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# If $S$ is a square of unit area. Consider any quadrilateral which has one vertex on each side of $S$. If $a,b,c$ and $d$ denote the length of the sides of the quadrilateral, then prove that $2≤a_{2}+b_{2}+d_{2}≤4$.

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## Text solutionVerified

Let the square $S$ is to be bounded by the lines $x=±1/2$ and $y=+1/2$
We have,
$a_{2}=(x_{1}−21 )_{2}+(21 −y_{1})_{2}$$b_{2}c_{2}d_{2} =x_{1}−y_{1}−x_{1}−y_{1}+21 =x_{2}−y_{1}−x_{2}+y_{1}+21 =x_{2}−y_{2}+x_{2}+y_{2}+21 =x_{1}−y_{2}+x_{1}−y_{2}+21 $$∴a_{2}+b_{2}+c_{2}+d_{2}=2(x_{1}+y_{1}+x_{2}+y_{2})+2$
Therefore, $0≤x_{1},x_{2},y_{1},y_{2}≤41 $$0≤x_{1}+x_{2}+y_{1}+y_{2}≤10≤2(x_{1}+x_{2}+y_{1}+y_{2})≤22≤2(x_{1}+x_{2}+y_{1}+y_{2})+2≤4 $
Alternate Solution
$c_{2}=x_{2}+y_{2}$$b_{2}a_{2}d_{2} =(1−x_{2})_{2}+y_{1}=(1−y_{1})_{2}+(1−x_{1})_{2}=x_{1}+(1−y_{2})_{2} $
On adding Eqs, (i), (ii), (iii) and (iv), we get
$a_{2}+b_{2}+c_{2}+d_{2}={x_{1}+(1−x_{1})_{2}}+{y_{1}+(1−y_{1})_{2}}+{x_{2}+(1−x_{2})_{2}}+{y_{2}+(1−y_{2})_{2}} $
where $x_{1},y_{1},x_{2},y_{2}$ all vary in the interval $[0,1]$.
Now, consider the function $y=x_{2}+(1−x)_{2},0≤x≤1$ differentiating $⇒dxdy =2x−2(1−x)$. For maximum or minimum $dxdy =0$.
$⇒2x−2(1−x)=0⇒⇒4x=2⇒Again, 2x−2+2x=0x=1/2dx_{2}d_{2}y =2+2=4 $
Hence, $y$ is minimum at $x=21 $ and its minimum value is 1/4. Clearly, value is maximum when $x=1$.
$∴$ Minimum value of $a_{2}+b_{2}+c_{2}+d_{2}=21 +21 +21 +21 =2$ and maximum value is $1+1+1+1=4$

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**LIVE**classesQuestion Text | If $S$ is a square of unit area. Consider any quadrilateral which has one vertex on each side of $S$. If $a,b,c$ and $d$ denote the length of the sides of the quadrilateral, then prove that $2≤a_{2}+b_{2}+d_{2}≤4$. |

Updated On | Feb 19, 2024 |

Topic | Application of Derivatives |

Subject | Mathematics |

Class | Class 12 |

Answer Type | Text solution:1 Video solution: 1 |

Upvotes | 255 |

Avg. Video Duration | 2 min |