Given :
In ∆ABC,
- ∠A = 60°
- ∠B = 60°
- BC = 12 units
- ∠BDC = 90°
To Find :
Solution :
As, we have :
So, By angle sum property of triangle :
∠A + ∠B + ∠C = 180°
[tex] \tt : \implies 60\degree + 60\degree + \angle C = 180\degree[/tex]
[tex] \tt : \implies 120\degree + \angle C = 180\degree[/tex]
[tex] \tt : \implies \angle C = 180\degree - 120\degree[/tex]
[tex] \tt : \implies \angle C = 60\degree[/tex]
Now, we have :
As, all angles are of same length, therefore it is a equilateral triangle.
We know that sides of equilateral triangle are equal.
[tex] \tt : \implies AB = BC = AC[/tex]
Now, we have BC = 12 units.
Hence, all sides are of 12 units.
Now, we know that area of equilateral triangle is :
[tex] \large \underline{\boxed{\bf{Area_{(equilateral \: triangle)} = \dfrac{\sqrt{3}}{4} side^{2}}}}[/tex]
[tex] \tt : \implies Area = \dfrac{\sqrt{3}}{4} \times (12 \: units)^{2}[/tex]
[tex] \tt : \implies Area = \dfrac{\sqrt{3}}{\cancel{4}} \times \cancel{144} \: units^{2}[/tex]
[tex] \tt : \implies Area = \sqrt{3} \times 36 \: units^{2}[/tex]
[tex] \tt : \implies Area = 36\sqrt{3} units^{2}[/tex]
So, Area of given triangle is 36√3 units².
