Question

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

Answer 1

The question asks us to find the area of a triangle. First, before we do anything, we need to state the formula for the area of a triangle, which is $A=\frac{1}{2}bh$, where $b$ is the base and $h$ is the height. In this scenario, we are given the length of the height, which is $9$. We need to find the length of KL, which is the base. We can see that this triangle is a 30, 60, 90 degree triangle, meaning that it has some special properties. First off, it means that the hypotenuse is two times the side opposite to it, which in this case is that AL (the hypotenuse) is two times the length of AK, which is the side opposite to it. Thus, the length of the hypotenuse is $9*2=18$. Now, we can use the Pythagorean Theorem to find the length of the base. 
$a^2+b^2=c^2$
$9^2+b^2=18^2$
$81+b^2=324$
$b^2=243$
$b= \sqrt{243}$
$b\approx15.59$
Now that we found all the required values for the formula, just plug it in and solve. When we do so, we get that the area is $\boxed{70.16}$. Hope this helped!