**Question:**

Represent $\sqrt{3.5}, \sqrt{9.4}, \sqrt{10.5}$ on the real number line.

**Solution:**

We are asked to represent the real numbers $\sqrt{3.5}, \sqrt{9.4}$ and $\sqrt{10.5}$ on the real number line

We will follow a certain algorithm to represent these numbers on real number line

(a) $\sqrt{3.5}$

We will take *A* as reference point to measure the distance

(1) Draw a sufficiently large line and mark a point *A* on it

(2) Take a point *B* on the line such that

(3) Mark a point *C* on the line such that

(4) Find mid point of *AB* and let it be *O*

(5) Take *O* as center and *OC* as radius and draw a semi circle. Draw a perpendicular *BD* which cuts the semi circle at *D*

(6) Take *B* as the center and *BD* as radius, draw an arc which cuts the horizontal line at *E*

(7) Point $E$ is the representation of $\sqrt{3.5}$

(b) $\sqrt{9.4}$

We will take *A* as reference point to measure the distance. We will follow the same figure in the part (a)

(1) Draw a sufficiently large line and mark a point *A* on it

(2) Take a point *B* on the line such that

(3) Mark a point *C* on the line such that

(4) Find mid point of *AB* and let it be *O*

(5) Take *O* as center and *OC* as radius and draw a semi circle. Draw a perpendicular *BC* which cuts the semi circle at *D*

(6) Take *B* as the center and *BD* as radius, draw an arc which cuts the horizontal line at *E*

(7) Point $E$ is the representation of $\sqrt{9.4}$

(c) $\sqrt{10.5}$

We will take *A* as reference point to measure the distance. We will follow the same figure in the part (a)

(1) Draw a sufficiently large line and mark a point *A* on

(2) Take a point *B* on the line such that

(3) Mark a point *C* on the line such that

(4) Find mid point of *AB* and let it be *O*

(5) Take *O* as center and *OC* as radius and draw a semi circle. Draw a perpendicular *BC* which cuts the semi circle at *D*

(6) Take *B* as the center and *BD* as radius, draw an arc which cuts the horizontal line at *E*

(7) Point $E$ is the representation of $\sqrt{10.5}

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