Question

How to represent \[\sqrt {9.3} \] on a number line?

Answer 1

Hint: First we will mark \[9.3cm\] on a number line. Then we will extend it \[1cm\] more. Then we will draw a perpendicular bisector. After that we will draw the semicircle. Then we will draw a perpendicular which will turn out to be \[\sqrt {9.3} \] when we calculate.

Complete answer:
The number is \[\sqrt {9.3} \]. We have to represent this number on a number line. First, we will draw a line segment called \[AO\]. Then we will take \[9.3cm\] measurement with the help of a scale and compass. We will measure \[9.3cm\] from the scale with the help of a compass, we will keep the compass needle at the starting point of the scale which is \[0\], and then stretch the other part of the compass containing pencil till \[9.3cm\].
After this, we will keep the compass needle at the starting point of the line segment \[AO\]. We will draw an arc on the line and cut the line segment with the help of the pencil on the other part of the compass. The point where the arc and the line segment intersect will be \[9.3cm\]. We will mark that point as \[B\]. This way we will get \[9.3cm\].
Now, we will take another point \[C\] on the line segment which is \[1cm\] more than \[9.3cm\]. We know that \[1cm\]more to \[9.3cm\] will be \[10.3cm\]. So, we need to mark \[1cm\] more than \[9.3cm\] that is \[10.3cm\]. We will apply the same procedure as we did earlier to mark \[9.3cm\]. After marking the \[C\] point the line looks like this:

Now, we will draw a perpendicular bisector of \[AC\]. We will take the compass needle and place it on \[A\], assuming it as the center. We will assume the radius to be more than the half of \[AC\] and draw an arc on both the upper and lower side of the line. Now, keeping the shape of the compass constant, that means the radius is the same, we will assume \[C\] to be the center and similarly, the radius will be more than the half of \[AC\]. Now, we will draw arcs on the upper and lower side of the line. The point where the arcs on the upper side and lower side intersect are the points for the perpendicular bisector. We will join the points through a straight line \[PQ\]. Now, \[PQ\] is the perpendicular bisector of \[AC\]. The point from line \[PQ\] which cuts \[AC\] will be the point \[X\]. Here, we can say that \[X\] is the central point of line \[AC\].
Now, with \[X\]as the center and \[XA\] as the radius, we will draw a semicircle. Since \[X\] as a center of \[AC\], \[XA\] and \[XC\] are the radius.
Now, we will draw a perpendicular at point \[B\]. We can do this either with the help of a compass or with the help of a protractor. Suppose the perpendicular intersects the semicircle at a point \[D\]. Thus, the length of the line \[BD\] is \[\sqrt {9.3} \].
 But we have to represent it on the number line. So, we will extend the line \[AO\]. Now, with \[B\] as center and radius equal to \[BD\], we will draw a big arc which cuts the line \[AO\]. The point which cuts the line is the point \[E\]. We know that the length of \[BD\] is equal to the length of \[BE\]. So, if \[BD\] is of length \[\sqrt {9.3} \], then \[BE\] is also \[\sqrt {9.3} \]. When we draw it, it looks like:

Note:
We have to always mark \[1cm\] more from the given measurement. This is done because the question contains a square root. Whatever be the length, we always have to mark \[1cm\] more according to the given question.