Answer:
[tex]28\sqrt{3}[/tex]
Step-by-step explanation:
The area of the big triangle is 1/2 b h = 1/2*6*(12^2 = 6^2 + x^2)
that ends up being [tex]\sqrt{108} = 36\sqrt{3}[/tex]
the small triangle are needs to be subtracted....
[tex]\frac{\left(4\cdot \:sin\left(90\right)\right)}{sin\left(30\right)}[/tex] that is the length of the unknown side...
1/2 B * h of that triangle get you to [tex]8\sqrt{3}[/tex]
just subtract the two areas
Answer:
(B) 28√3
Step-by-step explanation:
The area of quadrilateral ABED is equal to the area of triangle CDE subtracted from the area of triangle ABC.
Area of triangle CDE:
Triangle ABC is equilateral. All sides have length 12.
AB = BC = AC = 12
BE = 8
BE + EC = BC
8 + EC = 12
EC = 4
In an equilateral triangle, all angles measure 60°.
m<C = 60°
m<CDE = 30°
Triangle CDE is a 30-60-90 triangle.
DE = EC√3
DE = 4√3
area of triangle CDE = bh/2
area of triangle CDE = (EC)(DE)/2
area of triangle CDE = (4)(4√3)/2
area of triangle CDE = 8√3
Area of triangle ABC:
Side AC is a base of triangle ABC.
AC = 12
(1/2)AC = 6
The altitude of triangle ABC from side AC to vertex B measures
h = 6√3
area of triangle ABC = bh/2
area of triangle ABC = (AC)(h)/2
area of triangle ABC = (12)(6√3)/2
area of triangle ABC = 36√3
area of quadrilateral ABED = area of triangle ABC - area of triangle CDE
area of quadrilateral ABED = 36√3 - 8√3
area of quadrilateral ABED = 28√3