A rectangle is 6 cm long and 5 cm wide. When each dimension is increased by x cm, the area is tripled. Find the value of x.
(6+x)(5+x) = 3*(6*5)
x^2 + 11x + 30 = 90
x^2 + 11x - 60 = 0
(x+15)(x-4) = 0
x = 4 -- we like positive dimensions
check: (6+4)(5+4) = 90 = 3*30
To solve this problem, we can set up an equation using the given information.
The original dimensions of the rectangle are 6 cm long and 5 cm wide, which means the original area is 6 cm * 5 cm = 30 cm^2.
When each dimension is increased by x cm, the new length becomes (6 + x) cm and the new width becomes (5 + x) cm.
According to the problem, the new area is tripled, which means the new area is 3 times the original area.
So, the equation we can set up is:
(6 + x) cm * (5 + x) cm = 3 * 30 cm^2
Now let's solve this equation to find the value of x.
Expand the equation:
(6 + x) * (5 + x) = 90
Multiply the terms using FOIL (First, Outer, Inner, Last):
30 + 11x + x^2 = 90
Rearrange the equation and combine like terms:
x^2 + 11x + 30 = 90
Subtract 90 from both sides:
x^2 + 11x + 30 - 90 = 0
x^2 + 11x - 60 = 0
Now we have a quadratic equation. We can solve it by factoring or using the quadratic formula. Let's use factoring:
(x + 15)(x - 4) = 0
This equation will be true if x + 15 = 0 or x - 4 = 0.
Solving for x in each case:
Case 1: x + 15 = 0
x = -15
Case 2: x - 4 = 0
x = 4
So, there are two possible values for x: -15 and 4. However, since we are dealing with dimensions of a rectangle, a negative value does not make sense in this context.
Therefore, the value of x is 4 cm.
To solve this problem, let's start by finding the original area of the rectangle.
The original length of the rectangle is 6 cm, and the original width is 5 cm. Therefore, the original area, A₁, is given by the formula:
A₁ = length × width
Substituting the given values into the formula, we have:
A₁ = 6 cm × 5 cm
A₁ = 30 cm²
Next, let's find the new dimensions of the rectangle after each dimension is increased by x cm. The new length will be 6 cm + x cm, and the new width will be 5 cm + x cm.
The new area, A₂, is then given by:
A₂ = (6 cm + x cm) × (5 cm + x cm)
A₂ = (6 + x) cm × (5 + x) cm
A₂ = (30 + 11x + x²) cm²
According to the problem, when we increase each dimension by x cm, the area is tripled. Therefore, we can write the equation:
A₂ = 3A₁
Substituting the values of A₂ and A₁, we have:
30 + 11x + x² = 3 × 30
30 + 11x + x² = 90
Rearranging the equation in standard quadratic form, we get:
x² + 11x - 60 = 0
Now we can solve this quadratic equation. Factoring or using the quadratic formula will help us find the values of x.
Factoring the equation, we obtain:
(x + 15)(x - 4) = 0
From the factored form, we have two possible solutions for x:
x + 15 = 0 → x = -15
x - 4 = 0 → x = 4
Since dimensions cannot be negative in this context, we discard x = -15. Therefore, the valid solution for x is:
x = 4
Hence, the value of x is 4 cm.