In the previous section we completed the discussion on Construction of quadrilaterals. In this chapter, we will discuss about Area of quadrilaterals. First we will find the method to calculate the area of a Parallelogram.
We are going to make some modifications to this rectangle: Mark a point E, 2 cm from D as shown in fig.13.1(b). Join A to E. We get a triangle AED. Cut this triangle out, and move it to the right side. Attach it to the original piece in such a way that the vertical edge of the triangle coincides with the vertical edge BC. This is shown in the animation in fig.(c).
What do we get? We get a parallelogram ABFE as shown in fig.(d). In this new parallelogram, the longer parallel sides have a length of 6 cm, which is same as the length of the original rectangle. The height of the new parallelogram is also the same as the height of the original rectangle, which is 3 cm
Now we will discuss about a peculiarity in the areas of the original rectangle and the newly obtained parallelogram. We can see that nothing has been added to the original rectangle. What we did was, cut out a triangle, and attach it back in another position. Since nothing new is added, the area will remain the same. That is., the area of the newly obtained parallelogram is same as the area of the original rectangle.
In the above experiment, we marked E at a distance of 2 cm from D. Instead of 2 cm, we can use different values like, 3 cm, 1 cm, 4 cm etc., In all cases, we will find that, the area of the final parallelogram will be equal to 18 cm2. This is shown in the fig.13.2 below:
The following two points have to be specially noted:
■ The base AB is same for the original rectangle and the newly obtained parallelogram
■ The height AD is also same for the original rectangle and the newly obtained parallelogram
■ In all the three cases, the area of the newly obtained parallelogram is equal to the area of the original rectangle because, nothing is taken away and, nothing new is added.
The above discussion gives us an easy method to find the area of any parallelogram:
• We took out a triangle from the left side of the rectangle, and attached it to the right side. Such a modification gave us a parallelogram
• We can think in the reverse direction: If we remove a triangle from the right side of a parallelogram, and attach it to the left side, we will get a rectangle having the same area
• So any parallelogram has a triangle hidden inside it. Once we understand this basic property, we can appreciate the following equivalence of area:
The area of any given parallelogram having base ‘b’ and height ‘h’ will be equal to the area of a rectangle having the same base and height. This is shown in the fig.13.3 below:
ΔBCE is formed by drawing a line BE perpendicular to DC. If we move this triangle to the left side, a rectangle will be obtained. The rectangle will have the same base b, and height h as the parallelogram. The area of this newly obtained rectangle will be same as the original parallelogram because, nothing is taken away and, nothing new is added.
This makes the 'area calculation' of any parallelogram easy. All we have to do is to find the area of a rectangle having the same base b and height h as the given parallelogram. Area of a rectangle can be calculated very easily. It is equal to b × h
So we get: Area of the parallelogram = b × h. Thus we can write:
Eq.13.1: A = b × h (Where A is the area, b is the base and h is the height)
Solved example 13.1:
A parallelogram has a base of 5 cm and height of 2 cm. What is it's area?
Solution: A = b × h. In this problem, b = 5 cm, h = 2 cm
∴ A = 5 × 2 = 10 cm2
Another direction:
In the above discussions, we have seen the following points:
• From a given rectangle, we can cut various triangles from the left side, and attach to the right side, to get various parallelograms. The areas of all these parallelograms will be same as the original rectangle. This was shown earlier in fig.13.2
• We also saw that we can work in reverse, to get a rectangle from a parallelogram. That is., We can remove a triangle from the right side of a parallelogram, and attach it to the left side to get a rectangle. This was shown in fig.13.3
• There is an important difference when we work in this reverse order: Only one unique triangle (ΔBCE in fig.13.3) can be removed from a parallelogram.
♦ The triangle have a fixed base, fixed height, and fixed hypotenuse.
♦ It's altitude (height) passes through the corner of the parallelogram.
♦ It's hypotenuse is equal to the sloping side of the parallelogram
• For a given parallelogram, this triangle is unique. If we try to take out any other triangle, we will not get the rectangle having the same area.
Based on the above information, we can now do the calculations in 'another direction'. We Know that, a parallelogram has two sets of parallel sides. One set (eg: AB and CD in fig.13.3) is in the horizontal direction, and the other set (BC and AD) is in a sloping direction. So far, we have been doing the calculations in the horizontal direction. Now we will see the sloping direction.
