Math, asked by ItzMeMukku, 1 month ago

Represent  \sqrt{9.3} on number line!!

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Answers

Answered by user0888
13

Solution:-

(In the first attachment)

Let's consider a semi-circle. If we choose C from the circumference, \triangle ABC is always a right triangle.

Let's draw a leg from C. It will be \overline{CH}.

After producing the leg we see three right triangles, and all three are similar to each other.

Here, we will choose two right triangles below.

\rightarrow \triangle ACH\sim \triangle CBH\ \mathrm{(\because AA\ Postulate)}

The corresponding sides are equal in ratio.

\rightarrow \overline{AH}:\overline{CH}=\overline{CH}:\overline{BH}

\rightarrow \overline{CH}^2=\overline{AH}\cdot \overline{BH} [Eqn. 1]

This is the relation of the three sides.

If we choose \overline{AH}=9.3 and \overline{BH}=1, according to [Eqn. 1] we have \overline{CH}=\sqrt{9.3}. So, it is possible to represent it on a number line.

Method:-

We see that \sqrt{9.3} is the length of one side. To represent it, we first put H at the origin.

Then draw \overline{AH}=9.3 on to the left and \overline{BH}=1 on to the right of the origin. And then let \overline{AB} be the diameter of the semi-circle.

After we draw a semi-circle, let's produce a leg from H. The length of the leg inside the circle is \overline{CH}=\sqrt{9.3}.

Let's draw it on to the right of the origin. We're done!

Learn More:-

  • Geometric Mean

(In the second attachment)

If a circle's radius is divided into a,b the length of the green line will be \sqrt{ab}. We call this the geometric mean, for short the G.M.

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Attachments:
Answered by 070307
3

Answer:

How can I plot root 9.3 on the number line?

Draw a line AB by measuring 9.3cm.

From the point, B add 1cm and mark it as C.

Mark the point of bisection by a compass and say it as 'O'

Measure AO which is the radius and draws a semi-circle.

From B draw a perpendicular AB touching the semi-circle and mark as D.

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