The length of a rectangle is 10 feet longer than it is wide. if each side is increased 10 feet, then the area is multiplied by 4. what was the size of the original rectangle?

Suppose:
width=x ft
length=(x+10) ft
Area:
x(x+10) ft²

Given that the sides are increased by 10, then the new dimensions will be:
width=(x+10) ft
length=(x+20) ft
thus the area will be:
(x+10)(x+20)=4×(x^2+10x)
solving for x we get:
x=-10 or x=20/3
thus the original width=20/3 and length=(10+20/3)=50/3 ft

129 Feet

Step-by-step explanation:

You can make an equation where Height=h, and Base=b. So going backwards with the first question, the base will equal 4 times the height: b=4h, and 11 less. b=4h-11. The the other equation will be just that the base plus the height is 49. b+h=49. Then you put that into y=mx+b form. The first equation already is, and the second equation would be b=49-h. So, since the equations equal the same things, you throw away the b. Then you make a new equation by having 4h-11 equal 49-h. Since this is true seeing as they are the same triangle. Then you just use propreties of equality for the equation. You end up with 5h=60. The. You divide 60 by 5 to make 5h just equal h. So you get h=6. Then, like the qeuation said, the base (unknown) plus the height (6) is 49. So to get from 6 to 49, the answer is 43. The formula for that area of a triangle is (base x height) /2 = area. So 43 times 6 is 258. Then divided by 2 is 129. So the area of the triangle is 129.