Math, asked by vikky156, 1 year ago

In traiangle ABC, F and E are the points on sides AB and AC, respectively, such that FE || BC and FE divides the triangle in parts of equal area. If AD perpendicular to BC ....; AD intersects FE at G, then GD:AG ?

Answers

Answered by abhi569
11

Answer:

Ratio GD : AG is 1 : ( √2 + 1 ).

Step-by-step explanation:

Given,

FE divides the triangle in parts of equal area.

It means :

= > Area of ∆AFE = Area of remaining part

= > Area of ∆AFE = Area of whole figure - Area of ∆AFE

= > Area of ∆AFE = Area of ∆ABC - Area of ∆AFE

= > Area of ∆AFE + Area of∆AFE = Area of ∆ABC

= > 2 x Area of ∆AFE = Area of ∆ABC

From the properties of triangles,

  • 1 / 2 x base x height = Area of triangle

= > 2 x 1 / 2 x AG x FE = 1 / 2 x AD x BC

= > AG x FE = 1 / 2 x AD x BC

= > 2 x FE / BC = AD / AG

= > 2 x FE / BC = ( AG + GD ) / AG { AD = AG + GD }

= > 2 x FE / BC = AG / AF + GD / AG

= > 2 x FE / BC = 1 + GD / AG

= > [ 2 x FE / BC ] - 1 = GD / AG ... ( 1 )

As we can observe, quadrilateral FECB is a trapezium. Therefore :

= > Area of ∆AFE = Area of trapezium FECB

= > 1 / 2 x AG x FE = 1 / 2 x ( FE + BC ) x GD { Area of trapezium = 1 / 2 x sum of || sides x height }

= > AG x FE = ( FE + BC ) x GD

= > FE / ( FE + BC ) = GD / AG ... ( 2 )

Comparing the value of GD / AG from ( 2 ) :

= > 2 x [ FE / BC ] - 1 = FE / ( FE + BC )

= > ( 2 FE - BC ) / BC = FE / ( FE + BC )

= > ( 2FE - BC ) ( FE + BC ) = FE.BC

= > 2FE^2 + 2BC.FE - BC.FE - BC^2 = FE.BC

= > 2FE^2 + BC.FE - BC^2 = FE.BC

= > 2EF^2 = BC^2

= > √2 EF = BC

Therefore,

= > GD / AG = FE / ( FE + BC )

= > GD / AG = FE / ( FE + √2 FE )

= > GD / AG = 1 / ( √2 + 1 )

Hence the required ratio GD : AG is 1 : ( √2 + 1 ).

Answered by avman08
0

Answer:

In traiangle ABC, F and E are the points on sides AB and AC, respectively, such that FE || BC and FE divides the triangle in parts of equal area. If AD perpendicular to BC ....; AD intersects FE at G, then GD:AG ?

Step-by-step explanation:

here

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