Given: ∆AKL, AK = 9, m∠K = 90°, m∠A = 60°. Find: The perimeter of ∆AKL, The area of ∆AKL.

Answer:Perimeter of ΔAKL is, [tex]27 +9 \sqrt{3}[/tex] units.Area of  ΔAKL is, [tex]\frac{81\sqrt{3} }{2}[/tex] square unitsStep-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