- A$\sqrt{65}$
- B$\sqrt{7} \overline{4}$
- ✓$\sqrt{8 \overline{3}}$
- D$\sqrt{9} \overline{7}$
Given,
$A=\left( a _1, a_2\right), B=\left(b_1, b_2\right)$, where $a_1, a_2, b_1, b_2$ are integers.
Distance between $A B$
$=\sqrt{\left(b_1-a_1\right)^2+\left(b_2-a_2\right)^2}$
$\therefore A B=\sqrt{\text { sum of squares of two number }}$
$\sqrt{65}=\sqrt{64+1}$, it is possible.
$\sqrt{74}=\overline{\sqrt{49+2}}$, it is also possible.
$\sqrt{97}=\sqrt{81+16}$, it is also possible.
$83$ is not a sum of squares of two number.
Hence, option $(c)$ is correct.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
where $A = {\sin ^2}\alpha - \sin \alpha + \frac{1}{4}$
and $B = {\tan ^2}\alpha + \frac{2}{{\sqrt 3 }}\tan \alpha + \frac{1}{3}$ , then the number of value $(s)$ of $\alpha $ in $\left[ { - \frac{{3\pi }}{2},2\pi } \right]$ is - (where $sgnx$ denotes signum function of $x$ )