Question

find the perimeter as a radical expression in simplest form

Answer 1

Answer:

The perimeter of the figure, in it's simplest form, is 2(1 + 5*sqrt(6)) feet

Step-by-step explanation:

The perimeter of a figure is the sum of all dimensions of the figure.

In this question, the perimeter is:

$2+\sqrt{54}+\sqrt{54}+2\sqrt{24}$

Adding like terms:

$2+\sqrt{54}+\sqrt{54}+2\sqrt{24}=2+2\sqrt{54}+2\sqrt{24}$

Now we have to factorize 54 and 24

Factorization of 54:

We go dividing by prime numbers.

54|2

27|3

9|3

3|3

1

So: 54 = 2*3*3*3 = 2*3³

Factorization of 24:

24|2

12|2

6|2

3|3

1

So 24 = 2*2*2*3=2³*3

Simplifying:

$2+2\sqrt{54}+2\sqrt{24}=2+2\sqrt{2\ast3^3}+2\sqrt{2^3\ast3}^{}$

Now we continue working to place in the simplest form

$2+2\sqrt{2\ast3^3}+2\sqrt{2^3\ast3}=2+2\sqrt{2\ast3\ast3^2}+2\sqrt{2^2\ast2\ast3}$

Now, for the radicals, we have that:

$\sqrt{2\ast3\ast3^2}=\sqrt{6\ast9}=\sqrt{6}\ast\sqrt{9}=3\sqrt{6}$$\sqrt{2^2\ast2\ast3}=\sqrt{4\ast6}=\sqrt{4}\ast\sqrt{6}=2\sqrt{6}$

Then

$2+2\sqrt{2\ast3\ast3^2}+2\sqrt{2^2\ast2\ast3}=2+2\ast3\sqrt{6}+2\ast2\sqrt{6}$

Finally:

$2+2\ast3\sqrt{6}+2\ast2\sqrt{6}=2+6\sqrt{6}+4\sqrt{6}=2+10\sqrt{6}=2(1+5\sqrt{6})$

The perimeter of the figure, in it's simplest form, is 2(1 + 5*sqrt(6)) feet