Through C, draw a line MN, parallel to the diagonal BD. This is shown in fig.b.
• Extend the side AB towards the right. It will meet MN at E. This is shown in the fig.c
• Join B and D to E. We will get a new ΔEBD as shown in fig.d
• What is the peculiarity of this new ΔEBD ?
ΔEBD and ΔCBD has the same base BD. Their third vertices E and C lies on the same line MN parallel to the base BD, and passing through C. So ΔEBD has the same area as ΔCBD
• Now ignore the original ΔCBD. Because, we have it's area in another shape, which is ΔEBD.
We split the original quadrilateral into two triangles, and now, one of them has transformed into another shape with the same area. So the total area of the quadrilateral remains unchanged
• Besides 'keeping the area the same', we achieve one more thing in this transformation:
Look at the final shape in fig.e: AED is a triangle. This was achieved by marking 'E' in line with the side AB. When one vertex comes in line with another two vertices, one side will disappear. This is shown in the animation in fig.14.12 below:
In fig.b, the vertex 'C' comes in line with side AB. So two sides: AB and BC becomes one side AE
• So we transformed a '4 sided quadrilateral' into a '3 sided triangle' with the same area.
Figs. (a) to (e) in 14.11 were drawn to show each step of the transformation. In an actual construction, We can do all the steps in a single fig. This is shown in the fig. 14.13 below:
The fig.14.13 was drawn quickly by following the steps below:
• Draw diagonal BD
• Draw MN parallel to BD through C
• On MN, mark E, in line with AB
• Join B and D to E. This gives the transformed shape: ΔAED
So now we know how to transform a quadrilateral into a triangle of same area. The same method can be used to transform any polygon into a triangle of same area. But we may have to do the above process many number of times. The 'number of times' will depend on the number of sides of the polygon. For example, if we are given a pentagon, we will apply the method once to get a quadrilateral. So the five sides of the pentagon will become four sides of a quadrilateral. Then we apply the method a second time to change the quadrilateral into a triangle. Like wise, if we are given a hexagon, we will have to apply the method 3 times. Fig.14.14 below shows the transformation of a pentagon.
• Fig.14.14(a) shows the original pentagon
• In fig.(b), the pentagon is split into a quadrilateral and a triangle, by drawing the diagonal BD
• In fig(c), one side of the pentagon is reduced, by bringing the vertex C to F. Thus we get a quadrilateral AFDE. This quadrilateral has the same area as the original pentagon ABCDE
• In fig(d), the method is applied again. This time, on the left side of the quadrilateral. (This is because, if we continue working on the right side, the resulting triangle will be too elongated towards the right. The reader may try that too). In fig.(d), the quadrilateral is split into two triangles.
• In fig(e), the vertex E is brought down to G. We get the final ΔGFD
■ While doing the actual construction, the above steps can be done in a single fig. This is shown in fig(f). The steps involved are:
• Draw the diagonal BD. Draw MN through C, parallel to BD
• Extend AB towards the right, to meet MN at F. Join B and D to F
• Now work on the left side of the pentagon. Draw diagonal AD. Draw M'N' through E, parallel to AD
• Extend AB towards the left, to meet M'N' at G. Join A and D to G. ΔGFD is the required triangle
In the next section we will see some solved examples.
• Fig.14.11(a) below, shows a quadrilateral ABCD
 |
| Fig.14.11 |
• Extend the side AB towards the right. It will meet MN at E. This is shown in the fig.c
• Join B and D to E. We will get a new ΔEBD as shown in fig.d
• What is the peculiarity of this new ΔEBD ?
ΔEBD and ΔCBD has the same base BD. Their third vertices E and C lies on the same line MN parallel to the base BD, and passing through C. So ΔEBD has the same area as ΔCBD
• Now ignore the original ΔCBD. Because, we have it's area in another shape, which is ΔEBD.
We split the original quadrilateral into two triangles, and now, one of them has transformed into another shape with the same area. So the total area of the quadrilateral remains unchanged
• Besides 'keeping the area the same', we achieve one more thing in this transformation:
Look at the final shape in fig.e: AED is a triangle. This was achieved by marking 'E' in line with the side AB. When one vertex comes in line with another two vertices, one side will disappear. This is shown in the animation in fig.14.12 below:
 |
| Fig.14.12 |
• So we transformed a '4 sided quadrilateral' into a '3 sided triangle' with the same area.
Figs. (a) to (e) in 14.11 were drawn to show each step of the transformation. In an actual construction, We can do all the steps in a single fig. This is shown in the fig. 14.13 below:
 |
| Fig.14.13 |
• Draw diagonal BD
• Draw MN parallel to BD through C
• On MN, mark E, in line with AB
• Join B and D to E. This gives the transformed shape: ΔAED
So now we know how to transform a quadrilateral into a triangle of same area. The same method can be used to transform any polygon into a triangle of same area. But we may have to do the above process many number of times. The 'number of times' will depend on the number of sides of the polygon. For example, if we are given a pentagon, we will apply the method once to get a quadrilateral. So the five sides of the pentagon will become four sides of a quadrilateral. Then we apply the method a second time to change the quadrilateral into a triangle. Like wise, if we are given a hexagon, we will have to apply the method 3 times. Fig.14.14 below shows the transformation of a pentagon.
• Fig.14.14(a) shows the original pentagon
• In fig.(b), the pentagon is split into a quadrilateral and a triangle, by drawing the diagonal BD
• In fig(c), one side of the pentagon is reduced, by bringing the vertex C to F. Thus we get a quadrilateral AFDE. This quadrilateral has the same area as the original pentagon ABCDE
• In fig(d), the method is applied again. This time, on the left side of the quadrilateral. (This is because, if we continue working on the right side, the resulting triangle will be too elongated towards the right. The reader may try that too). In fig.(d), the quadrilateral is split into two triangles.
• In fig(e), the vertex E is brought down to G. We get the final ΔGFD
■ While doing the actual construction, the above steps can be done in a single fig. This is shown in fig(f). The steps involved are:
• Draw the diagonal BD. Draw MN through C, parallel to BD
• Extend AB towards the right, to meet MN at F. Join B and D to F
• Now work on the left side of the pentagon. Draw diagonal AD. Draw M'N' through E, parallel to AD
• Extend AB towards the left, to meet M'N' at G. Join A and D to G. ΔGFD is the required triangle
In the next section we will see some solved examples.
No comments:
Post a Comment