Gardener's path
A gardener has an 1 1 m × 2 9 m rectangular garden and wants to build a diagonal path which is exactly 1 meter wide all along except for the two wedge-shaped ends, as shown.
Find the area of the path (in grey) in square meters.
The answer is 28.6.
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1 solution
I think this is wrong - we can work it out in an easier way. The two green triangles both have an area of ( 2 9 − 1 ) × 1 1 / 2 . Therefore the green area = 3 0 8 m 3 . The whole rectangle has an area of 2 9 × 1 1 = 3 1 9 m 3 . 3 1 9 − 3 0 8 = 1 1 m 3 .
This can also be worked out using the formula for the area of a parallelogram b a s e × p e r p e n d i c u l a r h e i g h t = 1 × 1 1 = 1 1 m 3
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The width of the path is 1m. The parallelogram base is NOT the width of the path.
The path area is that of a large rectangle with two green triangles removed. A = 1 1 ∗ 2 9 − 2 ∗ 2 1 ∗ 1 1 ∗ ( 2 9 − D F )
The path is also a long parallelogram with width BF and height 1. By the pythagorean theorem we have A = 1 1 2 + ( 2 9 − D F ) 2
Set these equal and solve
1 1 ∗ 2 9 − 1 1 ∗ ( 2 9 − D F ) = 1 1 2 + ( 2 9 − D F ) 2
1 2 1 D F 2 = 1 1 2 + ( 2 9 − D F ) 2
1 2 0 D F 2 + 5 8 x − 9 6 2 = 0
The exact positive solution to this quadratic is D F = 2 . 6
So the area of the path is 2 . 6 ∗ 1 1 = 2 8 . 6 m 2
NOTE: The problem as it is currently written can be seen as a trick question (to a solver not paying close attention.) The path is a long parallelogram but the 1 meter width of the path is NOT DF. Who would look at a garden path and call that the width? Site masters advise on whether to keep wording or clarify or give more of a hint than italics on the word wide.