# Find the area of rectangle whose length is one more three time the width and the perimeter is 26.

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### 3 Answers

We have to find the area of rectangle whose length is one more three times the width and the perimeter is 26.

Let the width be W.

Now the length is 3W + 1.

The perimeter is 2L + 2W = 26

=> 2(3W + 1) + 2W = 26

=> 6W + 2 + 2W = 26

=> 8W = 24

=> W = 24/8

=> W = 3

The length is 3*3 +1 = 10

**Therefore the area of the rectangle is 10*3 = 30 units^2.**

We will assume that the length of the rectangle is L, and the width is W.

Given that the length is 1 more that 3 time the width.

==> L = 1 + 3*W

==> L = 3W + 1 ...............(1)

Also, given the perimeter is 26.

==> 2L + 2W = 26

We will divide by 2:

==> L + W = 13.............(2)

Now we will substitute (1) into (2).

==> ( 3W + 1) + W = 13

==> 3W + 1 + W = 13

==> 4W = 12

We will divide by 4:

==> W = 3

==> L + 3w + 1 = 3*3 + 1 = 10

==> L =10

Then, the area of the rectangle is given by:

A = L*W = 10*3 = 30

**Then, the area or the rectangle is 30 square units.**

Given length 1+3times width = 1+3x, where x is the assumed width.

Therefore the perimeter P = 2(length+width) = 2(1+3x+x) = 8x+2 which is equal to 26.

8x+2 = 26.

8x= 26-2 = 24.

x = 24/8 = 3.

So width = 3. Length = 3x+1 = 3*3+1 = 10.

Therefore the area of the rectangle = length *width = 10*3 = 30 sq units.