Question

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

Answer 1

Answer:

$\frac{49\sqrt{3}}{2}$

Step-by-step explanation:

This is a 30-60-90 triangle.  The sides of this triangle can be represented as t (across from the 30° angle since it is the smallest side), 2t (across from the 90° angle since it is the largest side), and t√3 (across from the 60° angle).

Since F is the right angle, this means that AD is across from F and is the hypotenuse.  The hypotenuse of a right triangle is the longest side; this means that 2t = 14, so t = 7.  This also tells us that t√3 = 7√3.

This means that the height and base of the triangle are 7 and 7√3.  Using the formula for the area of a triangle, we have

A = 1/2bh = 1/2(7)(7√3) = 1/2(49√3)

$=\frac{49\sqrt{3}}{2}$

Answer 2

The area of the triangle AFD is 49√2/2 square units

The area of the triangle AFD is expressed as:

• A = 1/2absin m ∠D

• a = AD
• b = FD

Get the length of FD using SOH CAH TOA identity

cos 30 = FD/14

FD = 14cos30

FD = 7√2

Get the required area

Area of the triangle AFD = 1/2(14)(7√2)sin30

Area of the triangle AFD = 49√2sin30

Area of the triangle AFD = 49√2/2 square units

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