Determine the length of AB
1 answer:
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<em>The</em><em> </em><em>length</em><em> </em><em>of</em><em> </em><em>other</em><em> </em><em>leg</em><em> </em><em>is</em><em> </em><em>3</em><em>0</em><em> </em><em>ft</em>
<em>Look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em>
<em>Hope</em><em> </em><em>it</em><em> </em><em>helps</em><em>.</em><em>.</em><em>.</em>
<em>Good</em><em> </em><em>luck</em><em> </em><em>on</em><em> </em><em>your</em><em> </em><em>assignment</em>
The path should be 6.38 feet wide
Let x be the width of the path.
Since the rectangular plot has an area with dimensions 15 ft by 10 ft, we add x to each dimension to give the dimensions of the rectangular plot plus uniform path.
So, the dimensions are 15 + x by 10 + x respectively.
So, the area of the rectangular plot plus uniform path is A
A = (15 + x)(10 + x)
Since the combined area of the path is 350 square feet, A = 350 ft²
So, (15 + x)(10 + x) = 350
Expanding the brackets, we have
150 + 15x + 10x + x² = 350
x² + 25x + 150 = 350
x² + 25x + 150 - 350 = 0
x² + 25x - 200 = 0
Using the quadratic formula, we find x, the width of the path
So,
Since x cannot be negative, we choose the positive answer.
x = 6.375 ft
x ≅ 6.38 ft
So, the path should be 6.38 feet wide
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