Prove that the circumcentre of an equilateral triangle is the same as its centroid.

i) Calculate the length of side of an equilateral triangle with vertices on a circle of diameter 1 centimetre.

ii) Calculate the perimeter of such a triangle.

The figure is given below:

Let ABC be a triangle

AD, BE and CF are medians intersecting at point O.

In triangles ABD and ACD,

BD = CD (AD is the median)

∠ABD = ∠ACD = 60°

AB = AC (Equilateral Triangle)

∴ triangles ABD and ACD are congruent.

∴ ∠BAD = ∠CAD = 30°

In triangle ABD

Sum of all angles of a triangle = 180°

∴ ∠ABD + ∠DAB + ∠ADB = 180°

∴ 60° + 30° + ∠ADB = 180°

90° + ∠ADB = 180°

∴ ∠ADB = 180° – 90° = 90°

∴ ∠ADC = 180 – ∠ADB (Supplementary angles)

= 180 – 90°

= 90°

Similarly,

∠OBD = ∠OCD = 30° …(eq)1

In triangles OBD and OCD,

∠OBD = ∠OCD = 30° (from eq 1)

∠ADB = ∠ADC = 90°

OD = OC is a common side

By RHS criterion,

Triangles OBD and OCD are congruent.

∴ OB = OC

Similarly, OC = OA and OA = OB

∴ OA = OB = OC

Hence, pt. O is a circumcentre and also a intersection of medians.

Hence, the circumcentre of an equilateral triangle is the same

as its centroid.

i)

Diameter of circle = 1 centimetre

∴ Radius of circle = cm = 0.5 cm

As proved above, the circumcentre of an equilateral triangle is the same as its centroid.

Also, centroid divides equilateral in the ratio 2:1

∴ radius

∴

∴ = × 0.5 =

But, for an equilateral triangle,

Height of triangle =

=

∴ side of triangle =

ii) Perimeter of triangle = Sum of all sides of triangle

As for an equilateral triangle, all sides are equal

Let a = side of triangle =

Perimeter of triangle = a + a + a = 3a = cm