In the previous section we saw some solved examples demonstrating the multiplication of irrational numbers. In this section, we will see a few more solved examples.
Solved example 16.15
All red triangles in the fig.16.31(a) are equilateral.
What is the ratio of the side of outer square to
side of inner square?
 |
| Fig.16.31 |
Solution:
If we analyse the fig.16.31(a) carefully, we will get the following points:
• Two red triangles are
placed face to face, and the gap is filled by two yellow triangles
• The above four
triangles make up one rectangular unit as shown in fig.b
• Four such rectangular
units are arranged to make up the outer larger square. This is shown
in fig.c
• Fig.d shows that, the
side of the outer square is the sum of the following two:
♦ length of the
rectangular unit
♦ width of the
rectangular unit
• Fig.d also shows that,
the side of the inner square is the difference of the following two:
♦ length of the
rectangular unit
♦ width of the
rectangular unit
Now we can do the
calculations:
1. Let the side of the
equilateral triangle be 'x'
2. Then the altitude = √[
x2 – (x⁄2)2] = √[x2 -(x2⁄4)] = √[ (4x2- x2)⁄4] = √[ 3x2⁄4] = √3x⁄2
3. Length of the 'rectangular unit' = 2 times altitude = 2 × √3x⁄2 = √3x
8. In the above problem, we see a particular pattern in fig.16.31(a).
9. There is an outer square and an inner square
10. There is only one criterion that determines the sizes of the squares. The criterion is: 'The size of the red equilateral triangle'.
11. To be more specific, the criterion is: 'The side of the red equilateral triangle'. Because, size of an equilateral triangle depends on it's side. The angles are always 60o
12. The sizes of the outer and inner squares depend on (11).
♦ We can increase the sizes of squares by increasing the side of the equilateral triangle
♦ We can decrease the sizes of squares by decreasing the side of the equilateral triangle
13. But whatever be the sizes of the squares, the ratio of the sides of those squares will always be that given in (7)
Solved example 16.16
From the above two problems, we can note the difference between (√3 × √12) and (√3 × √1.2). The first pair gives a natural number. But the second pair doesn't.
Another case:
(iii) Product = √5 × √8 = √(5×8) = √40 = √4 × √10 = 2 ×√10
3. Length of the 'rectangular unit' = 2 times altitude = 2 × √3x⁄2 = √3x
4. Width of the 'rectangular
unit' = base of the red triangle = x
5. So we get: Side of the
outer square = √3x + x = (√3 + 1)x
6. Side of the inner square =
√3x - x = (√3 -1)x
7. So ratio of sides =
(√3+1)x⁄(√3-1)x = (√3+1)⁄(√3-1)
8. In the above problem, we see a particular pattern in fig.16.31(a).
9. There is an outer square and an inner square
10. There is only one criterion that determines the sizes of the squares. The criterion is: 'The size of the red equilateral triangle'.
11. To be more specific, the criterion is: 'The side of the red equilateral triangle'. Because, size of an equilateral triangle depends on it's side. The angles are always 60o
12. The sizes of the outer and inner squares depend on (11).
♦ We can increase the sizes of squares by increasing the side of the equilateral triangle
♦ We can decrease the sizes of squares by decreasing the side of the equilateral triangle
13. But whatever be the sizes of the squares, the ratio of the sides of those squares will always be that given in (7)
Solved example 16.16
From the pairs of numbers
given below, pick out those, whose product is a natural number, or a
fraction
(i) √3, √12 (ii) √3,
√1.2 (iii) √5, √8 (iv) √0.5, √8 (v) √(71⁄2),
√(31⁄3)
Solution:
(i) Product = √3 × √12 =
√(3×12) = √36 = 6
‘6’ is a natural number.
So the product of √3 and √12 is a natural number
(ii) Product = √3 × √1.2
= √(3×1.2) = √3.6
√3.6 cannot be expressed as
a natural number or a fraction. So the product of √3 and √1.2 is
not a natural number or a fraction.
From the above two problems, we can note the difference between (√3 × √12) and (√3 × √1.2). The first pair gives a natural number. But the second pair doesn't.
Another case:
• (√3 × √1.2 × √1000) is
a natural number because:
√3 × √1.2 × √1000 =
√(3×1.2×1000) = √3600 = √36 × √100 = 6 ×10 = 60
(iii) Product = √5 × √8 = √(5×8) = √40 = √4 × √10 = 2 ×√10
√10 cannot be expressed as a
natural number or a fraction. So the product of √5 and √8 is not
a natural number or a fraction
(iv) Product = √0.5 × √8
= √(0.5×8) = √4 = 2
‘2’ is a natural number.
So the product of √0.5 and √8 is a natural number
(v) Product = √(71⁄2)
× √(31⁄3) = √[(15⁄2)
× √(10⁄3)] = √[150⁄6]
= √[50⁄2] = √25 = 5
‘5’ is a natural number.
So the product of √(71⁄2) and √(31⁄3)
is a natural number
Solved example 16.17
Calculate the perimeter of the
quadrilateral ABCD shown in fig.16.32 below
 |
| Fig.16.32 |
Solution:
1. Within the quadrilateral
ABCD, there is a right angled triangle ABD. Both it’s legs are 1 m
in length.
2. So the hypotenuse BD = √[12
+ 12] = √[1 + 1] = √2
3. Consider the ΔBCD. Two of
it’s angles are given as 60o. So the third angle must
also be equal to 60o (∵ sum of the interior angles of
any triangle is 180o )
4. When all three angles are
60o, it is an equilateral triangle. That is., all it’s
sides are the same.
5. So BC = CD = DB. But in
(2), we calculated BD as √2. So BC = CD = √2
6. Thus we now have all the
four sides of the quadrilateral ABCD. The perimeter =
1 + 1 + √2 + √2 = 2 +
√2 = 2 + 1.414 = 3.414 m (correct to 1 mm)
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