# For going to a city $\mathrm{B}$ from city $\mathrm{A}$, there is a route via city $\mathrm{C}$ such that $\mathrm{AC} \perp \mathrm{CB}$, $\mathrm{AC}=2 x \mathrm{~km}$ and $\mathrm{CB}=2(x+7) \mathrm{km}$. It is proposed to construct a $26 \mathrm{~km}$ highway which directly connects the two cities $\mathrm{A}$ and $\mathrm{B}$. Find how much distance will be saved in reaching city B from city A after the construction of the highway.

Given:

For going to a city $\mathrm{B}$ from city $\mathrm{A}$, there is a route via city $\mathrm{C}$ such that $\mathrm{AC} \perp \mathrm{CB}$, $\mathrm{AC}=2 x \mathrm{~km}$ and $\mathrm{CB}=2(x+7) \mathrm{km}$. It is proposed to construct a $26 \mathrm{~km}$ highway which directly connects the two cities $\mathrm{A}$ and $\mathrm{B}$.

To do:

We have to find how much distance will be saved in reaching city B from city A after the construction of the highway.

Solution:

In $\triangle \mathrm{ACB}$, by Pythagoras theorem,

$A B^{2} =A C^{2}+B C^{2}$

$(26)^{2}=(2 x)^{2}+[2(x+7)]^{2}$

$676 =4 x^{2}+4(x^{2}+49+14 x)$

$676 =4 x^{2}+4 x^{2}+196+56 x$

$676 =8 x^{2}+56 x+196$

$8 x^{2} +56 x-480=0$

$8(x^2+7x-60)=0$

$x^{2}+7 x-60=0$

$x^{2}+12 x-5 x-60=0$

$x(x+12)-5(x-12)=0$

$(x+12)(x-5)=0$

$x=5$ or $x=-12$ which is not possible since the distance cannot be negative.

$x=5$

Therefore,

$A C=2 x$

$=2(5)$

$=10 \mathrm{~km}$

$B C=2(x+7)$

$=2(5+7)$

$=24 \mathrm{~km} Hence, the distance covered to city$B$from city$A$via city$C=A C+B C=10+24=34 \mathrm{~km}$Distance covered to city$B$from city$A$after the highway is constructed$=B A=26 \mathrm{km}$Therefore, the distance saved$=34-26=8 \mathrm{~km}$Hence, the required saved distance is$8\ km\$.

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Updated on: 10-Oct-2022

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