Given: ∆ABC is an isosceles triangle.
G is the centroid.
AB = AC = 13 cm, BC = 10 cm.
To find: AG
Construction: Extend AG to intersect side BC at D, B – D – C.
Centroid G of ∆ABC lies on AD
∴ seg AD is the median. (i)
∴ D is the midpoint of side BC.
∴ DC = 1/2 BC
= 1/2 × 10 = 5
In ∆ABC, seg AD is the median. [From (i)]
∴ AB2 + AC2 = 2AD2 + 2DC2 [Apollonius theorem]
∴ 132 + 132 = 2AD2 + 2(5)2
∴ 2 × 132 = 2AD2 + 2 × 25
∴ 169 = AD2 + 25 [Dividing both sides by 2]
∴ AD2 = 169 – 25
∴ AD2 = 144
∴ AD = √144 [Taking square root of both sides]
= 12 cm We know that, the centroid divides the median in the ratio 2 : 1.
∴ AG/GD = 2/1
∴ GD/AG = 1/2 [By invertendo]
∴ (GD + AG)AG = (1 + 2)/2 [By componendo]
∴ AD/AG = 3/2 = [A – G – D]
∴ 12/AG = 3/2
∴ AG = (12 x 2)/3
= 8cm
∴ The distance between the vertex oppesite to the base and the centroid id 8 cm.