# Find the area of the rectangle if the length is three times the width and the perimeter is 16.

### 3 Answers | Add Yours

Let the length be L and the width be W.

Perimeter = 2(L+W). Area = L*W.

The perimeter is given as 16. Also the length is thrice the width.

This gives L = 3W

2(3W + W) = 16

=> 2*4W = 16

=> W = 16/8

=> W = 2

L = 3*W = 3*2 = 6

Therefore the area is 2*6 = 12.

**The required area of the rectangle is 12 square units.**

Let the length of the rectangle be L and the width be w.

Given that the length of the rectangle is 3 times the width.

Then we will write:

L = 3*w ..............(1).

Also, given that the perimeter of the rectangle is 16.

Then we will write:

==> 2L + 2W = 16

We will divide by 2.

==> L + w = 8...........(2)

Now we will substitute with L = 3w

==> 3w + w = 8

==> 4w = 8

==> w= 2

==> L = 3*2 = 6

Then, the length of the rectangle is 6, and the width is 2 .

Now we will calculate the area of the rectangle.

We know that the area of the rectangle is given by:

\A = L * w

==> A = 2*6 = 12

**Then, the area is 12 square units.**

Given that length is 3 times width.

So if x is width, then length = 3x.

We know that if p is the perimeter of the rectangle, then p = 2*(length+width).

So p = 2(3x+x)= 8x.

Actual perimeter given = 16.

So 8x = 16.

x= 16/8 = 2.

Therefore x = 2 is the width.

3x= 3*2 = 6 is the length.

So area of the rectangle = length*width = 6*2 = 12 sq units.