# D, E and F are respectively the mid-points of sides AB, BC and CA of ∆ ABC. Find the ratio of the areas of ∆ DEF and ∆ ABC

**Solution:**

As we know that the line segment joining the midpoints of two sides of a triangle is parallel to the third side and half of it-- (mid-point theorem).

In ΔABC,

From the data given in the figure, we can conclude that

F is the mid-point of AB and E is the mid-point of AC

So, by the mid-point theorem, we have,

FE || BC and FE = 1/2BC

⇒ FE || BC and FE || BD [BD = 1/2BC]

As we know that the opposite sides of the parallelogram are equal and parallel

Therefore, BDEF is a parallelogram.

Similarly, in ΔFBD and ΔDEF, we have

FB = DE (Opposite sides of parallelogram BDEF are equal)

FD = FD (Common)

BD = FE (Opposite sides of parallelogram BDEF are equal)

Therefore, ΔFBD ≅ ΔDEF

Similarly, we can prove that ΔAFE ≅ ΔDEF and ΔEDC ≅ ΔDEF

As we know, if two triangles are congruent, then their areas must be equal.

Hence,

Area(ΔFBD) = Area(ΔDEF) …**(1)**

Area(ΔAFE) = Area(ΔDEF) …….**(2)**

and,

Area(ΔEDC) = Area(ΔDEF) ……**(3)**

Now,

Area(ΔABC) = Area(ΔFBD) + Area(ΔDEF) + Area(ΔAFE) + Area(ΔEDC) ……**(4)**

Area(ΔABC) = Area(ΔDEF) + Area(ΔDEF) + Area(ΔDEF) + Area(ΔDEF) [ Area(ΔFBD) = Area(ΔDEF) and Area(ΔEDC) = Area(ΔDEF) ]

From equation (1), (2) and (3),

⇒ Area(ΔDEF) = (1/4)Area(ΔABC)

⇒ Area(ΔDEF)/Area(ΔABC) = 1/4

Therefore,

Area(ΔDEF): Area(ΔABC) = 1 / 4 = 1:4

**Video Solution:**

## D, E and F are respectively the mid-points of sides AB, BC and CA of ∆ ABC. Find the ratio of the areas of ∆ DEF and ∆ ABC

### NCERT Class 10 Maths Solutions - Chapter 6 Exercise 6.4 Question 5:

D, E and F are respectively the mid-points of sides AB, BC and CA of ∆ ABC. Find the ratio of the areas of ∆ DEF and ∆ ABC

D, E and F are respectively the mid-points of sides AB, BC and CA of ∆ABC. Then the ratio of the areas of ∆DEF and ∆ABC is 1:4