Asked by Jee-Anne
What is the area of a rectangle whose length is 1 unit longer than its width and whose diagonal is 2 unit longer than the width?
Answers
Answered by
Reiny
looks like Pythagoras.
width --- x
length --- x+1
diagonal --- x+2
x^2 + (x+1)^2 = (x+2)^2
by recognition:
the sides are one unit apart, the only Pythagorean triple like that is 3,4,5, since 3^2 + 4^2 = 5^2.
If you did not recognize that:
x^2 + x^2 + 2x + 1 = x^2 + 4x + 4
x^2 - 2x - 3 = 0
(x - 3)(x+1) = 0
x = 3 or x = -1, but a side can't be negative, so
x = 3
find the area using base*height
width --- x
length --- x+1
diagonal --- x+2
x^2 + (x+1)^2 = (x+2)^2
by recognition:
the sides are one unit apart, the only Pythagorean triple like that is 3,4,5, since 3^2 + 4^2 = 5^2.
If you did not recognize that:
x^2 + x^2 + 2x + 1 = x^2 + 4x + 4
x^2 - 2x - 3 = 0
(x - 3)(x+1) = 0
x = 3 or x = -1, but a side can't be negative, so
x = 3
find the area using base*height
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