# Ex.1.2 Q3 Number System Solution - NCERT Maths Class 9

## Question

Show how \(\begin{align}\sqrt 5 \end{align}\) represented on the number line.

## Text Solution

**What is known?**

Integer \(5\).

**What is unknown?**

Point representing \(\begin{align}\sqrt 5 \end{align}\) on the number line.

**Steps:**

We shall write **\(5\)** on the sum of two squares in the form \(\begin{align}5=2^{2}+1^{2}=4+1=5 \end{align}\). This shows we need to construct a right triangle with sides **\(2\)** and **\(1\)** units. So the hypotenuse becomes \(\begin{align} \sqrt{5} \end{align}\) units on the number line. We shall proceed as follows.

**Diagram**

On the number line take **\(2\)** units from **\(O\)** and represent the point as **\(A\)**. At **\(A\)** draw the perpendicular and mark **\(B\)** such that \(\begin{align} AB =1\end{align}\) unit with **\(O\)** as center and \(\begin{align} OB\end{align}\) as radius. Draw an arc to cut the number line at **\(C\)**. Now **\(C\)** represents \(\begin{align}\sqrt 5 \end{align}\).

In \(\Delta OAB,\)

\[\begin{align} O{{B}^{2}}&=O{{A}^{2}}+A{{B}^{2}} \\ &={{2}^{2}}+{{1}^{2}} \\ &=5 \\ \therefore OB &=\sqrt{5}=OC \\ \end{align}\]