Perimeter of the triangle ABC is
[tex]\frac{15\sqrt{15}}{3}+5\sqrt{5}[/tex]
Given :
In right △ABC , the right angle is at C, m∠A=30∘ , and AC=5√ 5 units.
we use the special triangle ratio.
Special triangle 30:60:90 ratio is [tex]x:x\sqrt{3} :2x[/tex]
the side opposite to 60 degree is x
[tex]AC=x\sqrt{3} =5\sqrt{5}\\x\sqrt{3} =5\sqrt{5}\\x=\frac{5\sqrt{5}}{\sqrt{3} } \\x=\frac{5\sqrt{15}}{3} \\[/tex]
The side opposite of 30 degree that is [tex]BC=\frac{5\sqrt{15} }{3}[/tex]
Now we find out hypotenuse
Hypotenuse is 2x
[tex]\\x=\frac{5\sqrt{15}}{3} \\\\AB=2x=2 \cdot \frac{5\sqrt{15}}{3}\\AB=\frac{1`0\sqrt{15}}{3}[/tex]
Now we add all the sides to find the perimeter of triangle ABC
[tex]AB+BC+AC\\\frac{10\sqrt{15}}{3}+\frac{5\sqrt{15}}{3}+5\sqrt{5} \\\frac{15\sqrt{15}}{3}+5\sqrt{5}[/tex]
Perimeter of the triangle ABC is [tex]\frac{15\sqrt{15}}{3}+5\sqrt{5}[/tex]
Learn more : brainly.com/question/14524498
