Question

The midpoints of the sides of a triangle along with any of the vertices as the fourth point makes a parallelogram of area equal to
(a) $\frac{1}{2}\left(\mathrm{ar}∆ABC\right)$
(b) $\frac{1}{3}\left(\mathrm{ar}∆ABC\right)$
(c) $\frac{1}{4}\left(\mathrm{ar}∆ABC\right)$
(d) $\mathrm{ar}\left(∆ABC\right)$

Answer 1

(a) $\frac{1}{2}$ ( ar ∆ ABC)

Join, FE.
∆ABC has been divided into 4 triangles of equal areas.
So, ​ar(∆AFE) = $\frac{1}{4}$​× ​(ar∆ABC)
∴ ar(∣​∣gm AFDE) = ar (∆AFE) + ar(∆FED)
= 2 × ​ar(∆AFE) = ​2 × ​$\frac{1}{4}$​ × ​(ar∆ABC) = ​$\frac{1}{2}$(ar∆ABC)
Hence, ​ar(∣​∣gm AFDE) = ​​$\frac{1}{2}$(ar∆ABC)