For going to a city B from the city A, there is 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 highway.

From figure,

△ABC is right triangle.

By pythagoras theorem,

⇒ AB² = AC² + BC²

⇒ 26² = (2x)² + [2(x + 7)]²

⇒ 676 = 4x² + (2x + 14)²

⇒ 676 = 4x² + 4x² + 196 + 56x

⇒ 676 - 196 = 8x² + 56x

⇒ 8x² + 56x = 480

⇒ 8(x² + 7x) = 480

⇒ x² + 7x = 60

⇒ x² + 7x - 60 = 0

⇒ x² + 12x - 5x - 60 = 0

⇒ x(x + 12) - 5(x + 12) = 0

⇒ (x - 5)(x + 12) = 0

⇒ x = 5 or x = -12.

Since, distance cannot be negative so, x ≠ -12.

Distance taken to reach from B to A without highway = BC + AC

= 2(x + 7) + 2x

= 2x + 14 + 2x

= 4x + 14

= 4(5) + 14

= 20 + 14 = 34 km

Distance taken to reach from B to A through highway = 26 km

Distance saved = 34 - 26 = 8 km.

Hence, 8 km will be saved in reaching city A from B after construction of highway.