Question

the length of a rectangle is 2m longer than its width. if the perimeter of the rectangle is 40m , find its area.

Answer 1

99

Explanation:

Let's call the width of the rectangle "w".

According to the problem, the length is "2m longer than its width", so the length must be:

w + 2

The perimeter of a rectangle is given by:

2(length + width)

In this case, we know the perimeter is 40 meters:

2(w + 2 + w) = 40

Simplify and solve for "w":

4w + 4 = 40

4w = 36

w = 9

So the width of the rectangle is 9 meters.

To find the length, we can use the equation we came up with earlier:

length = w + 2 = 9 + 2 = 11

Now we can find the area of the rectangle:

Area = length x width = 11 x 9 = 99

Therefore, the area of the rectangle is 99 square meters.

Answer 2

Answer: Area = 99 cm²

Explanation:

In order to find the area, we'll first find the side lengths of the rectangle by setting up an equation.

Use the perimeter formula:

P = 2(l + w)

40 = 2(x + 2 + x)

Solve for x.

40 = 2(2x + 2)

40 = 4x + 4

36 = 4x

9 = x

So, the rectangle's width is 9 cm; since the length is 2 m longer than the width, it means that the length is 11 m.

Now we know both the width and the length:

• width = 9 cm
• length = 11 cm

Use these sides to find the area.

• 
  `\sf{A=11*9}`
• 
  `\sf{A=99\;cm^2}`

Therefore, the area is 99 cm².