Area Ques 4
- Let $O(0,0), A(2,0)$ and $B\left(1, \frac{1}{\sqrt{3}}\right)$ be the vertices of a triangle. Let $R$ be the region consisting of all those points $P$ inside $\triangle O A B$ which satisfy $d(P, O A) \geq \mathrm{min}$ $\{d(P, O B), d(P, A B)\}$, where $d$ denotes the distance from the point to the corresponding line. Sketch the region $R$ and find its area.
(1997C, 5M)
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Answer:
Correct Answer: 4.$(2 - \sqrt{3})$ $ sq$ $ units$
Solution: Let the coordinates of $P$ be $(x, y)$.
Equation of line $O A$ be $y=0$.
Equation of line $O B$ be $\sqrt{3} y=x$.
Equation of line $A B$ be $\sqrt{3} y=2-x$.
$d(P, O A)=$ Distance of $P$ from line $O A=y$
$d(P, O B)=$ Distance of $P$ from line $O B=\frac{|\sqrt{3} y-x|}{2}$
$d(P, A B)=$ Distance of $P$ from line $A B=\frac{|\sqrt{3} y+x-2|}{2}$
Given, $d(P, O A) \leq \min \{d(P, O B), d(P, A B)\}$
$y \leq \min \left\{\frac{\sqrt{3} y-x \mid}{2}, \frac{|\sqrt{3} y+x-2|}{2}\right\}$
$\Rightarrow \quad y \leq \frac{|\sqrt{3} y-x|}{2}$ and $y \leq \frac{|\sqrt{3} y+x-2|}{2}$
Case I When $y \leq \frac{|\sqrt{3} y-x|}{2}$
[since, $\sqrt{3} y-x<0$ ]
$ y \leq \frac{x-\sqrt{3} y}{2} \Rightarrow(2+\sqrt{3}) y \leq x \Rightarrow y \leq x \tan 15^{\circ} $
Case II When $y \leq \frac{|\sqrt{3} y+x-2|}{2}$,
$2 y \leq 2-x-\sqrt{3} y \quad$
[since, $\sqrt{3 y}+x-2<0$ ]
$\Rightarrow \quad(2+\sqrt{3}) y \leq 2-x \Rightarrow y \leq \tan 15^{\circ} \cdot(2-x)$
From above discussion, $P$ moves inside the triangle as shown below :
$\Rightarrow \quad $ Area of shaded region
$ \begin{aligned} & =\text { Area of } \triangle O Q A \\ & \left.=\frac{1}{2} \text { (Base) } \times \text { (Height }\right) \\ & =\frac{1}{2}(2)\left(\tan 15^{\circ}\right)=\tan 15^{\circ}=(2-\sqrt{3}) \text { sq unit } \end{aligned} $
BETA
