SSC CGL 4 06 2019 Shift 1 Paper

Question : 55 of 100

Marks: +1, -0

In ΔABC, F and E are the points on sides AB and AC, respectively, such that FE ∥ BC and FE divides the triangle in two parts of equal area. If AD ⊥ BC and AD intersects FE at G, then GD : AG = ?

$√2:1$
$(√2-1):1$
$2√2:1$
$(√2+1):1$

Validate

Solution:

Let area of

Δ\AFE = 1

then area of

⬠\BCEF = 1

\ {\text "Area of "\ΔA F E}/{\text "Area of " \ΔA B C}=\ {1}/{2}=\ {A G^{2}}/{A D^{2}}

\ {A G}/{A D}=\ {1}/{\√{2}}

\ {G D}/{A G}=\ {\√{2}-1}/{1}

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