Question:
A rectangular pool of 20 m wide and 60 m long is surrounded by a walkway of uniform width. If the total area of the walkway is 516 \(m^2\), how wide, in metres, is the walkway?
A rectangular pool of 20 m wide and 60 m long is surrounded by a walkway of uniform width. If the total area of the walkway is 516 \(m^2\), how wide, in metres, is the walkway?
Show Hint
The pool plus walkway is a rectangle \((20+2w)\times(60+2w)\); subtract the pool area to get the walkway area.
Updated On: Nov 4, 2026
- 3 m
- 2 m
- 4 m
- 3.5 m
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Verified By Collegedunia
The Correct Option is A
Solution and Explanation
Step 1: Set up.
Let the walkway width be \(w\) m. The outer rectangle is \((20+2w)\) by \((60+2w)\).
Step 2: Area equation.
\[(20+2w)(60+2w)-20\times 60=516\]\[1200+160w+4w^2-1200=516\Rightarrow 4w^2+160w-516=0\Rightarrow w^2+40w-129=0\]Step 3: Solve.
\[w=\frac{-40\pm\sqrt{1600+516}}{2}=\frac{-40\pm 46}{2}\]so \(w=3\) (the other root, \(-43\), is rejected).
Step 4: Check.
Outer rectangle \(26\times 66=1716\); \(1716-1200=516\).
Final Answer:
The correct option is (1): 3 m.
Let the walkway width be \(w\) m. The outer rectangle is \((20+2w)\) by \((60+2w)\).
Step 2: Area equation.
\[(20+2w)(60+2w)-20\times 60=516\]\[1200+160w+4w^2-1200=516\Rightarrow 4w^2+160w-516=0\Rightarrow w^2+40w-129=0\]Step 3: Solve.
\[w=\frac{-40\pm\sqrt{1600+516}}{2}=\frac{-40\pm 46}{2}\]so \(w=3\) (the other root, \(-43\), is rejected).
Step 4: Check.
Outer rectangle \(26\times 66=1716\); \(1716-1200=516\).
Final Answer:
The correct option is (1): 3 m.
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