The area of a triangle ABC is 63 sq. units. Two parallel lines DE, FG. are drawn such that they divide the line segments AB and AC into three equal parts. What is the area of the quadrilateral DEGF?

We know that area of a triangle is proportional to square of the sides

Point D and F divides the line AB into 3 equal parts AD, DF and FB.

=> $$\frac{AD}{AB} = \frac{1}{3}    and \frac{AF}{AB} = \frac{2}{3}$$

Considering similar triangles $$\triangle ADE and \triangle ABC$$

$$\frac{Area of \triangle ADE}{Area of \triangle ABC}= \frac{AD^2}{AB^2}$$

=> $$Area of \triangle ADE/63 = 1/9$$

=> $$Area of  \triangle ADE = 63/9 = 7$$

Similarly considering similar triangles ΔAFG and ΔABC

$$\frac{Area of \triangle AFG}{Area of \triangle ABC} = \frac{AF^2}{AB^2}$$

=> $$Area of \triangle AFG/63 = 4/9$$

=> $$Area of \triangle AFG = \frac{63 \times 4}{9} = 28$$

Area of quadrilateral DEFG = Area of $$\triangle AFG - Area of \triangle ADE$$

= 28 - 7

= 21 sq.units

So, the answer would be option c)21 sq. units