Question

show how root 5 can be represented on number line pls do in note book I will mark as brainliest​

Answer 1

$\huge\rm\blue{ANSWER}$

We can represent $\sqrt{5}$ in the following ways-

• Draw a number line of two points (units) from an initial point say. 0 in the right/positive direction

• Let those two points represent A & B respectively.

• Now, draw a line of 1-unit perpendicular to B and let it be called as C

• Join OC to represent the hypotenuse of a triangle, right angled at B

• This hypotenuse of triangle ABC is showing a length of

• We can use a divider and with the length of the hypotenuse, we can make a cut upon the number line.

Answer 2

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


$\sf \: \bigg(OB { \bigg)}^{2} = \bigg(OA { \bigg)}^{2} + \bigg(OB { \bigg)}^{2}$


$\sf \: OB = \sqrt{ \bigg(OA { \bigg)}^{2} + \bigg( OB { \bigg)}^{2} }$


$\sf \: \sqrt{5} = \sqrt{ \bigg(2 { \bigg)}^{2} + \bigg(1 { \bigg)}^{2} }$