Question

the medians of a triangle ABC intersect each other at point G. if one it's median is ad prove that area of triangle ABC equals to three times area of triangle bgd

Answer 1

GIVEN: Triangle ABC, 3 medians intersecting at G.

Area(triangle ABC) = 27 cm²

G is the centroid of the triangle, which intersects each medion in the ratio 2:1.

Here, AG : GM = 2:1

So, if AG = 2a, GM = a

TO FIND: area(triangle BCG)

CONSTRUCTION: CX perpendicular to AM

CALCULATION:

area(triangle CAG) = 1/2 * AG * CX =

1/2 * 2a *CX……..(1)

area(triangle CGM) = 1/2 * GM * CX =

1/2 * a * CX……….(2)

So area (triangle CGM) = 1/2 of area(triangle CAG) ……..(3)

Similarly, ar(tri BGM) = 1/2 of ar(tri BAG)….(4)

By adding (3) & (4)

ar(tri BCG) = 1/2 {ar(triCAG) + ar(triBAG)}

So, if ar(tri BCG) = A ………….(5)

Then ar(triCAG) + ar(triBAG) = 2A……….(6)

Adding (5) & (6)

We get ar(tri ABC ) = 3A = 27

So, A= 27/3 = 9

=> ar( tri BCG) = 9 cm²

Answer 2

Answer

Answer:HOPES THIS HELPS U.

we know if the two triangle have same height and different base ,

then ratio of their area is equal to the ratio of their respective base .

since AD is a median

hence BD=CD

THEREFORE , area of triangle ABD= ACD ( both the triangle have same height and same base )

hence, area of ABD =1/2( AREA of ABC)———(1)

Also, area of ABP=1/3(Area of ABD)=1/3* 1/2(ABC)

HENCE Area of ABP=1/6( Area of ABC)