Question

triangle DEF=1/4 triangle ABC​

Answer 1

Answer:

Given that D, E and F are the mid-points of BC, CA and AB respectively, then from the mid-point theorem

EF = BC / 2, DE = AB / 2, DF = AC / 2.

But Δ ABC ~ Δ DEF

Ratio of areas of two similar triangles is equal to the square of the ratio of their corresponding sides

⇒ Area of triangle ABC / Area of triangle DEF = BC2 / EF2

⇒ Area of triangle ABC / Area of triangle DEF = (2EF)2 / EF2

⇒ Area of triangle ABC / Area of triangle DEF = 4 / 1. ∴Ratio of the areas of triangle DEF and triangle ABC is 1 : 4.

Answer 2

Parallelogram :A quadrilateral in which both pairs of opposite sides are parallel is called a parallelogram.

In a parallelogram diagonal divides it into two triangles of equal areas.

Proof:

Now, BDEF is a parallelogram so its diagonal FD divides its into two Triangles of equal areas.

∴ ar(ΔBDF) = ar(ΔDEF) — (i)

In Parallelogram AFDE

ar(ΔAFE) = ar(ΔDEF) (EF is a diagonal) — (ii)

In Parallelogram FDCE

ar(ΔCDE) = ar(ΔDEF) (DE is a diagonal) — (iii)

From (i), (ii) and (iii)

ar(ΔBDF) = ar(ΔAFE) = ar(ΔCDE) = ar(ΔDEF).....(iv)

ar(ΔBDF) + ar(ΔAFE) + ar(ΔCDE) + ar(ΔDEF) = ar(ΔABC)

4 ar(ΔDEF) = ar(ΔABC)(From eq iv)

ar(∆DEF) = 1/4 ar(∆ABC)

Hope it helps!