Question

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

Answer 1

(i) $\sqrt{3.5}$

= $\sqrt{{\left(\frac{3.5+1}{2}\right)}^2-{\left(\frac{3.5-1}{2}\right)}^2}=\sqrt{{\left(\frac{4.5}{2}\right)}^2-{\left(\frac{2.5}{2}\right)}^2}=(2.25{)}^2-(1.5{)}^2$

Steps of construction :

(a) Draw a line segment BC = 1.25 cm.

(b) At C, draw a ray CX making an angle of ${90}^{\circ }$

(c) With centre B and radius 2.25 cm, draw an arc which intersects CX at A.

(d) Join AB.

AC = $=\sqrt{A{B}^2-B{C}^2}=\sqrt{(2.25{)}^2-(1.25{)}^2}=\sqrt{5.0625-1.5625}=\sqrt{3.5}$

(ii) $\sqrt{9.4}$

$=\sqrt{{\left(\frac{9.4+1}{2}\right)}^2-{\left(\frac{9.4-1}{2}\right)}^2}=\sqrt{{\left(\frac{10.4}{2}\right)}^2-{\left(\frac{8.4}{2}\right)}^2}=\sqrt{(5.2{)}^2-(4.2{)}^2}$

Steps of construction :

(a) Draw a line segment BC = 4.2 cm.

(b) At C, draw a ray CX making an angle of ${90}^{\circ }$.

(c) With centre B and radius 5.2 cm draw an arc which intersects CX at A.

(d) Join AB.

AC = $=\sqrt{A{B}^2-B{C}^2}=\sqrt{(5.2{)}^2-(4.2{)}^2}=\sqrt{27.04-17.64}=\sqrt{9.40}=\sqrt{9.4}$

(iii) $\sqrt{10.5}$

$=\sqrt{{\left(\frac{10.5+1}{2}\right)}^2-{\left(\frac{10.5-1}{2}\right)}^2}=\sqrt{{\left(\frac{11.5}{2}\right)}^2-{\left(\frac{9.5}{2}\right)}^2}=\sqrt{(5.75{)}^2-(4.75{)}^2}$

Steps of construction :

(a) Draw a line segment BC = 5.75 cm.

(b) At C, draw a ray CX making an angle of ${90}^{\circ }$.

(c) With centre B and radius 5.75 cm, draw an arc intersecting CX at A.

(d) Join AB.

AB = $\sqrt{(5.75{)}^2-(4.75{)}^2}=\sqrt{33.0625-22.5625}=\sqrt{10.5000}=\sqrt{10.5}$