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

Given :-

$AC \perp CB, AC = 2x\; km, CB = 2 (x + 7) km$

$AB=26\; km$

In $\Delta ABC$ use, Pythagoras theorem

$AB^{2}=AC^{2}+BC^{2}$

$\left ( 26 \right )^{2}=\left ( 2x \right )^{2}+\left ( 2\left ( x+7 \right ) \right )^{2}$

$676=4x^{2}+4\left ( x^{2}+49+14x \right )$                         $(using \left ( a+b \right )^{2}=a^{2}+b^{2}+2ab)$

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

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

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

Dividing by 8 we get

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

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

$x\left ( x+12 \right )-5\left ( x+12 \right )=0$

$\left ( x+12 \right )\left ( x-5 \right )=0$

$x=-12 \; or\; x=5$

x = – 12 is not possible because distance cannot be negative.

$\therefore x=5$

Now

$AC = 2x = 2 \times 5 = 10 \; km$

$BC = 2(x + 7) = 2(5 + 7) = 2 \times 12 = 24 \; km$

Distance covered to reach city B from A via city C = AC + CB

$=10+24=34\; km$

Distance covered to reach city B from city A after the construction of highway

$BA = 26\; cm$

Saved distance $= 34 - 26 = 8 km.$