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Sagot :
the question stated that the lot figure is a rectangle because of the given of length and width.
P=2(l+w)--- where
P= perimeter --l= length -- w= width
58m= 2(17.5+w)
58= 35+2w
w= 11.5 meters
P=2(l+w)--- where
P= perimeter --l= length -- w= width
58m= 2(17.5+w)
58= 35+2w
w= 11.5 meters
What is the width of a lot with a perimeter of 58 meters and a length of 17.5 meters?
Perimeter = 2(length)+2(width)
Given: Length = 17.5 meters
Perimeter = 58 meters
Required: Width
Equation: Perimeter = 2(length)+2(width)
Solution: 58 meters = 2(17.5 meters) + 2n
58 meters = 35 meters + 2n
-2n = 35 meters - 58 meters
-2n = -23
[tex] \frac{-2n}{-2} = \frac{-23}{-2} [/tex]
n = 11.5
Answer: The width of the lot with a perimeter of 58 meters and a length of 17.5 meters is 11.5 meters.
Check: Perimeter = 2(length)+2(width)
58 meters = 2(17.5 meters) + 2(11.5 meters)
58 meters = 35 + 23 meters
58 meters = 58 meters
Perimeter = 2(length)+2(width)
Given: Length = 17.5 meters
Perimeter = 58 meters
Required: Width
Equation: Perimeter = 2(length)+2(width)
Solution: 58 meters = 2(17.5 meters) + 2n
58 meters = 35 meters + 2n
-2n = 35 meters - 58 meters
-2n = -23
[tex] \frac{-2n}{-2} = \frac{-23}{-2} [/tex]
n = 11.5
Answer: The width of the lot with a perimeter of 58 meters and a length of 17.5 meters is 11.5 meters.
Check: Perimeter = 2(length)+2(width)
58 meters = 2(17.5 meters) + 2(11.5 meters)
58 meters = 35 + 23 meters
58 meters = 58 meters
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