Consider the Parallelogram ABCD shown in fig.13.4(a) below:
In fig.(ii), the 'unique triangle' has been brought out. This is by drawing a line BE perpendicular to AD. (just as we drew BE perpendicular to CD in fig.13.3). So the unique triangle is ΔABE. In fig.(ii), this triangle is moved to the upper side of the parallelogram, and thus we get a rectangle of the same area (shown in fig.iii). The area of the rectangle is b × h, and so, the area of the original parallelogram is also b × h (because, nothing is taken away and, nothing new is added)
So, for a given parallelogram, a rectangle of equal area can be obtained in two directions. Now we will see some solved examples.
Solved example 13.2
Construct a parallelogram in which sides are 6 cm and 5 cm, and area is 25 square cm
Solution:
The parallelogram has to satisfy two conditions:
■ Sides must be 6 cm and 5 cm
■ Area must be 25 cm2
• We can first make a rectangle of 25 cm2 area, and then modify it to make a parallelogram having the same area. For that, we remove a triangle from the left side of the rectangle, and attach it to the right side.
• But the rectangle should have the same base as the parallelogram. Let the base of the rectangle be 5 cm. Then, for it to have the 25 cm2 area, the height must also be 5 cm (∵ 5 × 5 = 25)
• So we fix up a rectangle ABCD (or rather a square, because length and height are equal) having base 5 cm and height 5 cm. This is shown in fig.13.5(a)
♦ Note: If we take 6 cm as the base, there is no whole number which will give an area of 25 (∵ 25 ÷ 6 gives a decimal value, not a whole number)
• The side of the parallelogram is to be 6 cm. So, with A as center, draw an arc (shown in green colour in fig.b) of radius 6 cm, cutting CD at E. Join AE. The length of AE = 6 cm.
• Through B, draw a line BF', parallel to AE.
• Extend the line CD towards the right until it meets BF' at F.
• Change the firm lines AD, BC and ED into dashed lines. Then ABFE is the required Parallelogram.
Proof:
• ABFE is a parallelogram. It is evident because:
♦ Opposite sides AB and EF are already parallel as they are opposite sides of the original square ABCD
♦ Opposite sides AE and BF are parallel because BF is drawn parallel to AB
• In fig.c , we can see that, from the square ABCD, a triangle AED is removed from the left side, and another triangle BFC is attached at the right side. We have to prove that these two triangles are equal. For this, we can use SAS congruence:
• AD = BC = 5 cm (since opposite sides of a square)
• AE = BF = 6 cm (since opposite sides of a parallelogram)
• ∠DAE = ∠CBF (∵ AD∥BC and AE∥BF)
• So the two triangles are equal.
In the next section, we will see more solved examples.
Area of Parallelogram
Consider the rectangle ABCD shown in fig.13.1(a). It is a cardboard cut out, having a length of 6 cm and height of 3 cm. So it’s area is 6 × 3 = 18 cm2. |
| Fig.13.1 |
What do we get? We get a parallelogram ABFE as shown in fig.(d). In this new parallelogram, the longer parallel sides have a length of 6 cm, which is same as the length of the original rectangle. The height of the new parallelogram is also the same as the height of the original rectangle, which is 3 cm
Now we will discuss about a peculiarity in the areas of the original rectangle and the newly obtained parallelogram. We can see that nothing has been added to the original rectangle. What we did was, cut out a triangle, and attach it back in another position. Since nothing new is added, the area will remain the same. That is., the area of the newly obtained parallelogram is same as the area of the original rectangle.
In the above experiment, we marked E at a distance of 2 cm from D. Instead of 2 cm, we can use different values like, 3 cm, 1 cm, 4 cm etc., In all cases, we will find that, the area of the final parallelogram will be equal to 18 cm2. This is shown in the fig.13.2 below:
 |
| Fig.13.2 |
■ The base AB is same for the original rectangle and the newly obtained parallelogram
■ The height AD is also same for the original rectangle and the newly obtained parallelogram
■ In all the three cases, the area of the newly obtained parallelogram is equal to the area of the original rectangle because, nothing is taken away and, nothing new is added.
