# What is the area of the rectangle whose length is 3 times the width and the perimeter is 24.

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### 3 Answers

To find the area of the rectangle, first we need to determine the length and the width.

Let the length be L and the width be W.

Given that the length = 3 times the width.

Then we will write:

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

Also, given that the perimeter (P) = 24

==> 2L +2W = 24

Divide by 2:

==> L + W = 12...........(2)

Now let us substitute (1) in (2).

==> 3W + W = 12

==> 4*W = 12\

Now we will divide by 4.

==> W = 12/4 = 3

Then, W = 3 and L = 3*3 = 9

Then the length of the rectangle = 9, and the width = 3

Now we will calculate the area.

Area = L*W = 3* 9 = 27

**Then, the area of the rectangle = 27 square units.**

We'll start by writing the area of a rectangle.

A = length*width

But, from enunciation, length = 3width

A = 3width*width

A = 3width^2

We also know, from enunciation, that the perimeter is P= 24.

P = 2length + 2width

P = 2*(3width) + 2width

P = 8 width

24 = 8 width

We'll divide by 8:

width = 24/8

width = 3

length = 3width

length = 9

A = 3*9

**A = 27 square units**

What is the area of the rectangle whose length is 3 times the width and the perimeter is 24.

Let the width = x

Then by data , lenth = 3 times width = 3x.

So the perimeter p = 2(length+width) = 2(3x+x) = 8x.by definition.

p = 24 by data.

Therefore p = 8x= 24.

So 8x= 24.

x= 24/8 = 3.

Therefore length = 3x = 3*3 = 9.

Therefore area = length *width = 3x^2 .

Therefore Area = 3*4^2 = 48 sq unit.