Answer:
See below
Step-by-step explanation:
[tex] In\: \triangle AKL, \:\\ AK = 9,\: \\m\angle K = 90\degree, \: \\m\angle A= 60\degree \\
\therefore m\angle L = 30\degree\\
By \:30°-60°-90° \:\triangle \:theorem:\\
AK = \frac{1}{2} \times AL\\(SIDE\: OPPOSITE \:TO \:30\degree) \\\\
9 = \frac{1}{2} \times AL\\\\
AL = 9\times 2\\\\
\huge\red{AL = 18}\\\\
KL = \frac{\sqrt 3}{2} \times AL\\(SIDE\: OPPOSITE \:TO \:60\degree) \\\\
KL = \frac{\sqrt 3}{2}\times 18\\\\
KL = 9\times \sqrt 3\\\\
\huge\purple{KL = 9\sqrt 3}\\\\
P(\triangle AKL) = AK + KL + AL\\
P(\triangle AKL)= 9 + 9\sqrt 3+18\\
\huge\orange{P(\triangle AKL)=(27 + 9\sqrt 3) \: units} \\\\
A(\triangle AKL) = \frac{1}{2} \times KL\times AK\\
A(\triangle AKL) = \frac{1}{2} \times 9\sqrt 3\times 9\\
A(\triangle AKL) = \frac{81\sqrt 3}{2} \\
\huge\pink{A(\triangle AKL) = 40.5\sqrt 3\: units^2} [/tex]
