For going to a city B from city A there is a route via city C such that ACCB, 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. find how much distance will be saved in reaching city B from city A after the construction of the highway.


Given,

ACCB,


AC = 2x km,


CB = 2 (x + 7) km and AB = 26 km


On drawing the figure, we get the right angled ∆ ACB right angled at C.


Now, in ∆ACB,


AB2 = AC2 + BC2 by Pythagoras theorem,


(26)2 = (2x)2 + {2(x + 7)}2


676 = 4x2 + 4(x2 + 196 + 14x)


676 = 4x2 + 4x2 + 196 + 56x


676 = 82 + 56x + 196


8x2 + 56x – 480 = 0



On dividing by 8, we get


X2 + 7x – 60 = 0


x2 + 12x - 5x - 60 = 0


x(x + 12)-5(x + 12) = 0


(x + 12)(x - 5) = 0


x = -12, x = 5


As the distance can’t be negative.


x = 5 [ x ≠ -12]


Now, AC = 2x = 10km


And BC = 2(x + 7) = 2(5 + 7) = 24 km


The distance covered to each city B from city A via city C


= AC + BC


= 10 + 24


= 34 km


Distance covered to each city B from city A after the construction of the highway = BA = 26 km


Hence, the required saved distance is 34 – 26 = 8 km.


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