Question

State true or false:

The medians of a triangle $ABC$ intersect each other at point $G$ . If one of its medians is $AD$ .prove that:

Area $(△ABD)=3\times$ Area $(△BGD)$

• A. True
• B. False

Answer 1

The correct option is A True

Given, the medians of a triangle ABC intersect each other at point G . One of its medians is AD
Using property when the medians of triangles intersect each other, it divides into ratio of 2:1.
So, AG:GD$=$ 2:1
or, GD$=\frac{1}{3}$ AD (1)
Drawing a altitude BF from B to AD at F.
So,Area of $△ABD$$=\frac{1}{2}\times base\times altitude=\frac{1}{2}\times AD\times BF$
Area of $△BGD$$=\frac{1}{2}\times base\times altitude=\frac{1}{2}\times GD\times BF$
$=\frac{1}{2}\times \frac{1}{3}AD\times BF$ ( GD$=\frac{1}{3}$ AD, From 1)
Taking the ratio of both the areas ,
$\frac{Areaof△ABD}{Areaof△BGD}=\frac{\frac{1}{2}\times AD\times BF}{\frac{1}{2}\times \frac{1}{3}AD\times BF}=3$
or, Area of $△ABD$$=3\times$ Area of $△BGD$