the length of a rectangle is 2cm longer than the width. if its length and width are both increased by 4cm, its area is increased by 72 cm^2. find the dimensions of the original rectangle.

Why are you calling this trigonometry?

width -- x
length -- x+2
area = x(x+2)

new width = x+4
new length = x+6
new area = (x+4)(x+6)

(x+4)(x+6) - x(x+4) = 72

solve for x

8 and 6

To solve this problem, let's break it down step by step.

Let's assume the width of the original rectangle is "x" cm.

According to the given information, the length of the rectangle is 2 cm longer than the width, so the length would be "x + 2" cm.

The area of a rectangle is calculated by multiplying its length and width. So, the original area of the rectangle would be (x + 2) * x cm^2.

Now, if both the length and width are increased by 4 cm, the new dimensions would be (x + 2 + 4) cm and (x + 4) cm.

The new area of the rectangle would be (x + 2 + 4) * (x + 4) cm^2, which is equal to (x + 6) * (x + 4) cm^2.

According to the problem, the new area is increased by 72 cm^2. So, we can set up the following equation:

(x + 6) * (x + 4) - (x + 2) * x = 72

Expanding the equation:

(x^2 + 10x + 24) - (x^2 + 2x) = 72

Simplifying further:

x^2 + 10x + 24 - x^2 - 2x = 72

Combining like terms:

8x + 24 = 72

Subtracting 24 from both sides:

8x = 72 - 24

8x = 48

Dividing both sides by 8:

x = 6

Therefore, the width of the original rectangle is 6 cm.

The length of the rectangle is 2 cm longer than the width, so the length would be x + 2 = 6 + 2 = 8 cm.

Hence, the dimensions of the original rectangle are 6 cm (width) and 8 cm (length).