In the previous section we completed the discussion on the multiplication of irrational numbers. In this section, we will see division.
1. Consider the product: 2 ×
3 = 6. It can be written as division also:
6⁄3 =
2 AND 6⁄2 = 3
In general, for any two
natural numbers or fractions x and y:
■ We can write the product
as x × y = z
■ We can write the division
as z⁄x = y AND z⁄y
= x
2. Now consider the product:
√2 × √3 = √6. It can be written as division also:
(√6⁄√3)
= √2 AND (√6⁄√2) = √3
In general, for any two
positive numbers x and y:
■ We can write the product
as √x × √y = √z
■ We can write the division
as (√z⁄√x) = √y AND
(√z⁄√y) = √x
3. Now consider the following
equations that we saw in (1):
6⁄3 =
2 and 6⁄2 = 3
3. Taking square root on both
sides, we get √(6⁄3) = √2 and = √(6⁄2)
= √3
4. We have seen in (2) above
that (√6⁄√3) = √2.
5. So √(6⁄3)
and (√6⁄√3), are both equal to √2
That means: √(6⁄3)
= (√6⁄√3)
The square root ‘applied to
the whole fraction’ is equivalent to applying the square root
‘separately to numerator and denominator’
5. Similarly, consider the
second part in (2): √(6⁄2) = √3
6. We have seen in (2) above
that (√6⁄√2) = √3.
7. So √(6⁄2)
and (√6⁄√2), are both equal to √3
That means: √(6⁄2)
= (√6⁄√2)
■ The square root ‘applied to
the whole fraction’ is equivalent to applying the square root
‘separately to numerator and denominator’
In general, for any two
positive numbers x and y: √(x⁄y) = (√x⁄√y)
Let us see a practical
application of this property:
Compute: √(1⁄2)
1. We have √(1⁄2)
= √1⁄√2 = 1⁄√2)
2. Putting approximate value
of √2, we get: 1⁄1.414 = 0.707
3. So we can write: Square
root of 1⁄2 = √(1⁄2)
= √1⁄√2 = 1⁄√2
= 1⁄1.414 = 0.707
There is an easier
method. For that, we use the fact that 1⁄2 =
2⁄4
So we get: √(1⁄2)
= √(2⁄4) = √2⁄√4
= √2⁄2 = 1.414⁄2 =
0.707
• The advantage of using
this method is that, unlike 1⁄1.414 , we do
not need a calculator to find 1.414⁄2
• From the above, we get
another interesting result: √(1⁄2) = √2⁄2
■ That is., Square root of
1⁄2 = half of √2
This can be verified
geometrically also:
1. Consider ⊿ABC in fig.16.33(a) below:
 |
| Fig.16.33 |
Both it’s legs AB and BC are 1 m long. So the hypotenuse BC will be
√2 m
2. Now draw perpendicular
bisectors to all the three sides as shown in fig.b
(a) The red arcs are for the
perpendicular bisector of AB. This bisector cuts AB at E. So AE = BE
= 1⁄2
m
(b) The blue arcs are for the
perpendicular bisector of AC. This bisector cuts AC at F. So AF = CF
= 1⁄2
m
(c) The green arcs are for the
perpendicular bisector of BC. This bisector cuts BC at D. So CD = BD
= √2⁄2
m
• Note that, all the three
bisectors intersect at ‘D’, which is the midpoint of the
hypotenuse.
3. Consider the inner triangle
EBD. As DE is the perpendicular bisector of AB, The ∠DEB = 90o .
So triangle DEB is a right triangle.
4. FD is parallel to AB. Also
∠FAB = ∠DEB = 90o. So AF [from 2(b)] = DE = 1⁄2
m
5. Thus we get the two legs of
the inner triangle EBD:
• DE = 1⁄2
m [from (4)] • BE = 1⁄2
m [from 2(a)]
6. Applying Pythagoras
theorem, we get: Hypotenuse BD = √[(1⁄2)2
+ (1⁄2)2]
= √[1⁄4
+ 1⁄4]
= √[1⁄2]
= √(1⁄2)
7. Thus, by doing calculations
on the inner triangle EBD, we obtain BD = √(1⁄2)
8. Now, considering the whole
triangle ABC, we obtain BD = half of BC = √2⁄2
m
9. Comparing (7) and (8), we
get: √(1⁄2)
= √2⁄2
Another example: Compute:
√(1⁄3)
√(1⁄3)
= √(3⁄9) = √3⁄√9
= √3⁄3 = 1.732⁄3 =
0.5774
Now we will see a solved
example:
Solved example 16.18
Calculate the length of side
of the equilateral triangle shown in fig.16.34(a) below:
 |
| Fig.16.34 |
Solution:
We are given a triangle. The
only details available are: (i) it is an equilateral triangle, and
(ii) it’s altitude is 4 cm long
1. Let us name the triangle as
ABC, as shown in fig.b. The altitude is CD
2. All angles of an equilateral
triangle are 60o.
