Question

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

Answer 1

Perimeter of the isosceles triangle = 13+13+10 = 36 cm
Thus, s = 36/2 = 18cm.
Area of triangle = [18(18–13)(18–13)(18–10)}^0.5 = [18*5*5*8]^0.5 = 60 sq cm.
Altitude of the triangle = 60*2/10 = 12 cm.
Since the centroid is two thirds of the way from the vertex to any base, the answer to your question is 8 cm.

Answer 2

The distance between the vertex opposite the base and the centroid is 8 cm.

Given

In an isosceles triangle, ABC it has sides and base.

Where, AB = AC = 13 cm

                     BC = 10 cm

                     AM is the median on BC

                     P is the centroid on median BC.

BM = CM = 1/2 BC = 1/2×10 = 5 cm

Using, Apollonius theorem,

           In ΔABC, M is the mid point of BC.

               $AB^2+AC^2 = 2AM+2BM^2$

               13² + 13² = 2AM² + 2(5)²

                169 + 169 = 2AM² + 2(25)

                 338 = 2AM² + 50

                 2AM² = 338 - 50

                 2AM² = 288

                   AM² = 288 / 2 = 144

                   AM = 12 cm

"P" is the centroid which divides the median in the ratio of 2 : 1

                  AP : PM = 2 : 1

                  AP = 2 PM

                  AM = AP + PM

                  AM = AP + AP/2

                         = 3AP/2

                 AP = 2/3 AM

                       = 2/3 × 12

                AP = 8 cm

Therefore, The distance between the vertex opposite the base and the centroid is 8 cm.

To learn more...

1. brainly.in/question/3655994

2. brainly.in/question/4089643