In ∆ABC, AD is the median and G is the centroid point on AD such that AG : GD = 2 : 1. Then ar (∆BDG) : ar (∆ABC) is equal to:

Question

Question

This question was previously asked in

SSC CGL Previous Paper 53 (Held On: 10 June 2019 Shift 2)

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Answer 1

Concepts Used:

1. For two triangles, if their base are on the same line and the third vertex is common, then the heights of both triangles become equal

Area(Δ1) : Area(Δ2) = Base1 : Base2

2. Centroid divided the median in the ration 2:1

Calculations:

F1 V.G Deepak 17.10.2019 D 4

AD is median, so AD divides the triangle into two equal part

Area (ΔADB) = Area (ΔADC)

In ΔBDG and ΔBGA, the point B is common and their base are on the same line

⇒ Heights (H) of ΔBDG and ΔBGA will be equal

⇒ Area(ΔBDG)/Area(ΔBGA) = [(1/2)DG × H] / [(1/2)AG × H]

⇒ Area(ΔBDG)/Area(ΔBGA) = DG/AG

⇒ Area(ΔBDG)/Area(ΔBGA) = 1/2

Let the Area(ΔBDG) = 1 unit

⇒ Area(ΔBGA) will be = 2 units

Area (ΔBAD) = Area (ΔBDG) + Area (ΔBGA) = 1 + 2 = 3 units

Area (ΔABC) = 2 × Area (ΔBAD) = 2 × 3 = 6 units

⇒ Area (ΔBDG) : Area (ΔABC) = 1 : 6