Mathematics
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.
Answer
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.
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