Question

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

Answer 1

Perimeter = 27 + 9*sqrt(3), Approximately 42.58845727 
Area = 40.5*sqrt(3). Approximately 70.14805771  
Since all triangles have a total of 180 degrees and we've been given 2 of those angles, the remaining angle is 180 - 90 - 60 = 30 degrees. So we have a 30,60,90 degree right triangle. Drawing the triangle and assigning the proper angle to each vertex shows that AK is the short leg of the triangle. And since it's a 30,60,90 triangle, the hypotenuse is AL and it will be twice the length of AK, so it's 18. And finally, we can use the Pythagorean theorem to calculate the length of KL. So 
KL = sqrt(18^2 - 9^2) = sqrt(324 - 81) = sqrt(243) = sqrt(81*3) = 9*sqrt(3) 
 So the perimeter is 
P = 9 + 18 + 9*sqrt(3) = 27 + 9*sqrt(3). Which is approximately 42.58845727 
 The area is base times height divided by 2. And we have a base of 9 and a height of 9*sqrt(3). So
 A = 9 * 9*sqrt(3)/2 = 81*sqrt(3)/2 = 40.5*sqrt(3). Which is approximately 70.14805771