Question

Given: ∆AFD, m ∠F = 90° AD = 14, m ∠D = 30° Find: Area of ∆AFD

Answer 1

This is a 30-60-90 special triangle. in a 30-60-90 right triangle, the shortest leg is half of the hypotenuse, and the other leg is (√3)/2 of the hypotenuse, 

so FA=(1/2)AD=7, DF=(√3)/2)*14=7√3

the area is (1/2)*7*7√3≈42.4

Answer 2

Answer: Area of ∆ADF is 42.44 m² .

Step-by-step explanation:

Since we have given that

AD = 14 m

m∠F = 90°

m∠D = 30°

We need to find the area of ∆ADF,

As we know the formula for " Area of triangle ":

$Area=\frac{1}{2}\times b\times h$

For this we need to find the base and height of the given triangle.

We will use the "Trigonometric Ratio ":

$\sin 60\textdegree=\frac{FA}{AD}\\\\\frac{\sqrt{3}}{2}=\frac{FA}{14}\\\\\frac{14}{2}\times \sqrt{3}=FA\\\\7\sqrt{3}=FA\\\\12.12\ m=FA$

Similarly, we will find the base.

$\cos 60\textdegree=\frac{DF}{AD}\\\\\frac{1}{2}=\frac{DF}{14}\\\\DF=\frac{14}{2}=7$

Now, we will apply the formula for Area of triangle :

$Area=\frac{1}{2}\times FA\times DF\\\\Area=\frac{1}{2}\times 12.12\times 7\\\\Area=42.435\\\\Area=42.44\ m^2$

Hence, Area of ∆ADF is 42.44 m² .