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(A) Given,$AC \perp CB$,$AC = 2x \, km$,$CB = 2(x+7) \, km$ and $AB = 26 \, km$.From the figure,we have a right-angled $	riangle ACB$ with the right

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$8$

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(A) Given,$AC \perp CB$,$AC = 2x \, km$,$CB = 2(x+7) \, km$ and $AB = 26 \, km$.From the figure,we have a right-angled $	riangle ACB$ with the right angle at $C$.In $	riangle ACB$,by the Pythagoras theorem:$AB^2 = AC^2 + BC^2$$(26)^2 = (2x)^2 + \{2(x+7)\}^2$$676 = 4x^2 + 4(x^2 + 49 + 14x)$$676 = 4x^2 + 4x^2 + 196 + 56x$$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$$x(x + 12) - 5(x + 12) = 0$$(x + 12)(x - 5) = 0$$x = -12$ or $x = 5$.Since distance cannot be negative,$x = 5$ (as $x 
eq -12$).Now,$AC = 2x = 2(5) = 10 \, km$ and $BC = 2(x+7) = 2(5+7) = 24 \, km$.The distance covered to reach city $B$ from city $A$ via city $C$ is $AC + BC = 10 + 24 = 34 \, km$.The distance covered to reach city $B$ from city $A$ after the construction of the highway is $AB = 26 \, km$.Therefore,the distance saved is $34 - 26 = 8 \, km$.

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