In an isosceles triangle, length of the congruent sides is 13 em and i

1.	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» 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.

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»

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.