Question

Consider the triangle having vertices $O(0,0),A(2,0)$ and $B(1,\sqrt{3}).$ Also $b\le$ min $\{a_1,a_2,a_3,.......,a_n\}$ means $b\le a_1$ when $a_1$ is least; $b\le a_2$ when $a_2$ is least and so on. From this we can say $b\le a_2,b\le a_2,..........,b\le a_n$.

Let $R$ be the region consisting of all those points $P$ inside $\Delta OAB$ which satisfy $d(P,OA)\le$ min $\left[d\right(P,OB),d(P,AB\left)\right],$ where denotes the distance from the point to the corresponding line. Then the area of the region $R$ is

• A. $\sqrt{3}$ sq. units
• B. $(2+\sqrt{3})$ sq. units
• C. $\frac{\sqrt{3}}{2}$ sq. units
• D. $\frac{1}{\sqrt{3}}$ sq. units

Answer 1

The correct option is D $\frac{1}{\sqrt{3}}$ sq. units

$d(P,OA)\le$ min $\left[d\right(P,OB),d(P,AB\left)\right]$
Which implies
$d(P,OA)\le d(P,OB)$ and $d(P,OA)\le d(P,AB)$
When $d(P,OA)=d(P,OB),P$ is equidistant from $OA$ and $OB,$ or $P$ lies on angular bisector of lines $OA$ and $OB.$
Hence, when $d(P,OA)<d(P,OB)$ point $P$ is nearer to $OA$ than $OB$ or lies below bisector of $OA$ and $OB.$ Similarly, when $d(P,OA)d(P,AB),P$ is nearer to $OA$ than $AB,$ or lies below bisector of $OA$ and $AB.$ Therefore, the required area is equal to the area of $OIA.$
Now,
$tanBOA=\sqrt{3}$ this implies that
$BOA={60}^{\circ }$
Thus the triangle is equilateral of side $2$ units and the centroid as well the incentre and median will coincide at $(1,\frac{1}{\sqrt{3}})$
Hence the area will be
$\frac{bh}{2}$
$=\frac{1}{2}\left(2\right)\left(\frac{1}{\sqrt{3}}\right)$
$=\frac{1}{\sqrt{3}}$