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Create line segment BM so that it is perpendicular to AC (creating a 90° angle).  Now you have created ΔAMB which is a 30°-60°-90° triangle and ΔCMB which is a 45°-45°-90° triangle.

ΔAMB

hypotenuse: AB = 12 (given) = 2x  ⇒ x = 6

60°:  BM = x√3  ⇒ BM =  6√3

30°:  AM = x   ⇒ AM = 6

ΔCMB

45°: BM =  6√3  (solved from ΔAMB)  = x

45°:  MC = x  ⇒ MC =  6√3

hypotenuse:  BC = x√2   ⇒ BC = 6√6

Perimeter (P)

P = AB + BC + AC   (AC = AM + MC per segment addition postulate)

P =  12 + 6√6 + 6 +  6√3

P = 18 + 6√6 + 6√3

Answer:

Perimeter ≈ 44.78 units

Step-by-step explanation:

In ΔABC first draw a median on line AB from vertex A and name that point as M . Now in ΔAMB,

[tex]\sin 30 =\frac{perpendicular}{hypotenuse}\\ \sin 30=\frac{AM}{AB}\\\frac{1}{2}=\frac{AM}{12}\\AM=6[/tex]

Now, By Pythagoras Theorem in ΔAMB ,

[tex]AB^{2}=AM^{2}+ MB^{2}\\144=36+MB^{2}\\MB=6\sqrt{3}[/tex]

As median divides the line segment in two equal parts so BC =

[tex]2\cdot MB[/tex]

BC = 12√3


Now in ΔAMC,

[tex]\sin 30 =\frac{perpendicular}{hypotenuse}\\ \sin 30=\frac{AM}{CA}\\\frac{1}{2}=\frac{6}{CA}\\CA=12[/tex]

So, perimeter = AB+BC+CA

= 12+12+12√3 = 44.78


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