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