Question

For going to a city $B$ from city $A$, there is a route via city $C$ such that $AC\perp CB$, $AC=2x\mathrm{km}$ and $CB=2\left(x+7\right)\mathrm{km}$. It is proposed to construct a $26\mathrm{km}$ 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.

Answer 1

Step 1: State given data and draw a suitable diagram

It is given that, $AC=2x\mathrm{km}$

$CB=2\left(x+7\right)\mathrm{km}$

$AB=26\mathrm{km}$

Also given, $AC\perp CB$

Let us draw a suitable diagram.

Step 2: Use Pythagoras Theorem,

Applying Pythagoras Theorem in $∆ABC$, we get,

${\left(AC\right)}^2+{\left(CB\right)}^2={\left(AB\right)}^2$

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

$\Rightarrow$ ${\left(2x\right)}^2+4{\left(x+7\right)}^2={\left(26\right)}^2$ $\left[\because {\left(\mathrm{ab}\right)}^m={\left(a\right)}^m{\left(b\right)}^m\right]$

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

$\Rightarrow$ $4x^2+4x^2+56x+196=676$

$\Rightarrow$ $8x^2+56x+196=676$

$\Rightarrow$ $8x^2+56x+196-676=0$

$\Rightarrow$ $8x^2+56x-480=0$

$\Rightarrow$ $8\left(x^2+7x-60\right)=0$

$\Rightarrow$ $x^2+7x-60=0$ $\left[\because 0÷8=0\right]$

Step 3: Calculate $x$ using factorization method

Factorizing the equation $x^2+7x-60=0$, we get,

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

$\Rightarrow$ $x^2+12x-5x-60=0$

$\Rightarrow$ $\left(x^2+12x\right)-\left(5x+60\right)=0$

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

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

So, $x-5=0$ or $x+12=0$

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

Since distance cannot be negative.

Therefore, $x=-12$ cannot be the required value.

So, $x=5$

Step 4: Calculation of distance saved

When, a route via city $C$ was being used to go to city $B$ from city $A$,

Initial distance travelled $\left(d_i\right)=AC+CB$

$\Rightarrow$ $d_i=2x+2\left(x+7\right)\mathrm{km}$

$\Rightarrow$ $d_i=2x+2\left(x+7\right)\mathrm{km}$

$\Rightarrow$ $d_i=2x+2x+14\mathrm{km}$

$\Rightarrow$ $d_i=4x+14\mathrm{km}$

$\Rightarrow$ $d_i=4\times 5+14\mathrm{km}$ $\left[\because x=5\right]$

$\Rightarrow$ $d_i=20+14\mathrm{km}$

$\Rightarrow$ $d_i=34\mathrm{km}$

And, the distance traveled by highway $\left(d_h\right)=26\mathrm{km}$.

Thus, the distance saved $=d_i-d_h$

$\Rightarrow$ the distance saved $=34-26\mathrm{km}$

$\Rightarrow$ the distance saved $=8\mathrm{km}$

Hence, a distance of $8\mathrm{kilometers}$ will be saved in reaching city $B$ from city $A$ after the construction of the highway.