Name: Inverse Function: Theorem
Uploaded: 2019-07-19
Duration: 4 min 27 s
Description: Did you know that only one-one and onto functions have an inverse? Play this video to learn more about the existence of inverse functions. Are you ready to e...

Question

In the square $A B C D$ with side $A B = 2$ a straight line $M N$ is drawn perpendicularly to $A C$ . Denoting the distance from the vertex $A$ to the line $M N$ as $x$ , express through $x$ the area $S$ of the triangle $A M N$ cut off from the square by the straight line $M N$ . Find this area at $x = \frac{\sqrt{2}}{2}$ and at $x = 2$

Accepted Answer

Given:- $A B C D$ is square in which side length $A B = 2$ .

A straight line $M N$ is drawn & the distance from $A$ is $x$

As, $A C = \sqrt{2} A B = 2 \sqrt{2}$

Hence, $0 \leq x \leq 2 \sqrt{2}$

If $x \leq \sqrt{2}$

$∵ A E = x$

Also, $A M = A N$

Also, in an isosceles right angle triangle perpendicular drawn from the vertex divides the hypotenuse.

$∴ M E = N E$

And $A E^{2} = M E \cdot N E$

$∴ M E = N E = x$

$∴ M N = 2 x as M E = N E = x$

$S (x) = S_{△ A M N} = \frac{1}{2} \times x \times 2 x = x^{2}$

If $x > \sqrt{2}$

Here, $A E^{'} = x$

$∴ C E^{'} = 2 \sqrt{2} - x$

$∴ A (△ C M^{'} N^{'}) = {(2 \sqrt{2} - x)}^{2}$

$∴ S (x) = A (square A B C D) - A (C M^{'} N^{'}) = 4 - {(2 \sqrt{2} - x)}^{2} = - x^{2} + 4 x \sqrt{2} - 4$

$∴ S (x) = \{\begin{cases} x^{2} 0 \leq x \leq \sqrt{2} \\ - x^{2} + 4 x \sqrt{2} - 4 \sqrt{2} < x \leq 2 \sqrt{2} \end{cases}$

$∴ S (\frac{\sqrt{2}}{2}) = {(\frac{\sqrt{2}}{2})}^{2} = \frac{2}{4} = \frac{1}{2}$

And $S (2) = - 4 + 8 \sqrt{2} - 4 = 8 (\sqrt{2} - 1)$ .

Important Questions on Introduction to Mathematical Analysis

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In the square A B C D with side A B=2 a straight line MN is drawn perpendicularly to A C. Denoting the distance from the vertex A to the line MN as x, express through x the area S of the triangle AMN cut off from the square by the straight line M N. Find this area at x=22 and at x=2

Important Questions on Introduction to Mathematical Analysis

Learn and Prepare for any exam you want !