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 .

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=(1)/(2)* b* h`

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

We will use the "Trigonometric Ratio ":

`\sin 60\textdegree=(FA)/(AD)\\\\(√(3))/(2)=(FA)/(14)\\\\(14)/(2)* √(3)=FA\\\\7√(3)=FA\\\\12.12\ m=FA`

Similarly, we will find the base.

`\cos 60\textdegree=(DF)/(AD)\\\\(1)/(2)=(DF)/(14)\\\\DF=(14)/(2)=7`

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

`Area=(1)/(2)* FA* DF\\\\Area=(1)/(2)* 12.12* 7\\\\Area=42.435\\\\Area=42.44\ m^2`

Hence, Area of ∆ADF is 42.44 m² .