Question

2\sqrt{27} in.
2\sqrt{12} in.
\sqrt{12} in.
3\sqrt{3} in.
find the perimeter of the quadrilateral in simplest form.
\left(3\sqrt{12} + 3\sqrt{3} + 2\sqrt{27}\
ight) in.
33\sqrt{3} in.
\left(9\sqrt{3} + 2\sqrt{27}\
ight) in.
15\sqrt{3} in.

Explanation:

Step1: Recall Perimeter Formula

Perimeter of a quadrilateral is the sum of all its sides. So, we add \(2\sqrt{12}\), \(2\sqrt{27}\), \(\sqrt{12}\), and \(3\sqrt{3}\).

Step2: Simplify Radicals

• Simplify \(2\sqrt{12}\): \(\sqrt{12}=\sqrt{4\times3}=2\sqrt{3}\), so \(2\sqrt{12}=2\times2\sqrt{3}=4\sqrt{3}\).
• Simplify \(\sqrt{12}\): As above, \(\sqrt{12}=2\sqrt{3}\).
• Simplify \(2\sqrt{27}\): \(\sqrt{27}=\sqrt{9\times3}=3\sqrt{3}\), so \(2\sqrt{27}=2\times3\sqrt{3}=6\sqrt{3}\).

Step3: Sum the Sides

Now, sum the simplified sides: \(4\sqrt{3}+6\sqrt{3}+2\sqrt{3}+3\sqrt{3}\)? Wait, no, original sides are \(2\sqrt{12}\), \(2\sqrt{27}\), \(\sqrt{12}\), \(3\sqrt{3}\). Wait, correct sum: \(2\sqrt{12}+2\sqrt{27}+\sqrt{12}+3\sqrt{3}\). Combine like terms for \(\sqrt{12}\): \(2\sqrt{12}+\sqrt{12}=3\sqrt{12}\). Now, simplify \(3\sqrt{12}\): \(3\times2\sqrt{3}=6\sqrt{3}\). Then \(2\sqrt{27}=6\sqrt{3}\), and \(3\sqrt{3}\) remains. Wait, no, let's do again:

Wait, original sides:

• Side1: \(2\sqrt{12}\)
• Side2: \(2\sqrt{27}\)
• Side3: \(\sqrt{12}\)
• Side4: \(3\sqrt{3}\)

So sum: \(2\sqrt{12}+2\sqrt{27}+\sqrt{12}+3\sqrt{3}=(2\sqrt{12}+\sqrt{12})+2\sqrt{27}+3\sqrt{3}=3\sqrt{12}+2\sqrt{27}+3\sqrt{3}\). But simplify \(3\sqrt{12}\): \(3\times2\sqrt{3}=6\sqrt{3}\), \(2\sqrt{27}=6\sqrt{3}\), so total: \(6\sqrt{3}+6\sqrt{3}+3\sqrt{3}=15\sqrt{3}\)? Wait, no, wait: \(2\sqrt{12}=4\sqrt{3}\), \(\sqrt{12}=2\sqrt{3}\), so \(4\sqrt{3}+2\sqrt{3}=6\sqrt{3}\). Then \(2\sqrt{27}=6\sqrt{3}\), and \(3\sqrt{3}\). So total: \(6\sqrt{3}+6\sqrt{3}+3\sqrt{3}=15\sqrt{3}\). Wait, let's check the options. One of the options is \(15\sqrt{3}\) in.

Wait, let's re - calculate:

Perimeter \(P = 2\sqrt{12}+2\sqrt{27}+\sqrt{12}+3\sqrt{3}\)

Combine \(2\sqrt{12}\) and \(\sqrt{12}\): \( (2 + 1)\sqrt{12}=3\sqrt{12}\)

Now, \(3\sqrt{12}=3\times2\sqrt{3}=6\sqrt{3}\) (since \(\sqrt{12}=2\sqrt{3}\))

\(2\sqrt{27}=2\times3\sqrt{3}=6\sqrt{3}\) (since \(\sqrt{27}=3\sqrt{3}\))

And we have \(3\sqrt{3}\)

Now sum \(6\sqrt{3}+6\sqrt{3}+3\sqrt{3}=15\sqrt{3}\)

Answer:

\(15\sqrt{3}\) in.