Let vertices of triangle are at points A(3,2),B(-3, 2) and C(0,2-√3).
Then by using distance formula between two points, we get sides of a triangle.
AB= {√(-3-3)²+(2-2)²} =√(-6)²= √36 = 6 unit.
BC={√(0+3)²+(2-√3 -2)²}={√9+(-√3)²}=√(9+3)=√12
CA ={√(3-0)²+(2-(2-√3)²}={√9+(-√3)²}=√(9+3)=√12
Here, BC = CA = √12 unit,so triangle is isosceles.
Then, as we know that area of an isosceles triangle, if it's equal sides be a and base be b.
1/2 × b ×√(a² - b²/4).
Therefore area of an isosceles triangle= 1/2×6√{(√12)²-3²} = 3×√(12–9)= 3√3 sq unit,Ans.
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Let vertices of triangle are at points A(3,2),B(-3, 2) and C(0,2-√3).
Then by using distance formula between two points, we get sides of a triangle.
AB= {√(-3-3)²+(2-2)²} =√(-6)²= √36 = 6 unit.
BC={√(0+3)²+(2-√3 -2)²}={√9+(-√3)²}=√(9+3)=√12
CA ={√(3-0)²+(2-(2-√3)²}={√9+(-√3)²}=√(9+3)=√12
Here, BC = CA = √12 unit,so triangle is isosceles.
Then, as we know that area of an isosceles triangle, if it's equal sides be a and base be b.
1/2 × b ×√(a² - b²/4).
Therefore area of an isosceles triangle= 1/2×6√{(√12)²-3²} = 3×√(12–9)= 3√3 sq unit,Ans.