Question

Represent √5 on no. line​

Answer 1

$\bf{\underline{Show \: That:-}}$

▣ $\sf \red{\sqrt{5}}$ on a no. line.

$\bf{\underline{Solution:-}}$

Let, a ∆AOB

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\linethickness{0.3mm} \qbezier( - 0.4,2)(2.8,4)(2.8,4)\qbezier(2.8,2)(2.8,4)(2.8,4)\put(0.7,3.2){\huge?}\put(2.9,2.8){ \bf 1 }\put(1.4,1.7){ \bf 2 }\put(2.7,1.7){ \bf A }\put(2.7,4.1){ \bf B }\put( - 0.7,1.7){ \bf O } \qbezier( -0.4,2)(2.8,2)(2.8,2) \end{picture}$

we, know that

$\huge{\boxed{\sf H^2 = P^2 + B^2}} \quad \bigg\lgroup{ \sf Pythagoras\: Theorem} \bigg\rgroup$

$\sf \dashrightarrow OB^2 = BA^2 + OA^2$

where,

• BA = 1
• OA = 2

So,

$\sf \dashrightarrow OB^2 = BA^2 + OA^2$

$\sf \dashrightarrow OB^2 = 1^2 + 2^2$

$\sf \dashrightarrow OB^2 = 1+ 4$

$\sf \dashrightarrow OB^2 = 5$

$\sf \dashrightarrow OB = \sqrt{5}$

$\small{\sf \therefore \underline{OB = \sqrt{5}}}$

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\linethickness{0.3mm} \qbezier( - 0.4,2)(2.8,4)(2.8,4)\qbezier(2.8,2)(2.8,4)(2.8,4)\put(0.7,3.2){ \bf \sqrt{5} }\put(2.9,2.8){ \bf 1 }\put(1.4,1.7){ \bf 2 }\put(2.7,1.7){ \bf A }\put(2.7,4.1){ \bf B }\put( - 0.7,1.7){ \bf O } \qbezier( -0.4,2)(2.8,2)(2.8,2) \end{picture}$

Now, Let's represent it on no. line

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\linethickness{0.3mm} \put(2,2){\vector(1, 0){3}} \multiput( - 0.4,1.8)(1.6,0){4}{\line(0,1){0.4}}\put(2,2){\vector(-1, 0){3}}\qbezier( - 0.4,2)(2.8,4)(2.8,4)\qbezier(2.8,2)(2.8,4)(2.8,4)\qbezier(2.8,4)(3.6,3.2)(3.4,2)\put( - 0.5,1.5){ \bf 0 }\put(1.1,1.5){ \bf 1 }\put(2.7,1.5){ \bf 2 }\put(4.3,1.5){ \bf 3 }\put(0.7,3.2){ \bf \sqrt{5} } \thicklines\put(3.9,2.7){\vector(-1,-2){0.3}}\put(3.7,2.8){ \bf \sqrt{5} }\put(2.5,2.8){ \bf 1 }\put(1.4,2.1){ \bf 2 }\put(3.3,1.6){ \bf P }\put(2.9,2.1){ \bf A }\put(2.7,4.1){ \bf B }\put( - 0.7,2.2){ \bf O } \put(3.4,2){\circle*{0.2}} \end{picture}$

Construction:-

$\underline{\pink{\sf Step :1}}$ Draw a no. line marked from 0 to 3.

$\underline{\pink{\sf Step :2}}$ Draw an arc of 1cm from point 2

$\underline{\pink{\sf Step :3}}$ Join A and B.

$\underline{\pink{\sf Step :4}}$ Join B and O also.

$\underline{\pink{\sf Step :5}}$ Open compass from 0 to B and draw an arc.

$\underline{\pink{\sf Step :6}}$ Where the arc meets on line name that point as P.

$\underline{\pink{\sf Step :7}}$ Hence, √5 was represented on number line.

Answer 2

Answer:-

Let, a ∆AOB

⠀⠀⠀H²=P²+B² (Pythagoras theorem)

=> OB²=BA²+OA²

where,

⠀⠀• BA=7

⠀⠀•OA=2

So,

=> OB² = BA²+OA²

=> OB² = 1²+2²

=> OB² = 1+4

=> OB² = 5

=> OB = √5

Construction:-

• Draw a number line marked 0 to 3.

• Draw an Arc from 1cm to 2cm.

• Then join A and B.

• Later join B and O.

• Open compass from 0 to B and then draw an Arc.

• Now where both are met name that point P.

• hence √5 is represented on a number line.