Answer:
Given: In triangle AKL, AK = 9 units, [tex]m\angle K = 90^{\circ}[/tex] [tex]m\angle A = 60^{\circ}[/tex]
In triangle AKL
Sum of interior angles of the triangle is 180 degree;
[tex]m\angle A +m\angle K + m\angle L = 180^{\circ}[/tex]
or
[tex]m\angle L = 180^{\circ}-(\angle A +\angle K)[/tex]
Substitute the given values we get;
[tex]m\angle L = 180^{\circ}-(60^{\circ} + 90^{\circ}) = 180^{\circ} -150^{\circ} = 30^{\circ}[/tex]
Now, in a 30°-60°-90° triangle, the length of the hypotenuse is twice the length of the shorter leg, and the length of the longer leg is [tex]\sqrt{3}[/tex] times the length of the shorter leg.
Length of shorter leg (AK) = 9 units
From the above definition:
[tex]\text{Length of hypotenuse(AL)}=2 \times \text{Length of Shorter leg}[/tex]
⇒ [tex]AL = 2 \times 9 = 18[/tex] units
and [tex]\text{Length of longer leg(KL)}= \sqrt{3} \times \text{Length of Shorter leg}[/tex]
⇒ [tex]KL = \sqrt{3} \times 9 = 9\sqrt{3}[/tex] units.
To find the perimeter of ∆AKL;
Perimeter(P) is the sum of all the sides of a triangle.
⇒[tex]P=AK+KL+AL[/tex]
Substitute the given values we get;
[tex]P=9+9\sqrt{3}+18= 27+9\sqrt{3}[/tex] units.
Now, to find the area of triangle AKL;
[tex]\text{Area of triangle(AKL)} = \frac{1}{2}(KL)(AK)[/tex]
Substitute the given values we get;
[tex]\text{Area of triangle(AKL)} = \frac{1}{2}(9\sqrt{3})(9)=\frac{81\sqrt{3}}{2}[/tex] square units
Therefore, the perimeter of ∆AKL is, [tex]27+9\sqrt{3}[/tex] units anmd area of triangle is, [tex]\frac{81\sqrt{3}}{2}[/tex] square units
