Steps:
So before we jump into these problems, we must keep this particular rule in mind:
- Product Rule of Radicals: √ab = √a × √b. Additionally, remember that √a × √a = a
9.
So remember that the perimeter is the sum of all the sides. In this case:
[tex]P=3\sqrt{12} +\sqrt{27} +2\sqrt{48}[/tex]
So firstly, using the product rule of radicals we need to simplify these radicals as such:
[tex]3\sqrt{12} =3*\sqrt{4}*\sqrt{3} =3*2*\sqrt{3}=6\sqrt{3}\\\sqrt{27} =\sqrt{9} *\sqrt{3} =3\sqrt{3} \\2\sqrt{48}=2*\sqrt{8}*\sqrt{6}=2*\sqrt{4}*\sqrt{2}*\sqrt{2}*\sqrt{3}=2*2*2*\sqrt{3}=8\sqrt{3}\\\\P=6\sqrt{3}+3\sqrt{3}+8\sqrt{3}[/tex]
Now, combine like terms (since they all have the same like term √3, they can all be added up):
[tex]P=6\sqrt{3}+3\sqrt{3}+8\sqrt{3}\\P=17\sqrt{3}[/tex]
Your final answer is 17√3 in.
10a.
For this, we will be using the pythagorean theorem, which is [tex]a^2+b^2=c^2[/tex] , where a and b are the legs of the right triangle and c is the hypotenuse of the triangle. In this case, √18 and √32 are our legs and we need to find the hypotenuse. Set up our equation as such:
[tex](\sqrt{18} )^2+(\sqrt{32})^2=c^2[/tex]
From here we can solve for the hypotenuse. Firstly, solve the exponents (remember that square roots and squared power cancel each other out):
[tex]18+32=c^2[/tex]
Next, add up the left side:
[tex]50=c^2[/tex]
Lastly, square root both sides of the equation:
[tex]\sqrt{50}=c[/tex]
The hypotenuse is √50 cm.
10b.
Now, the process is similar to that of 9 so I will just show the steps to the final answer.
[tex]P=\sqrt{18} +\sqrt{32} +\sqrt{50} \\\\\sqrt{18}=\sqrt{9}*\sqrt{2}=3\sqrt{2}\\\sqrt{32}=\sqrt{16}*\sqrt{2}=4\sqrt{2}\\\sqrt{50}=\sqrt{25}*\sqrt{2}=5\sqrt{2}\\\\P=3\sqrt{2}+4\sqrt{2}+5\sqrt{2}\\P=12\sqrt{2}[/tex]
The perimeter is 12√2 in.
11.
Now, the process is still similar to question 9 but remember that this time we are working with cube roots.
[tex]P=3\sqrt[3]{24} +\sqrt[3]{27}+2\sqrt[3]{81}\\\\3\sqrt[3]{24}=3*\sqrt[3]{8}*\sqrt[3]{3}=3*2*\sqrt[3]{3}=6\sqrt[3]{3}\\\sqrt[3]{27}=3\\2\sqrt[3]{81}=2*\sqrt[3]{27}*\sqrt[3]{3}=2*3*\sqrt[3]{3}=6\sqrt[3]{3}\\\\P=6\sqrt[3]{3}+3+6\sqrt[3]{3}\\P=3+12\sqrt[3]{3}[/tex]
Note that when adding the numbers together, 3 isn't a like term to the other 2 terms because it doesn't have ∛3 multiplied with it.
The perimeter is 3 + 12∛3 in.
