SATHEE: Area Question 41

Area Question 41

Question 41

Consider a square with vertices at $(1, 1), (- 1, 1), (- 1, - 1)$ and $(1, - 1)$ . If $S$ is the region consisting of all points inside the square which are nearer to the origin than to any edge. Then, sketch the region $S$ and find its area.

$(1995, 5$ M)

Show Answer

Solution:

The equations of the sides of the square are as follow :

$A B : y = 1, B C : x = - 1, C D : y = - 1, D A : x = 1$

Let the region be $S$ and $(x, y)$ is any point inside it.

Then, according to given conditions,

$\begin{array}{lc} \sqrt{x^{2} + y^{2}} < | 1 - x |, | 1 + x |, | 1 - y |, | 1 + y | \Rightarrow & x^{2} + y^{2} < (1 - x)^{2}, (1 + x)^{2}, (1 - y)^{2}, (1 + y)^{2} \Rightarrow & x^{2} + y^{2} < x^{2} - 2 x + 1, x^{2} + 2 x + 1, & y^{2} - 2 y + 1, y^{2} + 2 y + 1 \Rightarrow & y^{2} < 1 - 2 x, y^{2} < 1 + 2 x, x^{2} < 1 - 2 y and x^{2} < 2 y + 1 \end{array}$

Now, in $y^{2} = 1 - 2 x$ and $y^{2} = 1 + 2 x$ , the first equation represents a parabola with vertex at $(1 / 2, 0)$ and second equation represents a parabola with vertex $(- 1 / 2, 0)$ and in $x^{2} = 1 - 2 y$ and $x^{2} = 1 + 2 y$ , the first equation represents a parabola with vertex at $(0, 1 / 2)$ and second equation represents a parabola with vertex at $(0, - 1 / 2)$ . Therefore, the region $S$ is lying inside the four parabolas

$y^{2} = 1 - 2 x, y^{2} = 1 + 2 x, x^{2} = 1 + 2 y, x^{2} = 1 - 2 y$

where, $S$ is the shaded region.

Now, $S$ is symmetrical in all four quadrants, therefore $S = 4 \times$ Area lying in the first quadrant.

Now, $y^{2} = 1 - 2 x$ and $x^{2} = 1 - 2 y$ intersect on the line $y = x$ . The point of intersection is $E (\sqrt{2} - 1, \sqrt{2} - 1)$ .

Area of the region $O E F O$

=Area of $△ O E H +$ Area of $H E F H$

$\begin{aligned} = \frac{1}{2} (\sqrt{2} - 1)^{2} + \int_{\sqrt{2} - 1}^{1 / 2} \sqrt{1 - 2 x} d x & = \frac{1}{2} (\sqrt{2} - 1)^{2} + (1 - 2 x)^{3 / 2} \frac{2}{3} \cdot \frac{1}{2} (- 1)^{1 / 2} & = \frac{1}{2} (2 + 1 - 2 \sqrt{2}) + \frac{1}{3} (1 + 2 - 2 \sqrt{2})^{3 / 2} \end{aligned}$

$\begin{aligned} = \frac{1}{2} (3 - 2 \sqrt{2}) + \frac{1}{3} (3 - 2 \sqrt{2})^{3 / 2} & = \frac{1}{2} (3 - 2 \sqrt{2}) + \frac{1}{3} {[(\sqrt{2} - 1)^{2}]}^{3 / 2} & = \frac{1}{2} (3 - 2 \sqrt{2}) + \frac{1}{3} (\sqrt{2} - 1)^{3} & = \frac{1}{2} (3 - \sqrt{2}) + \frac{1}{3} [2 \sqrt{2} - 1 - 3 \sqrt{2} (\sqrt{2} - 1)] & = \frac{1}{2} (3 - 2 \sqrt{2}) + \frac{1}{3} [5 \sqrt{2} - 7] & = \frac{1}{6} [9 - 6 \sqrt{2} + 10 \sqrt{2} - 14] = \frac{1}{6} [4 \sqrt{2} - 5] sq units \end{aligned}$

Similarly, area $O E G O = \frac{1}{6} (4 \sqrt{2} - 5)$ sq units

Therefore, area of $S$ lying in first quadrant

$= \frac{2}{6} (4 \sqrt{2} - 5) = \frac{1}{3} (4 \sqrt{2} - 5) sq units$

Hence, $S = \frac{4}{3} (4 \sqrt{2} - 5) = \frac{1}{3} (16 \sqrt{2} - 20)$ sq units

Area Question 41

Question 41

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Area Question 41

Question 41

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

Menu