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Ex.1.5 Q4 Number System Solution - NCERT Maths Class 9

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Question

Represent \(\begin{align}\sqrt {9.3} \end{align}\) on the number line.

 

 Video Solution
Number Systems
Ex 1.5 | Question 4

Text Solution

Steps:

Draw a line and take \(AB = 9.3\) units on it.

From \(B\) measure a distance of \(1\) unit and mark \(C\) on the number line. Make the midpoint of \(AC\) as \(O.\)

With \(‘O’\) as center and \(OC\) as radius, draw a semicircle.

At \(B\), draw a perpendicular to cut the semicircle at \(D\), with \(B\) as center and \(BD\) as radius draw an arc to cut the number line at \(E\). Taking \(B\) as the origin the distance \({BE}=\sqrt{9.3}\) and hence \(E\) represents \(\sqrt{9.3}\) .

Proof:

\[\begin{align}AB &=9.3, BC=1 \\ AC &=A B+BC=10.3 \\ OC &=\frac{AC}{2}=\frac{10.3}{2}=5.15 \\ OC&=OD=5.15\\O B &=OC-BC\\&=5.15-1=4.15 \end{align}\]

In right angled\(\begin{align}\Delta OBD,\end{align}\)

\[\begin{align} {BD}^{2} &={OD}^{2}-{OB}^{2}\\ &=(5.15)^{2}-(4.15)^{2} \\ &=(5.15+4.15)(5.15-4.15)\\&\rm{Using} \quad a^{2}-b^{2}=(a+b)(a-b) \\  &=9.3 \times 1\\  &=9.3 \\ BD &=\sqrt{9.3}=BE \end{align}\]

 Video Solution
Number Systems
Ex 1.5 | Question 4
  
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