Question

If ABCD is a rectangle, E and F are the mid-points of BC and AD respectively and G is any point on EF. Then ar(ΔGAB) : ar (rectangle ABCD) = _________.

Answer 1

Given:
ABCD is a rectangle
E and F are the mid-points of BC and AD respectively
G is any point on EF


Since, E and F are the mid-points of BC and AD respectively.
Therefore, ar(BEFA) = ar(ECDF) = $\frac{1}{2}$× ar(ABCD) ...(1)

We know, if a triangle and a rectangle are on the same base and between the same parallels, then the area of the triangle is half the area of the rectangle.
Thus, ar(ΔGAB) = $\frac{1}{2}$ar(BEFA) ...(2)

From (1) and (2),
ar(ΔGAB) = $\frac{1}{2}$×$\frac{1}{2}$× ar(ABCD)
⇒ ar(ΔGAB) = $\frac{1}{4}$× ar(ABCD)
⇒ $\frac{\mathrm{ar}\left(∆GAB\right)}{\mathrm{ar}\left(\mathrm{rectangle}ABCD\right)}=\frac{1}{4}$
​

Hence, ar(ΔGAB) : ar(rectangle ABCD) = 1 : 4.