Answer:
Perimeter of ΔAKL is, [tex]27 +9 \sqrt{3}[/tex] units.
Area of ΔAKL is, [tex]\frac{81\sqrt{3} }{2}[/tex] square units
Step-by-step explanation:
Given: In ΔAKL , AK = 9 units , [tex]m\angle K = 90^{\circ}[/tex] and [tex]m\angle A= 60^{\circ}[/tex].
In ΔAKL
[tex]\tan (A) = \frac{KL}{AK}[/tex]
Substitute the value AK = 9 units and [tex]m\angle A =60^{\circ}[/tex] to solve for KL ;
[tex]\tan(60^{\circ}) = \frac{KL}{9}[/tex]
[tex]\sqrt{3} = \frac{KL}{9}[/tex]
⇒[tex]KL = 9\sqrt{3} units[/tex]
In right angle ΔAKL,
Using Pythagoras theorem;
[tex]AL^2 = AK^2+KL^2[/tex] ......[1]
Substitute AK = 9 units and [tex]KL =9\sqrt{3} units[/tex] in [1] to solve for AL;
[tex]AL^2 = 9^2+(9\sqrt{3})^2[/tex]
[tex]AL^2 = 81+(81 \cdot 3)[/tex]
[tex]AL^2 = 81+243 = 324[/tex]
[tex]AL = \sqrt{324} = 18 units[/tex]
Perimeter of triangle is the sum of all the sides.
Perimeter of triangle AKL = AK +KL +AL = [tex]9 + 9\sqrt{3} + 18 = 27 +9 \sqrt{3}[/tex] units.
Formula for Area of right angle triangle is given by:
[tex]A = \frac{1}{2} Base \times Height[/tex]
Area of triangle AKL= [tex]\frac{1}{2} (9\sqrt{3}) \times (9)[/tex]
= [tex]\frac{81\sqrt{3} }{2}[/tex] square units
