Question

In an isosceles triangle, length of the congruent sides is 13 em and its.base is 10 cm. Find the distance between the vertex opposite to the base and the centroid. 

Answer 1

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.