SOLUTION: One dimension of a square is increased by 10 meters and the other dimension is increased by 5 meters. The area of the new rectangle is three times that of the original square. How
Question 2444: One dimension of a square is increased by 10 meters and the other dimension is increased by 5 meters. The area of the new rectangle is three times that of the original square. How long was one side of the square? Found 2 solutions by longjonsilver, gsmani_iyer:Answer by longjonsilver(2297) (Show Source):
You can put this solution on YOUR website! Let side of square be x
so, area of square = x^2
Rectangle will have length = x+10 and width = x+5
so, area of rectangle = (x+10)(x+5)
We are told that rectangle area = 3(square area), so...
(x+10)(x+5) = 3x^2
x^2 + 15x + 50 = 3x^2
2x^2 - 15x - 50 = 0
(2x + 5)(x - 10) = 0
so 2x+5=0 OR x-10=0
x=-5/2 or x=10
Physically, x=10m.
jon
You can put this solution on YOUR website! Let us say the side of the square = x
The area of the square = Sq.mtrs.
So after the increase, the sides of the rectangle is:
length = (x+10); width = (x+5);
So the area of the new rectangle is = (x+10)(x+5)
= = 3 Sq.mtrs. (Given.)
= 15x + 50 = 2 ...(By subtracting from both the sides)
= 2-15x-50 = 0 (By bringing all the terms to one side to make it
a quadratic equation.)
By factorising the above expression, we get,
= 2-20x+5x-50 = 0
= 2x(x-10)+5(x-10) = 0
= (x-10)*(2x+5) = 0
So either (x-10) or (2x+5) has to be equal to 0.
If (x-10) = 0, then x = 10;
If (2x+5) = 0, then x = -2.5; (because 2x should be = -5)
But the length cannot be negative.
Hence, the side of the square was equal to 10mtrs.
