Asked by Paps olarte
The diagonals of a rectangle is 8m longer than it's shorter side. If the area is 60m.sqr., what are its dimentions?
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Answers
Answered by
Reiny
short side --- x
diagonal --- x+8
other side --- y
x^2 + y^2 = (x+8)^2
x^2 + y^2 = x^2 + 16x + 64
y^2 = 16x + 64
y = √(16x+64)
area = xy = 60
x√(16x+64) = 60
square both sides
x^2(16x+64) = 3600
16x^3 + 64x^2=3600
x^3 + 4x^2 - 225 = 0
looks like x = 5 works
then y = √(16(5) + 64) = 12
the rectangle is 5 m by 12 m
If we had known the numbers came out that "nice" we could have done this by trial and error
e.g. factor of 60:
1x60 , 2x30, 3x20, 4x16, 5x12, 6x10
the only one which give yield a whole-number hypotenuse is
5x12 --->5^2 + 12^2 = 13^2 , and 13 is 8 more than 5
diagonal --- x+8
other side --- y
x^2 + y^2 = (x+8)^2
x^2 + y^2 = x^2 + 16x + 64
y^2 = 16x + 64
y = √(16x+64)
area = xy = 60
x√(16x+64) = 60
square both sides
x^2(16x+64) = 3600
16x^3 + 64x^2=3600
x^3 + 4x^2 - 225 = 0
looks like x = 5 works
then y = √(16(5) + 64) = 12
the rectangle is 5 m by 12 m
If we had known the numbers came out that "nice" we could have done this by trial and error
e.g. factor of 60:
1x60 , 2x30, 3x20, 4x16, 5x12, 6x10
the only one which give yield a whole-number hypotenuse is
5x12 --->5^2 + 12^2 = 13^2 , and 13 is 8 more than 5
Answered by
Renna
How did you find 13 as c?(a²+b²=c²)
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marylou
salamat😊
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