Question

In the diagram, AD = DB and AH = HD. Find the ratio of the area of the shaded portion to that of the triangle DEF, if DE || BC and HG || AE.

• A. 2 : 3
• B. 1 : 3
• C. 1 : 2
• D. $1:1\frac{1}{3}$

Answer 1

The correct option is C

1 : 2

Consider the above diagram, where the area of $\Delta ABC=A$
The heights of $\Delta AHI,\Delta HGI,\Delta GEI$ are same as they are between the same two parallel lines.
Area of $\Delta$ HGI + Areaof $\Delta$ AHI + Areaof $\Delta$ GEI
= $\frac{3}{4}\times \frac{1}{4}A\dots$(i)
Since HG=$\frac{1}{2}AE$
$\Delta$ HGI= $\frac{1}{2}$(sum of $\Delta$ AHI+$\Delta HGI)\dots (2)$
From (1) and (2)
Area of the shaded portion $=\frac{2}{3}\times \frac{3}{16}A=\frac{1}{8}A$
Area of $\Delta DEF=\frac{1}{3}$ of trapezium DEBC
$=\frac{1}{3}\times \frac{3}{4}ofA=\frac{1}{4}A$
Therefore required ratio $=\frac{1}{8}A:\frac{1}{4}A=1:2$