Question

Question 6
For going to a city B from city A, there is a route via city C such that AC $\perp$ CB, AC = 2x 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.Ffind how much distance will be saved in reaching city B from city A after the construction of the highway.

Answer 1

Given, AC $\perp$ CB, CB = 2(x+7)km and AB = 26 km
On drawing the figure, we get a right angled $\Delta ACB$ right angled at C.

Now, in $\Delta ACB$ by using Pythagoras theorem,
$A{B}^2=A{C}^2+B{C}^2$
$\Rightarrow (26{)}^2=(2x{)}^2+{\left\{2\right(x+7\left)\right\}}^2$
$\Rightarrow 676=4x^2+4(x^2+49+14x)$
$\Rightarrow 676=4x^2+4x^2+196+56x$
$\Rightarrow 676=8x^2+56x+196$
$\Rightarrow 8x^2+56x-480=0$
On dividing by 8, we get $x^2+7x-60=0$
$\Rightarrow x^2+12x-5x-60=0$
$\Rightarrow x(x+12)-5(x+12)=0$
$\Rightarrow (x+12)(x-5)=0$
$\therefore$ $x=-12,x=5$
Since, distance cannot be negative.
$\therefore$ $x=5$ $[\because x\ne -12]$
Now, $AC=2x=10km$
And $BC=2(x+7)=2(5+7)=24km$
The distance covered to reach city B from city A via city C
$=AC+BC$
$=10+24$
$=34km$
Distance covered to reach city B from city A after the construction of the highway
= BA = 26 km
Hence the distance saved in reaching city B from city A after the construction of the highway will be 34 – 26 i.e., 8 km.