Question

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

Answer 1

We have the triangle 30° - 60° - 90°.

The sides are in proportion: 2 : √3 : 1

(look at the picture).

$2a=14$       divide both sides by 2

$a=7$

$a\sqrt3=7\sqrt3$

The area of a trinagle AFD:

$A_{\Delta}=\dfrac{1}{2}(7)(7\sqrt3)=\boxed{\dfrac{49\sqrt3}{2}}$