Let there be an equilateral ABC
We need to find its area
We know that,
Area ∆ABC = 1/2 × Base × Height
Finding base & height of equilateral triangle ABC
Height is perpendicular from the vertex to the base.
Let us draw perpendicular from point A
So,
Height = AD
Base = BC = a
So, we need to find height AD
In equilateral triangle,
altitude is also the median
So, point D is also the mid-point of BC
Therefore,
BD = DC = a/2
Now, in ∆ADC
By Pythagoras theorem
AC ^{ 2 } = AD ^{ 2 } + DC ^{ 2 }
a ^{ 2 } = AD ^{ 2 } + (a/2) ^{ 2 }
^{ }
a ^{ 2 } = AD2 + a ^{ 2/4 }
AD^2 + a2/4 = a ^{ 2 }
AD^2 = a2- a ^{ 2/4 }
AD^2 = (4a ^{ 2 } - ^{ a2 } )/4
AD^2 = ^{ (3a2 } )/4
AD = √((3a ^{ 2 } )/4)
AD = a/2 √3
AD = (√3 a)/2
Now,
Height = AD
= √3/2 a
Base = BC
= a
Area of ∆ABC = 1/2 × Base × Height
= 1/2 × a × √3/2 a
= √3/4 a^2
∴ Area of equilateral triangle = √3/4 a ^{ 2 }
Find area of the following equilateral triangle whose sides are 2 cm
Side = a = 2 cm
Area of equilateral ∆ABC = √3/4 a ^{ 2 }
= √3/4 (2) ^{ 2 }
= √3/4 × 4
= √3 cm ^{ 2 }
∴ Area of equilateral triangle ∆ABC is √3 cm ^{ 2 }
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