Question

Represent the number √5 on the number line.​

Answer 1

I hope it will help you

Answer 2

Answer:

$\setlength{\unitlength}{14mm}</p><p>\begin{picture}(7,5)(0,0)</p><p>\thicklines</p><p>\put(0,0){\line(1,0){3}}</p><p>\put(3,0){\vector(0,1){3}}</p><p>\put(0.8,0.01){\line(5,3){2.2}}</p><p>\put(0,0){\vector(1,0){4.9}}</p><p>\put(2,0){\vector( -1,0){3}}</p><p>\multiput(-0.3, -0.1)(1.1,0){5}{\line(0,1){.2}}\put(-0.5,-0.5){- 1}\put(0.7,-0.5){0}\put(1.85,-0.5){1}\put(2.9,-0.5){2}\put(4,-0.5){3} \qbezier(2.5,1.3)(3,1.4)(3.5,1.2)\linethickness{.5}\qbezier(2.5,1.5)(3.5,1.4)(3.6,0)\put(2.7,0.18){\sf A}\put(3.1,1.5){\sf B}\put(.7,0.18){\sf O}\put( - 1, - 0.4){\sf X'}\put(4.8, - 0.4){\sf X}\put(3.4,- 0.35){\sf \sqrt{\sf5} }\put(3.67,0.1){\sf C}\end{picture}$

⠀

Draw number line as shown in the figure. Let the point O represent 0 and point A represents 2.

Draw perpendicular at A on the number line and cut-off arc AB = 1 unit

We have OA = 2 units and AB = 1 unit

⠀

$\underline{\bigstar\:\textsf{Using Pythagoras theorem in \triangle OAB :}}$

$:\implies\sf (Hypotenuse)^2=(Perpendicular)^2+(Base)^2\\\\\\:\implies\sf (OB)^2=(AB)^2+(OA)^2\\\\\\:\implies\sf (OB)^2=(2)^2+(1)^2\\\\\\:\implies\sf (OB)^2=4+1\\\\\\:\implies\sf (OB)^2=5\\\\\\:\implies\sf OB=\sqrt{5}$

⠀

Taking O as the centre and $\sf OB=\sqrt{5}$ as radius draw an arc cutting real line at C. Now $\sf OC=OB=\sqrt{5}$

⠀

$\therefore\:\underline{\textsf{Hence, C represents \sqrt{\sf5} on the number line}}.$