The above discussion gives us an easy method to find the area of any parallelogram:
• We took out a triangle from the left side of the rectangle, and attached it to the right side. Such a modification gave us a parallelogram
• We can think in the reverse direction: If we remove a triangle from the right side of a parallelogram, and attach it to the left side, we will get a rectangle having the same area
• So any parallelogram has a triangle hidden inside it. Once we understand this basic property, we can appreciate the following equivalence of area:
The area of any given parallelogram having base ‘b’ and height ‘h’ will be equal to the area of a rectangle having the same base and height. This is shown in the fig.13.3 below:
 |
| Fig.13.3 |
This makes the 'area calculation' of any parallelogram easy. All we have to do is to find the area of a rectangle having the same base b and height h as the given parallelogram. Area of a rectangle can be calculated very easily. It is equal to b × h
So we get: Area of the parallelogram = b × h. Thus we can write:
Eq.13.1: A = b × h (Where A is the area, b is the base and h is the height)
Solved example 13.1:
A parallelogram has a base of 5 cm and height of 2 cm. What is it's area?
Solution: A = b × h. In this problem, b = 5 cm, h = 2 cm
∴ A = 5 × 2 = 10 cm2
Another direction:
In the above discussions, we have seen the following points:
• From a given rectangle, we can cut various triangles from the left side, and attach to the right side, to get various parallelograms. The areas of all these parallelograms will be same as the original rectangle. This was shown earlier in fig.13.2
• We also saw that we can work in reverse, to get a rectangle from a parallelogram. That is., We can remove a triangle from the right side of a parallelogram, and attach it to the left side to get a rectangle. This was shown in fig.13.3
• There is an important difference when we work in this reverse order: Only one unique triangle (ΔBCE in fig.13.3) can be removed from a parallelogram.
♦ The triangle have a fixed base, fixed height, and fixed hypotenuse.
♦ It's altitude (height) passes through the corner of the parallelogram.
♦ It's hypotenuse is equal to the sloping side of the parallelogram
• For a given parallelogram, this triangle is unique. If we try to take out any other triangle, we will not get the rectangle having the same area.
Based on the above information, we can now do the calculations in 'another direction'. We Know that, a parallelogram has two sets of parallel sides. One set (eg: AB and CD in fig.13.3) is in the horizontal direction, and the other set (BC and AD) is in a sloping direction. So far, we have been doing the calculations in the horizontal direction. Now we will see the sloping direction.
Consider the Parallelogram ABCD shown in fig.13.4(a) below:
 |
| Fig.13.4 |
So, for a given parallelogram, a rectangle of equal area can be obtained in two directions. Now we will see some solved examples.
Solved example 13.2
Construct a parallelogram in which sides are 6 cm and 5 cm, and area is 25 square cm
Solution:
The parallelogram has to satisfy two conditions:
■ Sides must be 6 cm and 5 cm
■ Area must be 25 cm2
• We can first make a rectangle of 25 cm2 area, and then modify it to make a parallelogram having the same area. For that, we remove a triangle from the left side of the rectangle, and attach it to the right side.
• But the rectangle should have the same base as the parallelogram. Let the base of the rectangle be 5 cm. Then, for it to have the 25 cm2 area, the height must also be 5 cm (∵ 5 × 5 = 25)
• So we fix up a rectangle ABCD (or rather a square, because length and height are equal) having base 5 cm and height 5 cm. This is shown in fig.13.5(a)
♦ Note: If we take 6 cm as the base, there is no whole number which will give an area of 25 (∵ 25 ÷ 6 gives a decimal value, not a whole number)
 |
| Fig.13.5 |
• Through B, draw a line BF', parallel to AE.
• Extend the line CD towards the right until it meets BF' at F.
• Change the firm lines AD, BC and ED into dashed lines. Then ABFE is the required Parallelogram.
Proof:
• ABFE is a parallelogram. It is evident because:
♦ Opposite sides AB and EF are already parallel as they are opposite sides of the original square ABCD
♦ Opposite sides AE and BF are parallel because BF is drawn parallel to AB
• In fig.c , we can see that, from the square ABCD, a triangle AED is removed from the left side, and another triangle BFC is attached at the right side. We have to prove that these two triangles are equal. For this, we can use SAS congruence:
• AD = BC = 5 cm (since opposite sides of a square)
• AE = BF = 6 cm (since opposite sides of a parallelogram)
• ∠DAE = ∠CBF (∵ AD∥BC and AE∥BF)
• So the two triangles are equal.
In the next section, we will see more solved examples.
No comments:
Post a Comment