City route: \(AC\perp CB\), \(AC=2x\) km, \(CB=2(x+7)\) km. A direct highway \(AB=26\) km is planned. How many km are saved?
\(8\,\text{km}\)
Step 1: From the question, we know that \(AC \perp CB\). This means \(ACB\) is a right-angled triangle at point C.
Step 2: For a right-angled triangle, we can use the Pythagoras theorem: \(AB^2 = AC^2 + CB^2\).
Step 3: Substitute the given values: \(AC = 2x\) km, \(CB = 2(x+7)\) km, \(AB = 26\) km.
So, \(26^2 = (2x)^2 + (2(x+7))^2\)
Step 4: Simplify step by step: \(26^2 = 676\) \((2x)^2 = 4x^2\) \((2(x+7))^2 = 4(x+7)^2\)
Therefore, \(676 = 4x^2 + 4(x+7)^2\)
Step 5: Expand and simplify: \(676 = 4x^2 + 4(x^2 + 14x + 49)\) \(676 = 4x^2 + 4x^2 + 56x + 196\) \(676 = 8x^2 + 56x + 196\)
Step 6: Rearrange: \(8x^2 + 56x + 196 - 676 = 0\) \(8x^2 + 56x - 480 = 0\)
Divide by 8 to simplify: \(x^2 + 7x - 60 = 0\)
Step 7: Factorise: \((x+12)(x-5) = 0\)
So, \(x = -12\) or \(x = 5\). Since distance cannot be negative, we take \(x = 5\).
Step 8: Find lengths: \(AC = 2x = 2 \times 5 = 10\,\text{km}\) \(CB = 2(x+7) = 2 \times 12 = 24\,\text{km}\)
So the route via C is: \(AC + CB = 10 + 24 = 34\,\text{km}\)
Step 9: Direct route \(AB\) is given as \(26\,\text{km}\).
Step 10: Saving = (Distance via C) − (Direct distance) \(= 34\,\text{km} - 26\,\text{km} = 8\,\text{km}\).
Final Answer: The city saves 8 kilometres by using the direct highway.