For going to a city B from city A, there is a route via city C such that $\text{AC}\perp\text{CB},$ AC = 2x km and CB = 2(x + 7)km. It is proposed to construct a 26km 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.
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Distance saved by direct highway = (AC + BC) - AB
$\therefore\text{AC}\perp\text{CB}$ so by pythagoras theorem
$AC^2 + BC^2 = AB^2$
$\Rightarrow (2x)^2 + [2(x + 7)]^2 = 26^2$
$\Rightarrow 2^2x^2 + 2^2(x + 7)^2 = 676$
$\Rightarrow 4x^2 + 4(x^2 + 49 + 14x) = 676$
$\Rightarrow 4[x^2 + x^2 + 49 + 14x] = 676$
$\Rightarrow2\text{x}^2+14\text{x}+49=\frac{676}{4}$
$\Rightarrow 2x^2 + 14x + 49 = 169$
$\Rightarrow 2x^2 + 14x + 49 = 169$
$\Rightarrow 2x^2 + 14x + 49 - 169 = 0$
$\Rightarrow 2x^2 + 14x - 120 = 0$
$\Rightarrow x^2 + 7x - 60 = 0$
$\Rightarrow x^2 + 12x - 5x - 60 = 0$
$\Rightarrow x(x + 12) - 5(x + 12) = 0$
$\Rightarrow (x + 12)(x - 5) = 0$
$\Rightarrow x + 12 = 0 or x - 5$
$\Rightarrow x = -12 or x = 5$
$\Rightarrow x = -12$(rejected)
$\because\text{x}=5$
$\therefore$ The required distance $= AC + BC - AB$
$= 2x + 2x + 14 - 26$
$= 4x - 12 $$[\because\text{x}=5]$
$= 4 × 5 - 12$
$= 20 - 12$
= 8km
Hence, the distance saved by highway is 8km.
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