Length of rectanngle is 1 cm more than width. If length of rectangle is doubled the area is increased by 30cm2. find dimensions of original rectangle.
if the length and width are x,y respectively, then
x = y+1
2xy = xy+30
xy=30
x=6,y=5
check:
xy = 6*5 = 30
2xy = 12*5 = 60 = 30+30
To solve this problem, let's start by assigning variables to the dimensions of the rectangle. Let's say that the width of the rectangle is 'w' cm.
According to the given information, the length of the rectangle is 1 cm more than the width. So, the length would be 'w + 1' cm.
The area of a rectangle is calculated by multiplying its length and width. Therefore, the original area of the rectangle (before doubling the length) would be:
Area = length * width
Original Area = (w + 1) * w
Now, if we double the length, the new length would be '2(w + 1)' cm. And the new area would be:
New Area = new length * width
New Area = 2(w + 1) * w
According to the problem, the new area is increased by 30 cm²:
New Area - Original Area = 30
Substituting the expressions for the areas:
[2(w + 1) * w] - [(w + 1) * w] = 30
Now, let's solve the equation step by step:
2w² + 2w - w² - w = 30
w² + w = 30
Rearranging and setting the equation equal to zero:
w² + w - 30 = 0
Now, we need to factorize the quadratic equation:
(w + 6)(w - 5) = 0
Setting each factor equal to zero:
w + 6 = 0 or w - 5 = 0
Solving for 'w':
w = -6 or w = 5
Since width cannot be negative, we discard the negative value. Therefore, the width of the original rectangle is 5 cm.
The length of the original rectangle is 1 cm more than the width:
Length = width + 1
Length = 5 + 1
Length = 6 cm
So, the dimensions of the original rectangle are 5 cm (width) and 6 cm (length).