In the previous section we completed the discussion on Similar triangles. In this section we will learn about Polynomials.
Now we will consider area.
The original area = l × b = 3 × 2 = 6 cm (see fig.20.2.1 below)
1. When the length and width are
increased by 1 cm, new length = 4 cm, and new width = 3 cm. So new area =
4 × 3 = 12 cm2
We have seen perimeter and
area. Now let us see volume.
Consider a rectangle shown in
fig.20.1.1(a). It has a length (l) = 3 cm and width (b) = 2 cm.
 |
| Fig.20.1 |
• We can easily
calculate it's perimeter.
p = 2(l+b) = 2(3+2) = 2 × 5 = 10 cm
1. Suppose we increase the length
by 1 cm, and width also by the same 1 cm, the new length = 4 cm, and
new width = 3 cm
• Then new perimeter = 2(4+3) = 14
cm. This is shown in fig.20.1.1(c)
2. Suppose we increase the
original length by 2 cm, and original width also by the same 2 cm,
the new length = 3 +2 =5 cm, and new width = 2 + 2 = 4 cm
• Then new perimeter = 2(5+4) = 18
cm. This is shown in fig.20.1.2(c)
3. Suppose we increase the
original length by 11⁄2 cm, and original width also by the same 11⁄2 cm, the new length = 3 + 11⁄2 = 41⁄2 cm, and new width = 2 + 11⁄2 = 31⁄2 cm
• Then new perimeter = 2(41⁄2 + 31⁄2) = 16 cm
■ So we find that, when the
dimensions of the rectangle changes, the perimeter also changes. We
can write a general form (see fig.20.1.3):
• Let the increase in length be
'x'. Then the new length = (3+x)
• Let the increase in width also
be 'x'. Then the new width = (2+x)
• Then, the new perimeter = 2
[(3+x) + (2+x)] = 2 [5+2x] = 10+4x = 4x +10
We can say this:
• We have a rectangle with
length 3 cm and width 2 cm
• If the length and width are
increased by the same amount 'x', the new perimeter will be given by 4x+10
• So this is an easy method to
calculate new perimeter. Let us put some values for x. The new
perimeters are tabulated below:
Value of perimeter for different values of 'x':
| x (cm) | 1 | 11⁄2 | 2 | 21⁄4 | 3 |
|---|---|---|---|---|---|
| p (cm) | 14 | 16 | 18 | 19 | 22 |
The original area = l × b = 3 × 2 = 6 cm (see fig.20.2.1 below)
 |
| Fig.20.2 |
• In the fig.20.2.1.c, the additional
areas are marked by green rectangles. We can write:
• Original area = 6, additional
area = 3 +1 +2 = 6. So total area = 6 +6 = 12
2. When the original length and
width are increased by 2 cm, new length = 5 cm, and new width = 4 cm. So
new area = 5 × 4 = 20. This is shown in fig.20.2.2.c
• In the fig.20.2.2.c, the
additional areas are marked by green rectangles. We can write:
• Original area = 6, additional
area = 6 +4 +4 = 14. So total area = 6 +14 = 20
■ So we find that, when the dimensions of the rectangle changes, the area also changes. We can write a general form (see fig.20.2.3):
■ So we find that, when the dimensions of the rectangle changes, the area also changes. We can write a general form (see fig.20.2.3):
• Let the increase in length be 'x'. Then the new length = (3+x)
• Let the increase in width also be 'x'. Then the new width = (2+x)
• Then new area = [(3+x)(2+x)]. We
have learned how to multiply them, when we saw identities. We get: [(3+x)(2+x)] = 6 + 2x + 3x +
x2 = x2+ 5x + 6
[The value '6' in the above
result is the area of the red rectangle in fig.c. The others 2x, 3x
and x2 are the areas of additional green rectangles]
• We can write: New area = x2+ 5x + 6
• So this is an easy method to
calculate new areas. Let us put some values for x. The new areas are
tabulated below:
Value of area for different values of 'x':
| x (cm) | 1 | 11⁄2 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| a (cm) | 12 | 153⁄4 | 20 | 30 | 42 |
So we have seen two equations:
■ New perimeter = 4x+10
■ New area = x2+ 5x + 6
Apart from making our
calculations easier, such equations have wider applications.
• Instead of 'New perimeter',
let us write 'p(x)'. Then the equation will become: p(x) = 10 + 4x
• 'p(x)' indicates that we must
not put a 'permanent value' for perimeter.
• It is kind of a 'warning', which tells us that, in this particular problem, the perimeter, (which is denoted by 'p') will change based on the value of 'x'
• It is kind of a 'warning', which tells us that, in this particular problem, the perimeter, (which is denoted by 'p') will change based on the value of 'x'
• The manner in which the change takes
place, will be given on the right side of the '=' sign.
• Similarly, in the second
equation, instead of 'New area', let us write 'a(x)'. Then the
equation will become: a(x) = x2+ 5x + 6
• 'a(x)' indicates that we must not put a 'permanent value' for area.
• It is kind of a 'warning', which tells us that, in this particular problem, the area, (which is denoted by 'a') will change based on the value of 'x'
• The manner in which the change takes place, will be given on the right side of the '=' sign.
• It is kind of a 'warning', which tells us that, in this particular problem, the area, (which is denoted by 'a') will change based on the value of 'x'
• The manner in which the change takes place, will be given on the right side of the '=' sign.
1. In fig.20.3(a) below, we have a red box of length 3
cm, width 2 cm, and height 1 cm. So it's volume = 3 × 2 × 1 = 6 cm3
 |
| Fig.20.3 |
2. Its
length and width are increased by 'x' cm as indicated by the green
rectangles in fig.a. It's height is increased by the same 'x' cm, as indicated
by the yellow rectangle in fig.a.
3. If we give a height of 1 cm
for the green rectangles, they will reach the top level of the red box. This is shown in fig.b. Then, the additional volumes will
be:
♦ 3 ×x ×1 = 2x cm3
♦ 2 ×x ×1 = 3x cm3
♦ x ×x ×1 = x2 cm3
♦ 3 ×x ×1 = 2x cm3
♦ 2 ×x ×1 = 3x cm3
♦ x ×x ×1 = x2 cm3
• So the total volume in fig.b =
6 + 2x + 3x + x2 = x2+ 5x + 6
4. Now, in fig.c, the additional
height x cm is also given volume. The volume of this yellow box =
[(2+x)x(3+x)] = [(6+5x+x2)x] = x3+ 5x2 + 6x
5. Thus, the total volume in
fig.c = x2+ 5x + 6 + x3+ 5x2 + 6x
= x3+ 6x2 + 11x + 6
= x3+ 6x2 + 11x + 6
The same result can be
obtained by multiplying the new length, width and height together.
This is shown below:
(3+x)(2+x)(1+x) =
(6+5x+x2)(1+x)
= 6+5x+x2+6x+5x2+x3 = x3+6x2+11x+6
= 6+5x+x2+6x+5x2+x3 = x3+6x2+11x+6
We can write: v(x) = x3+6x2+11x+6
• 'v(x)' indicates that we must not put a 'permanent value' for volume.
• It is kind of a 'warning', which tells us that, in this particular problem, the volume, (which is denoted by 'v') will change based on the value of 'x'
• The manner in which the change takes place, will be given on the right side of the '=' sign
• It is kind of a 'warning', which tells us that, in this particular problem, the volume, (which is denoted by 'v') will change based on the value of 'x'
• The manner in which the change takes place, will be given on the right side of the '=' sign
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