Locate $\sqrt{13}$ on the number line.
Locate $\sqrt{13}$ on the number line.
We write $13$ as the sum of the squares of two natural numbers:
$13=9+4=3^{2}+2^{2}$
On the number line, take $OA =3$ units.
Draw $BA =2$ units, perpendicular
to $OA.$ Join $OB$ (see $Fig.$).
By Pythagoras theorem,
$OB =\sqrt{13}$
Using a compass with centre $O$ and radius $OB$, draw an arc which intersects the number line at the point $C$. Then, $C$ corresponds to $\sqrt{13}$
Similar Questions
For each question, select the proper option from four options given, to make the statement true : (Final answer only)
$(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})=\ldots \ldots \ldots$
For each question, select the proper option from four options given, to make the statement true : (Final answer only)
$(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})=\ldots \ldots \ldots$
Rationalise the denominator of the following:
$\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}$
Rationalise the denominator of the following:
$\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}$
Simplify:
$(\frac{3}{5})^4 + (\frac{8}{5})^{-12} + (\frac{32}{5})^{6}$
Simplify:
$(\frac{3}{5})^4 + (\frac{8}{5})^{-12} + (\frac{32}{5})^{6}$
Find five rational numbers between $-\frac{2}{3}$ and $\frac{1}{5}$.
Find five rational numbers between $-\frac{2}{3}$ and $\frac{1}{5}$.
Rationalise the denominator of the following:
$\frac{2+\sqrt{3}}{2-\sqrt{3}}$
Rationalise the denominator of the following:
$\frac{2+\sqrt{3}}{2-\sqrt{3}}$