3. The altitude CD bisects the
angle ACB. So ∠ACD = 30o
4. Also, the altitude is
perpendicular to AB. So ∠ADC = 90o
5. So the ⊿ADC is a 30, 60, 90
triangle. (Details here). If we know any one side of such a triangle,
we can calculate the other two sides
6. The one side available is CD = 4 cm. It is opposite to 60o, the intermediate angle.
So CD is the medium side
7. Medium side is √3 times the
smallest side. Here, the smallest side is AD (opposite to the
smallest ang1e 30o). So we can write:
8. CD = √3 × AD ⇒ 4 = √3
× AD ⇒ AD = 4⁄√3
cm
9. Now consider the largest side,
the hypotenuse AC. It is twice the smallest side. So we can write:
AC = 2 × AD ⇒ AC = 2 ×
4⁄√3 ⇒ AC = 8⁄√3
cm
10. So we have obtained the
required answer: One side = AC = 8⁄√3
cm. To find 8⁄√3 with out a calculator, we do some modifications:
11. Multiply both numerator and
denominator by √3. We get:
8⁄√3
= (8 × √3)⁄(√3
× √3) = 8√3⁄3
= (8 × 1.732)⁄3
= 4.619 cm
Solved example 16.19
Prove that (√2+1)(√2-1) =
1. Use this to compute 1⁄(√2-1)
correct to two decimal places
Solution:
1. (√2+1)(√2-1) is in the
form: (a+b)(a-b)
2. Now, (a+b)(a-b) is a part of a
particular identity.
• We have learned about identities in earlier
classes. We saw their applications in a previous chapter on
'solutions of equations'.
• (a+b)(a-b)= a2-b2
is the fifth identity mentioned at the beginning of chapter 15.3. Let
us see how we can apply this for our present problem:
3. We have: (√2+1)(√2-1) =
(√2)2
– 12
= 2 -1 = 1
4. So we easily proved that
(√2+1)(√2-1) = 1. From this we get 1⁄(√2-1)
= (√2+1)
5. ⇒1⁄(√2-1)
= 1.41 + 1 = 2.41
Solved example 16.20
Compute 1⁄(√2+1)
correct to 2 decimal places
Solution:
This can be done as a
continuation of the previous problem
6. We have proved [in (3)] that (√2+1)(√2-1) = 1. From this we get 1⁄(√2+1) = (√2-1)
7. ⇒1⁄(√2+1) = 1.41 - 1 = 0.41
Solved example 16.21
Prove that (i) √[22⁄3]
= 2√[2⁄3]
(ii) √[33⁄8]
= 3√[3⁄8]
Solution:
(i) 1. We have 22⁄3
= 8⁄3
= (2
× 4)⁄3
2. So √[22⁄3]
= √[(4 × 2)⁄3]
= √(4×2)⁄√3
= (√4×√2)⁄√3
= (2×√2)⁄√3
= 2 × [√2⁄√3]
= 2√[2⁄3]
(ii) 1. We have 33⁄8
= 27⁄8
= (3
× 9)⁄8
2. So √[33⁄8]
= √[(9 × 3)⁄8]
= √(9×3)⁄√8
= (√9×√3)⁄√8
= (3×√3)⁄√8
= 3 × [√3⁄√8]
= 3√[3⁄8]
Solved example 16.22
Find √45 + √180 + √80
Solution:
1. We have √45 = √[9×5] =
√9 × √5 = 3√5
2. We have √180 = √[36×5]
= √36 × √5 = 6√5
3. We have √80 = √[16×5]
= √16 × √5 = 4√5
4. Adding (1), (2) and (3), we
get: √45 + √180 + √80 = 3√5 + 6√5 + 4√5 = (3+6+4)√5 =
13√5
Solved example 16.23
If √10 = 3.16, calculate the
value of 2√2⁄√5
+ 3√5⁄√2
Solution:
1. Let us take each term
separately. 2√2⁄√5
= 2√2⁄√5
× √5⁄√5
(multiplying both numerator and denominator by √5)
⇒ 2√2⁄√5
= (2×√2×√5)⁄(√5 × √5) = 2√10⁄5
= 4√10⁄10 (multiplying both numerator and
denominator by 2)
⇒ 2√2⁄√5
= (4×3.16)⁄10 = 12.64⁄10
= 1.264
2. Similarly, for the second
term:3√5⁄√2 = 3√5⁄√2
× √2⁄√2 (multiplying both numerator and
denominator by √2)
⇒ 3√5⁄√2
= (3×√5×√2)⁄(√2 × √2) = 3√10⁄2
= (3×3.16)⁄2 = 9.48⁄2
= 4.74
3. Adding (1) and (2) we get
2√2⁄√5
+ 3√5⁄√2
= 1.264 + 4.74 = 6.004
Solved example 16.24
Which is the largest among
2√6, 3√3 and 4√2
Solution:
2√6 = √2 × √2 × √6 =
√[2×2×6] = √24
3√3 = √3 × √3 × √3 =
√[3×3×3] = √27
4√2 = √4 × √4 × √2 =
√[4×4×2] = √32
So 4√2 is the largest
Solved example 16.25
compute √12 - 1⁄√3
upto two decimal places
Solution: 1. Let us modify √12:
√12 = √12⁄√1
× √3⁄√3
= √36⁄√3
(multiplying both numerator and denominator by √3)
2. So √12 - 1⁄√3
= √36⁄√3
- 1⁄√3
= (√36-1)⁄√3
= (6-1)⁄√3
= 5⁄√3
3. Now, 5⁄√3
= 5√3⁄3
= (5×1.732)⁄3
= 2.89 (multiplying both numerator and denominator by √3)
Solved example 16.26
Simplify: 5√3 + 3√3 - √12
Solution: 1. Let us modify √12:
√12 = √[4×3]
= √4 × √3 = 2 × √3 = 2√3
∴ 5√3 + 3√3 - √12 = 5√3
+ 3√3 – 2√3 = (5 + 3 -2)√3 = 6√3
Rationalising the denominator
Many often it is easy to
handle the problems if the denominator is a rational number. We have
done it in some solved examples that we saw above in this section.
For example, in the solved example 16.23, we converted √5 and √2
in the denominators into 5 and 2 respectively.
To convert more complex
denominators, the identity: (a+b)(a-b)= a2-b2 is very helpful. Two examples are given
below:
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