Question

Question 4
Represent $\sqrt{9.3}$ on the number line.

Answer 1

Step 1: Draw a line segment of unit 9.3. Extend it to C such that BC is 1 unit.

Step 2: Now, AC = 10.3 units. Find the centre of AC and name it as O.

Step 3: Draw a semi-circle with radius OC and centre O.

Step 4: Draw a perpendicular line BD to AC at point B which intersects the semicircle at D. Also, Join OD.

Step 5: Now, OBD is a right-angled triangle.

Here, $OD=\frac{10.3}{2}$ (radius of semi-circle), $OC=\frac{10.3}{2}$, BC = 1

OB = OC – BC = $\left(\frac{10.3}{2}\right)$ – 1 = $\frac{8.3}{2}$

Using Pythagoras theorem,

$O{D}^2=B{D}^2+O{B}^2$

$\Rightarrow {\left(\frac{10.3}{2}\right)}^2=B{D}^2+{\left(\frac{8.3}{2}\right)}^2\Rightarrow B{D}^2={\left(\frac{10.3}{2}\right)}^2-{\left(\frac{8.3}{2}\right)}^2\Rightarrow B{D}^2=(\frac{10.3}{2}-\frac{8.3}{2})(\frac{10.3}{2}+\frac{8.3}{2})\Rightarrow B{D}^2=9.3\Rightarrow BD=\sqrt{9}.3$

Thus, the length of BD is $\sqrt{9.3}$.

Step 6: Taking BD as radius and B as centre draw an arc which touches the line segment. The point where it touches the line segment is at a distance of $\sqrt{9}.3$ from B as shown in the figure.