Help needed =]?

Question:if the lenghth of a rectangle is 5 more than the width and the area is 266 sq. feet find the dimensions.

if the hypotenuse of a right triangle is 50 feet and one leg is 40 feet, find the lenghth of the other leg.

thank you guys so muchh =]

Answers:
1. l = w + 5
a = lw
a = w(w + 5)
a = w^2 + 5w
266 = w^2 + 5w
w^2 + 5w - 266 = 0
(w + 19)(w - 14) = 0
w = -19, 14
Since a negative number is meaningless when describing area, the answer is:
l = 19
w = 14

2. a^2 + 40^2 = 50^2
a^2 + 1600 = 2500
a^2 = 900
sqrt(a^2) = sqrt(900)
a = +-30; disregard negative when dealing with lengths
a = 30
1. L = 5 + x
W = x
Area = 266

You multiply the Length and the Width together.
x(5+x) and you get 5x + x (squared)

266 = x (squared) + 5x

x(squared) + 5x - 266 = 0

From there on, I hope you get it.


for 2. the answer is 30. It's a 3,4,5 triangle.

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