Question

Prove that area of triangle formed by the medians of any triangle is 3/4th

of the area of the given
triangle​

Answer 1

Proved that area of triangle formed by the medians of any triangle is 3/4th

of the area of the given  triangle​.

To prove :

Area of triangle formed by the medians of any triangle is $\frac{3}{4}^{th}$of the area of the given triangle.

Given :

Area of triangle formed by the medians of any triangle is $\frac{3}{4}^{th}$.

From the figure,

BG + GC > BC   -----> ( 1 )

AG + BG > AB   ------> ( 2 )

AG + GC > AC   ------> ( 3 )

Adding the above equations ( 1 ), ( 2 ) and ( 3 ), we get

2AG + 2BG + 2GC > AB + BC + AC

Taking common number "2" from LHS, i.e.,

2 ( AG + BG + GC ) > AB + BC + AC   -----> ( 4 )

Here, AG = $\frac{2}{3}$ AD

         BG = $\frac{2}{3}$ BE

         GC = $\frac{2}{3}$ CF

Applying the above values in equation ( 4 ), it gives

2 (  $\frac{2}{3}$ AD + $\frac{2}{3}$ BE + $\frac{2}{3}$ CF ) > AB + BC + AC

Taking common term   $\frac{2}{3}$  from LHS, i.e.,

2 ×  $\frac{2}{3}$  ( AD + BE + CF ) > AB + BC +AC

$\frac{4}{3}$ (AD + BE + CF) > AB + BC + AC

4 (AD + BE + CF) > 3 (AB + BC + AC)

AD + BE + CF = Sum of the medians.

AB + BC + AC = Sum of the sides.

4 (Sum of the medians) > 3 (Sum of the sides)

Sum of the medians > $\frac{3}{4}$ (Sum of the sides)

Hence, it has been proved that area of triangle formed by the medians of any triangle is $\frac{3}{4}^{th}$ of the area of the given triangle.

To learn more...

1. If the medians of a ΔABC intersect at G show that ar (AGB) = ar (AGC) =ar (BGC) = 1/3 ar (ABC)      

brainly.in/question/83619

2. THE MEDIANS BE & CF of a triangle ABC intersect at G . prove that area of triangle GBC = area of quadrilateral AFGE .

brainly.in/question/83532

Answer 2

Step-by-step explanation:

जेड इक्वल टू थ्री एक्स इज इक्वल टू हाई क्वालिटी 30 डिग्री एंड शो इन द