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The area of a rectangular region enclosed by a wire is 1350 sq. yard
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17 Feb 2025, 01:11
17 Feb 2025, 01:11
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The area of a rectangular region enclosed by a wire is 1350 sq. yards. The length of the rectangular region is 15 yards more than the width of the region. What is the length of the wire in feet, required for fencing the rectangular region? (1 yard = 3 feet)
A. 250
B. 300
C. 350
D. 400
E. 450
A. 250
B. 300
C. 350
D. 400
E. 450
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The area of a rectangular region enclosed by a wire is 1350 sq. yard
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03 Mar 2025, 07:08
03 Mar 2025, 07:08
1
Let w = the width of the rectangle.
Given that the length is 15 yards more than the width, length = w+15
Since the area of a rectangle = width x length, we can set up the equation:
\(w * (w+15) = 1350\)
\(w^2 + 15w = 1350\)
\(w^2 + 15w - 1350 = 0\)
\((w+45)(w-30) = 0\)
\(w = -45, 30\)
Since the width can't be a negative value, -45 is ruled out. We find that width = 30 and length = 30+15 = 45.
To find the length required to fence the region, we need to calculate the perimeter and translate into feet.
perimeter in yards = 2 * (width+length) * 3 feet/yard = 2 (30 yards+ 45 yards) * 3 = 450 feet
Given that the length is 15 yards more than the width, length = w+15
Since the area of a rectangle = width x length, we can set up the equation:
\(w * (w+15) = 1350\)
\(w^2 + 15w = 1350\)
\(w^2 + 15w - 1350 = 0\)
\((w+45)(w-30) = 0\)
\(w = -45, 30\)
Since the width can't be a negative value, -45 is ruled out. We find that width = 30 and length = 30+15 = 45.
To find the length required to fence the region, we need to calculate the perimeter and translate into feet.
perimeter in yards = 2 * (width+length) * 3 feet/yard = 2 (30 yards+ 45 yards) * 3 = 450 feet
gmatclubot
The area of a rectangular region enclosed by a wire is 1350 sq. yard [#permalink]
03 Mar 2025, 07:08
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