Question

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संख्या रेखा पर निरूपित करें।
$\sf{(a) \: \sqrt{7.4}}$
$\sf{(b) \: \sqrt{3.5}}$
$\bf {Explain \: answer:}$
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Answer 1

Representing the given Irrational numbers on the number line -

$\red{(a)\sqrt{7.4}}$

Solution :

1. Mark the distance 7.4 cm from a fixed point A on a given line to obtain a point B such that AB = 7.4 cm.
2. From B mark a distance of 1 cm and mark the new point as C.
3. Find the midpoint of AC and mark the point as O.
4. Draw a semicircle with O of radius OC.
5. Draw a perpendicular to AC passing through B and intersecting the semicircle at D, then BD = $\red{\sqrt{7.4}}$.

$\rule{200}2$

$\blue{(b)\sqrt{3.5}}$

Solution :

1. Mark the distance 3.5 cm from a fixed point A on a given line to obtain a point B such that AB = 3.5 cm.
2. From B mark a distance of 1 cm and mark the new point as C.
3. Find the midpoint of AC and mark the point as O.
4. Draw a semicircle with O of radius OC.
5. Draw a perpendicular to AC passing through B and intersecting the semicircle at D, then BD = $\red{\sqrt{3.5}}$.

More Information -

To find $\sqrt{x}$, for any positive real number x, we mark B so that AB = x units. Also we mark C, such that BC = 1 Unit.

We get BD = $\red{\sqrt{x}}$.

It can be proved also by pythagoras theorem.

ΔABD is a right angled triangle.

The radius of the circle is $\bf{\frac{(x+1)}{2}}$ units.

$\bf Therefore, \\ \\ OC = OD = OA = \frac{(x+1)}{2} units \\ \\Now, \\ \\ OB = x-\frac{(x+1)}{2}=\frac{x-2}{2}$

So, by using Pythagoras Theorem we have,

$\bf BD^{2}=OD^{2}-OB^{2}\\ \\= {( \frac{x+1}{2} )}^{2}- {( \frac{x-1}{2} )}^{2}\\ \\=\frac{4x}{4}=4$

So,

$\bf BD = \sqrt{x}$

Then treat the BC line as number line, with B as 0, c as 1, and so on.

→ Draw a arc with centre B and radiu BD, which intersects the number line in E.

E represents √x.

Answer 2

Step-by-step explanation:

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